id int64 1 3.71k | title stringlengths 3 79 | difficulty stringclasses 3
values | description stringlengths 430 25.4k | tags stringlengths 0 131 | language stringclasses 19
values | solution stringlengths 47 20.6k |
|---|---|---|---|---|---|---|
10 | Regular Expression Matching | Hard | <p>Given an input string <code>s</code> and a pattern <code>p</code>, implement regular expression matching with support for <code>'.'</code> and <code>'*'</code> where:</p>
<ul>
<li><code>'.'</code> Matches any single character.</li>
<li><code>'*'</code> Matches zero or more... | Recursion; String; Dynamic Programming | C++ | class Solution {
public:
bool isMatch(string s, string p) {
int m = s.size(), n = p.size();
int f[m + 1][n + 1];
memset(f, 0, sizeof f);
function<bool(int, int)> dfs = [&](int i, int j) -> bool {
if (j >= n) {
return i == m;
}
if (f... |
10 | Regular Expression Matching | Hard | <p>Given an input string <code>s</code> and a pattern <code>p</code>, implement regular expression matching with support for <code>'.'</code> and <code>'*'</code> where:</p>
<ul>
<li><code>'.'</code> Matches any single character.</li>
<li><code>'*'</code> Matches zero or more... | Recursion; String; Dynamic Programming | C# | public class Solution {
private string s;
private string p;
private int m;
private int n;
private int[,] f;
public bool IsMatch(string s, string p) {
m = s.Length;
n = p.Length;
f = new int[m + 1, n + 1];
this.s = s;
this.p = p;
return dfs(0, 0);
... |
10 | Regular Expression Matching | Hard | <p>Given an input string <code>s</code> and a pattern <code>p</code>, implement regular expression matching with support for <code>'.'</code> and <code>'*'</code> where:</p>
<ul>
<li><code>'.'</code> Matches any single character.</li>
<li><code>'*'</code> Matches zero or more... | Recursion; String; Dynamic Programming | Go | func isMatch(s string, p string) bool {
m, n := len(s), len(p)
f := make([][]int, m+1)
for i := range f {
f[i] = make([]int, n+1)
}
var dfs func(i, j int) bool
dfs = func(i, j int) bool {
if j >= n {
return i == m
}
if f[i][j] != 0 {
return f[i][j] == 1
}
res := -1
if j+1 < n && p[j+1] == '*' ... |
10 | Regular Expression Matching | Hard | <p>Given an input string <code>s</code> and a pattern <code>p</code>, implement regular expression matching with support for <code>'.'</code> and <code>'*'</code> where:</p>
<ul>
<li><code>'.'</code> Matches any single character.</li>
<li><code>'*'</code> Matches zero or more... | Recursion; String; Dynamic Programming | Java | class Solution {
private Boolean[][] f;
private String s;
private String p;
private int m;
private int n;
public boolean isMatch(String s, String p) {
m = s.length();
n = p.length();
f = new Boolean[m + 1][n + 1];
this.s = s;
this.p = p;
return df... |
10 | Regular Expression Matching | Hard | <p>Given an input string <code>s</code> and a pattern <code>p</code>, implement regular expression matching with support for <code>'.'</code> and <code>'*'</code> where:</p>
<ul>
<li><code>'.'</code> Matches any single character.</li>
<li><code>'*'</code> Matches zero or more... | Recursion; String; Dynamic Programming | JavaScript | /**
* @param {string} s
* @param {string} p
* @return {boolean}
*/
var isMatch = function (s, p) {
const m = s.length;
const n = p.length;
const f = Array.from({ length: m + 1 }, () => Array(n + 1).fill(0));
const dfs = (i, j) => {
if (j >= n) {
return i === m;
}
... |
10 | Regular Expression Matching | Hard | <p>Given an input string <code>s</code> and a pattern <code>p</code>, implement regular expression matching with support for <code>'.'</code> and <code>'*'</code> where:</p>
<ul>
<li><code>'.'</code> Matches any single character.</li>
<li><code>'*'</code> Matches zero or more... | Recursion; String; Dynamic Programming | PHP | class Solution {
/**
* @param String $s
* @param String $p
* @return Boolean
*/
function isMatch($s, $p) {
$m = strlen($s);
$n = strlen($p);
$f = array_fill(0, $m + 1, array_fill(0, $n + 1, 0));
$dfs = function ($i, $j) use (&$s, &$p, $m, $n, &$f, &$dfs) {
... |
10 | Regular Expression Matching | Hard | <p>Given an input string <code>s</code> and a pattern <code>p</code>, implement regular expression matching with support for <code>'.'</code> and <code>'*'</code> where:</p>
<ul>
<li><code>'.'</code> Matches any single character.</li>
<li><code>'*'</code> Matches zero or more... | Recursion; String; Dynamic Programming | Python | class Solution:
def isMatch(self, s: str, p: str) -> bool:
@cache
def dfs(i, j):
if j >= n:
return i == m
if j + 1 < n and p[j + 1] == '*':
return dfs(i, j + 2) or (
i < m and (s[i] == p[j] or p[j] == '.') and dfs(i + 1, j)
... |
10 | Regular Expression Matching | Hard | <p>Given an input string <code>s</code> and a pattern <code>p</code>, implement regular expression matching with support for <code>'.'</code> and <code>'*'</code> where:</p>
<ul>
<li><code>'.'</code> Matches any single character.</li>
<li><code>'*'</code> Matches zero or more... | Recursion; String; Dynamic Programming | Rust | impl Solution {
pub fn is_match(s: String, p: String) -> bool {
let (m, n) = (s.len(), p.len());
let mut f = vec![vec![0; n + 1]; m + 1];
fn dfs(
s: &Vec<char>,
p: &Vec<char>,
f: &mut Vec<Vec<i32>>,
i: usize,
j: usize,
... |
11 | Container With Most Water | Medium | <p>You are given an integer array <code>height</code> of length <code>n</code>. There are <code>n</code> vertical lines drawn such that the two endpoints of the <code>i<sup>th</sup></code> line are <code>(i, 0)</code> and <code>(i, height[i])</code>.</p>
<p>Find two lines that together with the x-axis form a container... | Greedy; Array; Two Pointers | C | int min(int a, int b) {
return a < b ? a : b;
}
int max(int a, int b) {
return a > b ? a : b;
}
int maxArea(int* height, int heightSize) {
int l = 0, r = heightSize - 1;
int ans = 0;
while (l < r) {
int t = min(height[l], height[r]) * (r - l);
ans = max(ans, t);
if (height[... |
11 | Container With Most Water | Medium | <p>You are given an integer array <code>height</code> of length <code>n</code>. There are <code>n</code> vertical lines drawn such that the two endpoints of the <code>i<sup>th</sup></code> line are <code>(i, 0)</code> and <code>(i, height[i])</code>.</p>
<p>Find two lines that together with the x-axis form a container... | Greedy; Array; Two Pointers | C++ | class Solution {
public:
int maxArea(vector<int>& height) {
int l = 0, r = height.size() - 1;
int ans = 0;
while (l < r) {
int t = min(height[l], height[r]) * (r - l);
ans = max(ans, t);
if (height[l] < height[r]) {
++l;
} else ... |
11 | Container With Most Water | Medium | <p>You are given an integer array <code>height</code> of length <code>n</code>. There are <code>n</code> vertical lines drawn such that the two endpoints of the <code>i<sup>th</sup></code> line are <code>(i, 0)</code> and <code>(i, height[i])</code>.</p>
<p>Find two lines that together with the x-axis form a container... | Greedy; Array; Two Pointers | C# | public class Solution {
public int MaxArea(int[] height) {
int l = 0, r = height.Length - 1;
int ans = 0;
while (l < r) {
int t = Math.Min(height[l], height[r]) * (r - l);
ans = Math.Max(ans, t);
if (height[l] < height[r]) {
++l;
... |
11 | Container With Most Water | Medium | <p>You are given an integer array <code>height</code> of length <code>n</code>. There are <code>n</code> vertical lines drawn such that the two endpoints of the <code>i<sup>th</sup></code> line are <code>(i, 0)</code> and <code>(i, height[i])</code>.</p>
<p>Find two lines that together with the x-axis form a container... | Greedy; Array; Two Pointers | Go | func maxArea(height []int) (ans int) {
l, r := 0, len(height)-1
for l < r {
t := min(height[l], height[r]) * (r - l)
ans = max(ans, t)
if height[l] < height[r] {
l++
} else {
r--
}
}
return
}
|
11 | Container With Most Water | Medium | <p>You are given an integer array <code>height</code> of length <code>n</code>. There are <code>n</code> vertical lines drawn such that the two endpoints of the <code>i<sup>th</sup></code> line are <code>(i, 0)</code> and <code>(i, height[i])</code>.</p>
<p>Find two lines that together with the x-axis form a container... | Greedy; Array; Two Pointers | Java | class Solution {
public int maxArea(int[] height) {
int l = 0, r = height.length - 1;
int ans = 0;
while (l < r) {
int t = Math.min(height[l], height[r]) * (r - l);
ans = Math.max(ans, t);
if (height[l] < height[r]) {
++l;
} els... |
11 | Container With Most Water | Medium | <p>You are given an integer array <code>height</code> of length <code>n</code>. There are <code>n</code> vertical lines drawn such that the two endpoints of the <code>i<sup>th</sup></code> line are <code>(i, 0)</code> and <code>(i, height[i])</code>.</p>
<p>Find two lines that together with the x-axis form a container... | Greedy; Array; Two Pointers | JavaScript | /**
* @param {number[]} height
* @return {number}
*/
var maxArea = function (height) {
let [l, r] = [0, height.length - 1];
let ans = 0;
while (l < r) {
const t = Math.min(height[l], height[r]) * (r - l);
ans = Math.max(ans, t);
if (height[l] < height[r]) {
++l;
... |
11 | Container With Most Water | Medium | <p>You are given an integer array <code>height</code> of length <code>n</code>. There are <code>n</code> vertical lines drawn such that the two endpoints of the <code>i<sup>th</sup></code> line are <code>(i, 0)</code> and <code>(i, height[i])</code>.</p>
<p>Find two lines that together with the x-axis form a container... | Greedy; Array; Two Pointers | PHP | class Solution {
/**
* @param Integer[] $height
* @return Integer
*/
function maxArea($height) {
$l = 0;
$r = count($height) - 1;
$ans = 0;
while ($l < $r) {
$t = min($height[$l], $height[$r]) * ($r - $l);
$ans = max($ans, $t);
i... |
11 | Container With Most Water | Medium | <p>You are given an integer array <code>height</code> of length <code>n</code>. There are <code>n</code> vertical lines drawn such that the two endpoints of the <code>i<sup>th</sup></code> line are <code>(i, 0)</code> and <code>(i, height[i])</code>.</p>
<p>Find two lines that together with the x-axis form a container... | Greedy; Array; Two Pointers | Python | class Solution:
def maxArea(self, height: List[int]) -> int:
l, r = 0, len(height) - 1
ans = 0
while l < r:
t = min(height[l], height[r]) * (r - l)
ans = max(ans, t)
if height[l] < height[r]:
l += 1
else:
r -= 1
... |
11 | Container With Most Water | Medium | <p>You are given an integer array <code>height</code> of length <code>n</code>. There are <code>n</code> vertical lines drawn such that the two endpoints of the <code>i<sup>th</sup></code> line are <code>(i, 0)</code> and <code>(i, height[i])</code>.</p>
<p>Find two lines that together with the x-axis form a container... | Greedy; Array; Two Pointers | Rust | impl Solution {
pub fn max_area(height: Vec<i32>) -> i32 {
let mut l = 0;
let mut r = height.len() - 1;
let mut ans = 0;
while l < r {
ans = ans.max(height[l].min(height[r]) * ((r - l) as i32));
if height[l] < height[r] {
l += 1;
} ... |
11 | Container With Most Water | Medium | <p>You are given an integer array <code>height</code> of length <code>n</code>. There are <code>n</code> vertical lines drawn such that the two endpoints of the <code>i<sup>th</sup></code> line are <code>(i, 0)</code> and <code>(i, height[i])</code>.</p>
<p>Find two lines that together with the x-axis form a container... | Greedy; Array; Two Pointers | TypeScript | function maxArea(height: number[]): number {
let [l, r] = [0, height.length - 1];
let ans = 0;
while (l < r) {
const t = Math.min(height[l], height[r]) * (r - l);
ans = Math.max(ans, t);
if (height[l] < height[r]) {
++l;
} else {
--r;
}
}
... |
12 | Integer to Roman | Medium | <p>Seven different symbols represent Roman numerals with the following values:</p>
<table>
<thead>
<tr>
<th>Symbol</th>
<th>Value</th>
</tr>
</thead>
<tbody>
<tr>
<td>I</td>
<td>1</td>
</tr>
<tr>
<td>V</td>
<td>5</td>
</tr>
<tr>
<td>X</td>
<td>10</td>
</tr>
<tr>
<td>L</t... | Hash Table; Math; String | C | static const char* cs[] = {
"M", "CM", "D", "CD", "C", "XC",
"L", "XL", "X", "IX", "V", "IV", "I"};
static const int vs[] = {
1000, 900, 500, 400, 100, 90,
50, 40, 10, 9, 5, 4, 1};
char* intToRoman(int num) {
static char ans[20];
ans[0] = '\0';
for (int i = 0; i < 13; ++i) {
while ... |
12 | Integer to Roman | Medium | <p>Seven different symbols represent Roman numerals with the following values:</p>
<table>
<thead>
<tr>
<th>Symbol</th>
<th>Value</th>
</tr>
</thead>
<tbody>
<tr>
<td>I</td>
<td>1</td>
</tr>
<tr>
<td>V</td>
<td>5</td>
</tr>
<tr>
<td>X</td>
<td>10</td>
</tr>
<tr>
<td>L</t... | Hash Table; Math; String | C++ | class Solution {
public:
string intToRoman(int num) {
vector<string> cs = {"M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"};
vector<int> vs = {1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1};
string ans;
for (int i = 0; i < cs.size(); ++i) {
wh... |
12 | Integer to Roman | Medium | <p>Seven different symbols represent Roman numerals with the following values:</p>
<table>
<thead>
<tr>
<th>Symbol</th>
<th>Value</th>
</tr>
</thead>
<tbody>
<tr>
<td>I</td>
<td>1</td>
</tr>
<tr>
<td>V</td>
<td>5</td>
</tr>
<tr>
<td>X</td>
<td>10</td>
</tr>
<tr>
<td>L</t... | Hash Table; Math; String | C# | public class Solution {
public string IntToRoman(int num) {
List<string> cs = new List<string>{"M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"};
List<int> vs = new List<int>{1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1};
StringBuilder ans = new StringBuilder();
... |
12 | Integer to Roman | Medium | <p>Seven different symbols represent Roman numerals with the following values:</p>
<table>
<thead>
<tr>
<th>Symbol</th>
<th>Value</th>
</tr>
</thead>
<tbody>
<tr>
<td>I</td>
<td>1</td>
</tr>
<tr>
<td>V</td>
<td>5</td>
</tr>
<tr>
<td>X</td>
<td>10</td>
</tr>
<tr>
<td>L</t... | Hash Table; Math; String | Go | func intToRoman(num int) string {
cs := []string{"M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"}
vs := []int{1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1}
ans := &strings.Builder{}
for i, v := range vs {
for num >= v {
num -= v
ans.WriteString(cs[i])
}
}
return ans.String(... |
12 | Integer to Roman | Medium | <p>Seven different symbols represent Roman numerals with the following values:</p>
<table>
<thead>
<tr>
<th>Symbol</th>
<th>Value</th>
</tr>
</thead>
<tbody>
<tr>
<td>I</td>
<td>1</td>
</tr>
<tr>
<td>V</td>
<td>5</td>
</tr>
<tr>
<td>X</td>
<td>10</td>
</tr>
<tr>
<td>L</t... | Hash Table; Math; String | Java | class Solution {
public String intToRoman(int num) {
List<String> cs
= List.of("M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I");
List<Integer> vs = List.of(1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1);
StringBuilder ans = new StringBuilder();
... |
12 | Integer to Roman | Medium | <p>Seven different symbols represent Roman numerals with the following values:</p>
<table>
<thead>
<tr>
<th>Symbol</th>
<th>Value</th>
</tr>
</thead>
<tbody>
<tr>
<td>I</td>
<td>1</td>
</tr>
<tr>
<td>V</td>
<td>5</td>
</tr>
<tr>
<td>X</td>
<td>10</td>
</tr>
<tr>
<td>L</t... | Hash Table; Math; String | PHP | class Solution {
/**
* @param Integer $num
* @return String
*/
function intToRoman($num) {
$cs = ['M', 'CM', 'D', 'CD', 'C', 'XC', 'L', 'XL', 'X', 'IX', 'V', 'IV', 'I'];
$vs = [1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1];
$ans = '';
foreach ($vs as $i =>... |
12 | Integer to Roman | Medium | <p>Seven different symbols represent Roman numerals with the following values:</p>
<table>
<thead>
<tr>
<th>Symbol</th>
<th>Value</th>
</tr>
</thead>
<tbody>
<tr>
<td>I</td>
<td>1</td>
</tr>
<tr>
<td>V</td>
<td>5</td>
</tr>
<tr>
<td>X</td>
<td>10</td>
</tr>
<tr>
<td>L</t... | Hash Table; Math; String | Python | class Solution:
def intToRoman(self, num: int) -> str:
cs = ('M', 'CM', 'D', 'CD', 'C', 'XC', 'L', 'XL', 'X', 'IX', 'V', 'IV', 'I')
vs = (1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1)
ans = []
for c, v in zip(cs, vs):
while num >= v:
num -= v
... |
12 | Integer to Roman | Medium | <p>Seven different symbols represent Roman numerals with the following values:</p>
<table>
<thead>
<tr>
<th>Symbol</th>
<th>Value</th>
</tr>
</thead>
<tbody>
<tr>
<td>I</td>
<td>1</td>
</tr>
<tr>
<td>V</td>
<td>5</td>
</tr>
<tr>
<td>X</td>
<td>10</td>
</tr>
<tr>
<td>L</t... | Hash Table; Math; String | Rust | impl Solution {
pub fn int_to_roman(num: i32) -> String {
let cs = [
"M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I",
];
let vs = [1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1];
let mut num = num;
let mut ans = String::new();
... |
12 | Integer to Roman | Medium | <p>Seven different symbols represent Roman numerals with the following values:</p>
<table>
<thead>
<tr>
<th>Symbol</th>
<th>Value</th>
</tr>
</thead>
<tbody>
<tr>
<td>I</td>
<td>1</td>
</tr>
<tr>
<td>V</td>
<td>5</td>
</tr>
<tr>
<td>X</td>
<td>10</td>
</tr>
<tr>
<td>L</t... | Hash Table; Math; String | TypeScript | function intToRoman(num: number): string {
const cs: string[] = ['M', 'CM', 'D', 'CD', 'C', 'XC', 'L', 'XL', 'X', 'IX', 'V', 'IV', 'I'];
const vs: number[] = [1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1];
const ans: string[] = [];
for (let i = 0; i < vs.length; ++i) {
while (num >= vs[i... |
13 | Roman to Integer | Easy | <p>Roman numerals are represented by seven different symbols: <code>I</code>, <code>V</code>, <code>X</code>, <code>L</code>, <code>C</code>, <code>D</code> and <code>M</code>.</p>
<pre>
<strong>Symbol</strong> <strong>Value</strong>
I 1
V 5
X 10
L 50
C ... | Hash Table; Math; String | C | int nums(char c) {
switch (c) {
case 'I': return 1;
case 'V': return 5;
case 'X': return 10;
case 'L': return 50;
case 'C': return 100;
case 'D': return 500;
case 'M': return 1000;
default: return 0;
}
}
int romanToInt(char* s) {
int ans = nums(s[strlen(s) - 1]);
for (in... |
13 | Roman to Integer | Easy | <p>Roman numerals are represented by seven different symbols: <code>I</code>, <code>V</code>, <code>X</code>, <code>L</code>, <code>C</code>, <code>D</code> and <code>M</code>.</p>
<pre>
<strong>Symbol</strong> <strong>Value</strong>
I 1
V 5
X 10
L 50
C ... | Hash Table; Math; String | C++ | class Solution {
public:
int romanToInt(string s) {
unordered_map<char, int> nums{
{'I', 1},
{'V', 5},
{'X', 10},
{'L', 50},
{'C', 100},
{'D', 500},
{'M', 1000},
};
int ans = nums[s.back()];
for (int ... |
13 | Roman to Integer | Easy | <p>Roman numerals are represented by seven different symbols: <code>I</code>, <code>V</code>, <code>X</code>, <code>L</code>, <code>C</code>, <code>D</code> and <code>M</code>.</p>
<pre>
<strong>Symbol</strong> <strong>Value</strong>
I 1
V 5
X 10
L 50
C ... | Hash Table; Math; String | C# | public class Solution {
public int RomanToInt(string s) {
Dictionary<char, int> d = new Dictionary<char, int>();
d.Add('I', 1);
d.Add('V', 5);
d.Add('X', 10);
d.Add('L', 50);
d.Add('C', 100);
d.Add('D', 500);
d.Add('M', 1000);
int ans = d[s[s.L... |
13 | Roman to Integer | Easy | <p>Roman numerals are represented by seven different symbols: <code>I</code>, <code>V</code>, <code>X</code>, <code>L</code>, <code>C</code>, <code>D</code> and <code>M</code>.</p>
<pre>
<strong>Symbol</strong> <strong>Value</strong>
I 1
V 5
X 10
L 50
C ... | Hash Table; Math; String | Go | func romanToInt(s string) (ans int) {
d := map[byte]int{'I': 1, 'V': 5, 'X': 10, 'L': 50, 'C': 100, 'D': 500, 'M': 1000}
for i := 0; i < len(s)-1; i++ {
if d[s[i]] < d[s[i+1]] {
ans -= d[s[i]]
} else {
ans += d[s[i]]
}
}
ans += d[s[len(s)-1]]
return
} |
13 | Roman to Integer | Easy | <p>Roman numerals are represented by seven different symbols: <code>I</code>, <code>V</code>, <code>X</code>, <code>L</code>, <code>C</code>, <code>D</code> and <code>M</code>.</p>
<pre>
<strong>Symbol</strong> <strong>Value</strong>
I 1
V 5
X 10
L 50
C ... | Hash Table; Math; String | Java | class Solution {
public int romanToInt(String s) {
String cs = "IVXLCDM";
int[] vs = {1, 5, 10, 50, 100, 500, 1000};
Map<Character, Integer> d = new HashMap<>();
for (int i = 0; i < vs.length; ++i) {
d.put(cs.charAt(i), vs[i]);
}
int n = s.length();
... |
13 | Roman to Integer | Easy | <p>Roman numerals are represented by seven different symbols: <code>I</code>, <code>V</code>, <code>X</code>, <code>L</code>, <code>C</code>, <code>D</code> and <code>M</code>.</p>
<pre>
<strong>Symbol</strong> <strong>Value</strong>
I 1
V 5
X 10
L 50
C ... | Hash Table; Math; String | JavaScript | const romanToInt = function (s) {
const d = {
I: 1,
V: 5,
X: 10,
L: 50,
C: 100,
D: 500,
M: 1000,
};
let ans = d[s[s.length - 1]];
for (let i = 0; i < s.length - 1; ++i) {
const sign = d[s[i]] < d[s[i + 1]] ? -1 : 1;
ans += sign * d[... |
13 | Roman to Integer | Easy | <p>Roman numerals are represented by seven different symbols: <code>I</code>, <code>V</code>, <code>X</code>, <code>L</code>, <code>C</code>, <code>D</code> and <code>M</code>.</p>
<pre>
<strong>Symbol</strong> <strong>Value</strong>
I 1
V 5
X 10
L 50
C ... | Hash Table; Math; String | PHP | class Solution {
/**
* @param String $s
* @return Integer
*/
function romanToInt($s) {
$d = [
'I' => 1,
'V' => 5,
'X' => 10,
'L' => 50,
'C' => 100,
'D' => 500,
'M' => 1000,
];
$ans = 0;
... |
13 | Roman to Integer | Easy | <p>Roman numerals are represented by seven different symbols: <code>I</code>, <code>V</code>, <code>X</code>, <code>L</code>, <code>C</code>, <code>D</code> and <code>M</code>.</p>
<pre>
<strong>Symbol</strong> <strong>Value</strong>
I 1
V 5
X 10
L 50
C ... | Hash Table; Math; String | Python | class Solution:
def romanToInt(self, s: str) -> int:
d = {'I': 1, 'V': 5, 'X': 10, 'L': 50, 'C': 100, 'D': 500, 'M': 1000}
return sum((-1 if d[a] < d[b] else 1) * d[a] for a, b in pairwise(s)) + d[s[-1]]
|
13 | Roman to Integer | Easy | <p>Roman numerals are represented by seven different symbols: <code>I</code>, <code>V</code>, <code>X</code>, <code>L</code>, <code>C</code>, <code>D</code> and <code>M</code>.</p>
<pre>
<strong>Symbol</strong> <strong>Value</strong>
I 1
V 5
X 10
L 50
C ... | Hash Table; Math; String | Ruby | # @param {String} s
# @return {Integer}
def roman_to_int(s)
d = {
'I' => 1, 'V' => 5, 'X' => 10,
'L' => 50, 'C' => 100,
'D' => 500, 'M' => 1000
}
ans = 0
len = s.length
(0...len-1).each do |i|
if d[s[i]] < d[s[i + 1]]
ans -= d[s[i]]
else
ans += d[s[i]]
... |
13 | Roman to Integer | Easy | <p>Roman numerals are represented by seven different symbols: <code>I</code>, <code>V</code>, <code>X</code>, <code>L</code>, <code>C</code>, <code>D</code> and <code>M</code>.</p>
<pre>
<strong>Symbol</strong> <strong>Value</strong>
I 1
V 5
X 10
L 50
C ... | Hash Table; Math; String | Rust | impl Solution {
pub fn roman_to_int(s: String) -> i32 {
let d = vec![
('I', 1),
('V', 5),
('X', 10),
('L', 50),
('C', 100),
('D', 500),
('M', 1000),
]
.into_iter()
.collect::<std::collections::HashMap... |
13 | Roman to Integer | Easy | <p>Roman numerals are represented by seven different symbols: <code>I</code>, <code>V</code>, <code>X</code>, <code>L</code>, <code>C</code>, <code>D</code> and <code>M</code>.</p>
<pre>
<strong>Symbol</strong> <strong>Value</strong>
I 1
V 5
X 10
L 50
C ... | Hash Table; Math; String | TypeScript | function romanToInt(s: string): number {
const d: Map<string, number> = new Map([
['I', 1],
['V', 5],
['X', 10],
['L', 50],
['C', 100],
['D', 500],
['M', 1000],
]);
let ans: number = d.get(s[s.length - 1])!;
for (let i = 0; i < s.length - 1; ++i) {... |
14 | Longest Common Prefix | Easy | <p>Write a function to find the longest common prefix string amongst an array of strings.</p>
<p>If there is no common prefix, return an empty string <code>""</code>.</p>
<p> </p>
<p><strong class="example">Example 1:</strong></p>
<pre>
<strong>Input:</strong> strs = ["flower","flow&quo... | Trie; Array; String | C | char* longestCommonPrefix(char** strs, int strsSize) {
for (int i = 0; strs[0][i]; i++) {
for (int j = 1; j < strsSize; j++) {
if (strs[j][i] != strs[0][i]) {
strs[0][i] = '\0';
return strs[0];
}
}
}
return strs[0];
}
|
14 | Longest Common Prefix | Easy | <p>Write a function to find the longest common prefix string amongst an array of strings.</p>
<p>If there is no common prefix, return an empty string <code>""</code>.</p>
<p> </p>
<p><strong class="example">Example 1:</strong></p>
<pre>
<strong>Input:</strong> strs = ["flower","flow&quo... | Trie; Array; String | C++ | class Solution {
public:
string longestCommonPrefix(vector<string>& strs) {
int n = strs.size();
for (int i = 0; i < strs[0].size(); ++i) {
for (int j = 1; j < n; ++j) {
if (strs[j].size() <= i || strs[j][i] != strs[0][i]) {
return strs[0].substr(0, i)... |
14 | Longest Common Prefix | Easy | <p>Write a function to find the longest common prefix string amongst an array of strings.</p>
<p>If there is no common prefix, return an empty string <code>""</code>.</p>
<p> </p>
<p><strong class="example">Example 1:</strong></p>
<pre>
<strong>Input:</strong> strs = ["flower","flow&quo... | Trie; Array; String | C# | public class Solution {
public string LongestCommonPrefix(string[] strs) {
int n = strs.Length;
for (int i = 0; i < strs[0].Length; ++i) {
for (int j = 1; j < n; ++j) {
if (i >= strs[j].Length || strs[j][i] != strs[0][i]) {
return strs[0].Substring(0, ... |
14 | Longest Common Prefix | Easy | <p>Write a function to find the longest common prefix string amongst an array of strings.</p>
<p>If there is no common prefix, return an empty string <code>""</code>.</p>
<p> </p>
<p><strong class="example">Example 1:</strong></p>
<pre>
<strong>Input:</strong> strs = ["flower","flow&quo... | Trie; Array; String | Go | func longestCommonPrefix(strs []string) string {
n := len(strs)
for i := range strs[0] {
for j := 1; j < n; j++ {
if len(strs[j]) <= i || strs[j][i] != strs[0][i] {
return strs[0][:i]
}
}
}
return strs[0]
} |
14 | Longest Common Prefix | Easy | <p>Write a function to find the longest common prefix string amongst an array of strings.</p>
<p>If there is no common prefix, return an empty string <code>""</code>.</p>
<p> </p>
<p><strong class="example">Example 1:</strong></p>
<pre>
<strong>Input:</strong> strs = ["flower","flow&quo... | Trie; Array; String | Java | class Solution {
public String longestCommonPrefix(String[] strs) {
int n = strs.length;
for (int i = 0; i < strs[0].length(); ++i) {
for (int j = 1; j < n; ++j) {
if (strs[j].length() <= i || strs[j].charAt(i) != strs[0].charAt(i)) {
return strs[0].su... |
14 | Longest Common Prefix | Easy | <p>Write a function to find the longest common prefix string amongst an array of strings.</p>
<p>If there is no common prefix, return an empty string <code>""</code>.</p>
<p> </p>
<p><strong class="example">Example 1:</strong></p>
<pre>
<strong>Input:</strong> strs = ["flower","flow&quo... | Trie; Array; String | JavaScript | /**
* @param {string[]} strs
* @return {string}
*/
var longestCommonPrefix = function (strs) {
for (let j = 0; j < strs[0].length; j++) {
for (let i = 0; i < strs.length; i++) {
if (strs[0][j] !== strs[i][j]) {
return strs[0].substring(0, j);
}
}
}
... |
14 | Longest Common Prefix | Easy | <p>Write a function to find the longest common prefix string amongst an array of strings.</p>
<p>If there is no common prefix, return an empty string <code>""</code>.</p>
<p> </p>
<p><strong class="example">Example 1:</strong></p>
<pre>
<strong>Input:</strong> strs = ["flower","flow&quo... | Trie; Array; String | PHP | class Solution {
/**
* @param String[] $strs
* @return String
*/
function longestCommonPrefix($strs) {
$rs = '';
for ($i = 0; $i < strlen($strs[0]); $i++) {
for ($j = 1; $j < count($strs); $j++) {
if ($strs[0][$i] != $strs[$j][$i]) {
... |
14 | Longest Common Prefix | Easy | <p>Write a function to find the longest common prefix string amongst an array of strings.</p>
<p>If there is no common prefix, return an empty string <code>""</code>.</p>
<p> </p>
<p><strong class="example">Example 1:</strong></p>
<pre>
<strong>Input:</strong> strs = ["flower","flow&quo... | Trie; Array; String | Python | class Solution:
def longestCommonPrefix(self, strs: List[str]) -> str:
for i in range(len(strs[0])):
for s in strs[1:]:
if len(s) <= i or s[i] != strs[0][i]:
return s[:i]
return strs[0]
|
14 | Longest Common Prefix | Easy | <p>Write a function to find the longest common prefix string amongst an array of strings.</p>
<p>If there is no common prefix, return an empty string <code>""</code>.</p>
<p> </p>
<p><strong class="example">Example 1:</strong></p>
<pre>
<strong>Input:</strong> strs = ["flower","flow&quo... | Trie; Array; String | Ruby | # @param {String[]} strs
# @return {String}
def longest_common_prefix(strs)
return '' if strs.nil? || strs.length.zero?
return strs[0] if strs.length == 1
idx = 0
while idx < strs[0].length
cur_char = strs[0][idx]
str_idx = 1
while str_idx < strs.length
return idx > 0 ? strs[0][0..idx-1] : ... |
14 | Longest Common Prefix | Easy | <p>Write a function to find the longest common prefix string amongst an array of strings.</p>
<p>If there is no common prefix, return an empty string <code>""</code>.</p>
<p> </p>
<p><strong class="example">Example 1:</strong></p>
<pre>
<strong>Input:</strong> strs = ["flower","flow&quo... | Trie; Array; String | Rust | impl Solution {
pub fn longest_common_prefix(strs: Vec<String>) -> String {
let mut len = strs.iter().map(|s| s.len()).min().unwrap();
for i in (1..=len).rev() {
let mut is_equal = true;
let target = strs[0][0..i].to_string();
if strs.iter().all(|s| target == s[0.... |
14 | Longest Common Prefix | Easy | <p>Write a function to find the longest common prefix string amongst an array of strings.</p>
<p>If there is no common prefix, return an empty string <code>""</code>.</p>
<p> </p>
<p><strong class="example">Example 1:</strong></p>
<pre>
<strong>Input:</strong> strs = ["flower","flow&quo... | Trie; Array; String | TypeScript | function longestCommonPrefix(strs: string[]): string {
const len = strs.reduce((r, s) => Math.min(r, s.length), Infinity);
for (let i = len; i > 0; i--) {
const target = strs[0].slice(0, i);
if (strs.every(s => s.slice(0, i) === target)) {
return target;
}
}
return ''... |
15 | 3Sum | Medium | <p>Given an integer array nums, return all the triplets <code>[nums[i], nums[j], nums[k]]</code> such that <code>i != j</code>, <code>i != k</code>, and <code>j != k</code>, and <code>nums[i] + nums[j] + nums[k] == 0</code>.</p>
<p>Notice that the solution set must not contain duplicate triplets.</p>
<p> </p>
<p... | Array; Two Pointers; Sorting | C | int cmp(const void* a, const void* b) {
return *(int*) a - *(int*) b;
}
int** threeSum(int* nums, int numsSize, int* returnSize, int** returnColumnSizes) {
*returnSize = 0;
int cap = 1000;
int** ans = (int**) malloc(sizeof(int*) * cap);
*returnColumnSizes = (int*) malloc(sizeof(int) * cap);
qs... |
15 | 3Sum | Medium | <p>Given an integer array nums, return all the triplets <code>[nums[i], nums[j], nums[k]]</code> such that <code>i != j</code>, <code>i != k</code>, and <code>j != k</code>, and <code>nums[i] + nums[j] + nums[k] == 0</code>.</p>
<p>Notice that the solution set must not contain duplicate triplets.</p>
<p> </p>
<p... | Array; Two Pointers; Sorting | C++ | class Solution {
public:
vector<vector<int>> threeSum(vector<int>& nums) {
sort(nums.begin(), nums.end());
vector<vector<int>> ans;
int n = nums.size();
for (int i = 0; i < n - 2 && nums[i] <= 0; ++i) {
if (i && nums[i] == nums[i - 1]) {
continue;
... |
15 | 3Sum | Medium | <p>Given an integer array nums, return all the triplets <code>[nums[i], nums[j], nums[k]]</code> such that <code>i != j</code>, <code>i != k</code>, and <code>j != k</code>, and <code>nums[i] + nums[j] + nums[k] == 0</code>.</p>
<p>Notice that the solution set must not contain duplicate triplets.</p>
<p> </p>
<p... | Array; Two Pointers; Sorting | C# | public class Solution {
public IList<IList<int>> ThreeSum(int[] nums) {
Array.Sort(nums);
int n = nums.Length;
IList<IList<int>> ans = new List<IList<int>>();
for (int i = 0; i < n - 2 && nums[i] <= 0; ++i) {
if (i > 0 && nums[i] == nums[i - 1]) {
continue... |
15 | 3Sum | Medium | <p>Given an integer array nums, return all the triplets <code>[nums[i], nums[j], nums[k]]</code> such that <code>i != j</code>, <code>i != k</code>, and <code>j != k</code>, and <code>nums[i] + nums[j] + nums[k] == 0</code>.</p>
<p>Notice that the solution set must not contain duplicate triplets.</p>
<p> </p>
<p... | Array; Two Pointers; Sorting | Go | func threeSum(nums []int) (ans [][]int) {
sort.Ints(nums)
n := len(nums)
for i := 0; i < n-2 && nums[i] <= 0; i++ {
if i > 0 && nums[i] == nums[i-1] {
continue
}
j, k := i+1, n-1
for j < k {
x := nums[i] + nums[j] + nums[k]
if x < 0 {
j++
} else if x > 0 {
k--
} else {
ans = append... |
15 | 3Sum | Medium | <p>Given an integer array nums, return all the triplets <code>[nums[i], nums[j], nums[k]]</code> such that <code>i != j</code>, <code>i != k</code>, and <code>j != k</code>, and <code>nums[i] + nums[j] + nums[k] == 0</code>.</p>
<p>Notice that the solution set must not contain duplicate triplets.</p>
<p> </p>
<p... | Array; Two Pointers; Sorting | Java | class Solution {
public List<List<Integer>> threeSum(int[] nums) {
Arrays.sort(nums);
List<List<Integer>> ans = new ArrayList<>();
int n = nums.length;
for (int i = 0; i < n - 2 && nums[i] <= 0; ++i) {
if (i > 0 && nums[i] == nums[i - 1]) {
continue;
... |
15 | 3Sum | Medium | <p>Given an integer array nums, return all the triplets <code>[nums[i], nums[j], nums[k]]</code> such that <code>i != j</code>, <code>i != k</code>, and <code>j != k</code>, and <code>nums[i] + nums[j] + nums[k] == 0</code>.</p>
<p>Notice that the solution set must not contain duplicate triplets.</p>
<p> </p>
<p... | Array; Two Pointers; Sorting | JavaScript | /**
* @param {number[]} nums
* @return {number[][]}
*/
var threeSum = function (nums) {
const n = nums.length;
nums.sort((a, b) => a - b);
const ans = [];
for (let i = 0; i < n - 2 && nums[i] <= 0; ++i) {
if (i > 0 && nums[i] === nums[i - 1]) {
continue;
}
let j = ... |
15 | 3Sum | Medium | <p>Given an integer array nums, return all the triplets <code>[nums[i], nums[j], nums[k]]</code> such that <code>i != j</code>, <code>i != k</code>, and <code>j != k</code>, and <code>nums[i] + nums[j] + nums[k] == 0</code>.</p>
<p>Notice that the solution set must not contain duplicate triplets.</p>
<p> </p>
<p... | Array; Two Pointers; Sorting | PHP | class Solution {
/**
* @param Integer[] $nums
* @return Integer[][]
*/
function threeSum($nums) {
sort($nums);
$ans = [];
$n = count($nums);
for ($i = 0; $i < $n - 2 && $nums[$i] <= 0; ++$i) {
if ($i > 0 && $nums[$i] == $nums[$i - 1]) {
... |
15 | 3Sum | Medium | <p>Given an integer array nums, return all the triplets <code>[nums[i], nums[j], nums[k]]</code> such that <code>i != j</code>, <code>i != k</code>, and <code>j != k</code>, and <code>nums[i] + nums[j] + nums[k] == 0</code>.</p>
<p>Notice that the solution set must not contain duplicate triplets.</p>
<p> </p>
<p... | Array; Two Pointers; Sorting | Python | class Solution:
def threeSum(self, nums: List[int]) -> List[List[int]]:
nums.sort()
n = len(nums)
ans = []
for i in range(n - 2):
if nums[i] > 0:
break
if i and nums[i] == nums[i - 1]:
continue
j, k = i + 1, n - 1
... |
15 | 3Sum | Medium | <p>Given an integer array nums, return all the triplets <code>[nums[i], nums[j], nums[k]]</code> such that <code>i != j</code>, <code>i != k</code>, and <code>j != k</code>, and <code>nums[i] + nums[j] + nums[k] == 0</code>.</p>
<p>Notice that the solution set must not contain duplicate triplets.</p>
<p> </p>
<p... | Array; Two Pointers; Sorting | Ruby | # @param {Integer[]} nums
# @return {Integer[][]}
def three_sum(nums)
res = []
nums.sort!
for i in 0..(nums.length - 3)
next if i > 0 && nums[i - 1] == nums[i]
j = i + 1
k = nums.length - 1
while j < k do
sum = nums[i] + nums[j] + nums[k]
if sum < 0
j += 1
elsif sum > 0
... |
15 | 3Sum | Medium | <p>Given an integer array nums, return all the triplets <code>[nums[i], nums[j], nums[k]]</code> such that <code>i != j</code>, <code>i != k</code>, and <code>j != k</code>, and <code>nums[i] + nums[j] + nums[k] == 0</code>.</p>
<p>Notice that the solution set must not contain duplicate triplets.</p>
<p> </p>
<p... | Array; Two Pointers; Sorting | Rust | use std::cmp::Ordering;
impl Solution {
pub fn three_sum(mut nums: Vec<i32>) -> Vec<Vec<i32>> {
nums.sort();
let n = nums.len();
let mut res = vec![];
let mut i = 0;
while i < n - 2 && nums[i] <= 0 {
let mut l = i + 1;
let mut r = n - 1;
w... |
15 | 3Sum | Medium | <p>Given an integer array nums, return all the triplets <code>[nums[i], nums[j], nums[k]]</code> such that <code>i != j</code>, <code>i != k</code>, and <code>j != k</code>, and <code>nums[i] + nums[j] + nums[k] == 0</code>.</p>
<p>Notice that the solution set must not contain duplicate triplets.</p>
<p> </p>
<p... | Array; Two Pointers; Sorting | TypeScript | function threeSum(nums: number[]): number[][] {
nums.sort((a, b) => a - b);
const ans: number[][] = [];
const n = nums.length;
for (let i = 0; i < n - 2 && nums[i] <= 0; i++) {
if (i > 0 && nums[i] === nums[i - 1]) {
continue;
}
let j = i + 1;
let k = n - 1;
... |
16 | 3Sum Closest | Medium | <p>Given an integer array <code>nums</code> of length <code>n</code> and an integer <code>target</code>, find three integers in <code>nums</code> such that the sum is closest to <code>target</code>.</p>
<p>Return <em>the sum of the three integers</em>.</p>
<p>You may assume that each input would have exactly one solu... | Array; Two Pointers; Sorting | C | int cmp(const void* a, const void* b) {
return (*(int*) a - *(int*) b);
}
int threeSumClosest(int* nums, int numsSize, int target) {
qsort(nums, numsSize, sizeof(int), cmp);
int ans = 1 << 30;
for (int i = 0; i < numsSize; ++i) {
int j = i + 1, k = numsSize - 1;
while (j < k) {
... |
16 | 3Sum Closest | Medium | <p>Given an integer array <code>nums</code> of length <code>n</code> and an integer <code>target</code>, find three integers in <code>nums</code> such that the sum is closest to <code>target</code>.</p>
<p>Return <em>the sum of the three integers</em>.</p>
<p>You may assume that each input would have exactly one solu... | Array; Two Pointers; Sorting | C++ | class Solution {
public:
int threeSumClosest(vector<int>& nums, int target) {
sort(nums.begin(), nums.end());
int ans = 1 << 30;
int n = nums.size();
for (int i = 0; i < n; ++i) {
int j = i + 1, k = n - 1;
while (j < k) {
int t = nums[i] + nums... |
16 | 3Sum Closest | Medium | <p>Given an integer array <code>nums</code> of length <code>n</code> and an integer <code>target</code>, find three integers in <code>nums</code> such that the sum is closest to <code>target</code>.</p>
<p>Return <em>the sum of the three integers</em>.</p>
<p>You may assume that each input would have exactly one solu... | Array; Two Pointers; Sorting | C# | public class Solution {
public int ThreeSumClosest(int[] nums, int target) {
Array.Sort(nums);
int ans = 1 << 30;
int n = nums.Length;
for (int i = 0; i < n; ++i) {
int j = i + 1, k = n - 1;
while (j < k) {
int t = nums[i] + nums[j] + nums[k];
... |
16 | 3Sum Closest | Medium | <p>Given an integer array <code>nums</code> of length <code>n</code> and an integer <code>target</code>, find three integers in <code>nums</code> such that the sum is closest to <code>target</code>.</p>
<p>Return <em>the sum of the three integers</em>.</p>
<p>You may assume that each input would have exactly one solu... | Array; Two Pointers; Sorting | Go | func threeSumClosest(nums []int, target int) int {
sort.Ints(nums)
ans := 1 << 30
n := len(nums)
for i, v := range nums {
j, k := i+1, n-1
for j < k {
t := v + nums[j] + nums[k]
if t == target {
return t
}
if abs(t-target) < abs(ans-target) {
ans = t
}
if t > target {
k--
} else... |
16 | 3Sum Closest | Medium | <p>Given an integer array <code>nums</code> of length <code>n</code> and an integer <code>target</code>, find three integers in <code>nums</code> such that the sum is closest to <code>target</code>.</p>
<p>Return <em>the sum of the three integers</em>.</p>
<p>You may assume that each input would have exactly one solu... | Array; Two Pointers; Sorting | Java | class Solution {
public int threeSumClosest(int[] nums, int target) {
Arrays.sort(nums);
int ans = 1 << 30;
int n = nums.length;
for (int i = 0; i < n; ++i) {
int j = i + 1, k = n - 1;
while (j < k) {
int t = nums[i] + nums[j] + nums[k];
... |
16 | 3Sum Closest | Medium | <p>Given an integer array <code>nums</code> of length <code>n</code> and an integer <code>target</code>, find three integers in <code>nums</code> such that the sum is closest to <code>target</code>.</p>
<p>Return <em>the sum of the three integers</em>.</p>
<p>You may assume that each input would have exactly one solu... | Array; Two Pointers; Sorting | JavaScript | /**
* @param {number[]} nums
* @param {number} target
* @return {number}
*/
var threeSumClosest = function (nums, target) {
nums.sort((a, b) => a - b);
let ans = 1 << 30;
const n = nums.length;
for (let i = 0; i < n; ++i) {
let j = i + 1;
let k = n - 1;
while (j < k) {
... |
16 | 3Sum Closest | Medium | <p>Given an integer array <code>nums</code> of length <code>n</code> and an integer <code>target</code>, find three integers in <code>nums</code> such that the sum is closest to <code>target</code>.</p>
<p>Return <em>the sum of the three integers</em>.</p>
<p>You may assume that each input would have exactly one solu... | Array; Two Pointers; Sorting | PHP | class Solution {
/**
* @param int[] $nums
* @param int $target
* @return int
*/
function threeSumClosest($nums, $target) {
$n = count($nums);
$closestSum = $nums[0] + $nums[1] + $nums[2];
$minDiff = abs($closestSum - $target);
sort($nums);
for ($i =... |
16 | 3Sum Closest | Medium | <p>Given an integer array <code>nums</code> of length <code>n</code> and an integer <code>target</code>, find three integers in <code>nums</code> such that the sum is closest to <code>target</code>.</p>
<p>Return <em>the sum of the three integers</em>.</p>
<p>You may assume that each input would have exactly one solu... | Array; Two Pointers; Sorting | Python | class Solution:
def threeSumClosest(self, nums: List[int], target: int) -> int:
nums.sort()
n = len(nums)
ans = inf
for i, v in enumerate(nums):
j, k = i + 1, n - 1
while j < k:
t = v + nums[j] + nums[k]
if t == target:
... |
16 | 3Sum Closest | Medium | <p>Given an integer array <code>nums</code> of length <code>n</code> and an integer <code>target</code>, find three integers in <code>nums</code> such that the sum is closest to <code>target</code>.</p>
<p>Return <em>the sum of the three integers</em>.</p>
<p>You may assume that each input would have exactly one solu... | Array; Two Pointers; Sorting | TypeScript | function threeSumClosest(nums: number[], target: number): number {
nums.sort((a, b) => a - b);
let ans: number = Infinity;
const n = nums.length;
for (let i = 0; i < n; ++i) {
let j = i + 1;
let k = n - 1;
while (j < k) {
const t: number = nums[i] + nums[j] + nums[k];... |
17 | Letter Combinations of a Phone Number | Medium | <p>Given a string containing digits from <code>2-9</code> inclusive, return all possible letter combinations that the number could represent. Return the answer in <strong>any order</strong>.</p>
<p>A mapping of digits to letters (just like on the telephone buttons) is given below. Note that 1 does not map to any lette... | Hash Table; String; Backtracking | C | char* d[] = {"abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
char** letterCombinations(char* digits, int* returnSize) {
if (!*digits) {
*returnSize = 0;
return NULL;
}
int size = 1;
char** ans = (char**) malloc(sizeof(char*));
ans[0] = strdup("");
for (int x = 0; di... |
17 | Letter Combinations of a Phone Number | Medium | <p>Given a string containing digits from <code>2-9</code> inclusive, return all possible letter combinations that the number could represent. Return the answer in <strong>any order</strong>.</p>
<p>A mapping of digits to letters (just like on the telephone buttons) is given below. Note that 1 does not map to any lette... | Hash Table; String; Backtracking | C++ | class Solution {
public:
vector<string> letterCombinations(string digits) {
if (digits.empty()) {
return {};
}
vector<string> d = {"abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
vector<string> ans = {""};
for (auto& i : digits) {
string s =... |
17 | Letter Combinations of a Phone Number | Medium | <p>Given a string containing digits from <code>2-9</code> inclusive, return all possible letter combinations that the number could represent. Return the answer in <strong>any order</strong>.</p>
<p>A mapping of digits to letters (just like on the telephone buttons) is given below. Note that 1 does not map to any lette... | Hash Table; String; Backtracking | C# | public class Solution {
public IList<string> LetterCombinations(string digits) {
var ans = new List<string>();
if (digits.Length == 0) {
return ans;
}
ans.Add("");
string[] d = {"abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
foreach (char i in ... |
17 | Letter Combinations of a Phone Number | Medium | <p>Given a string containing digits from <code>2-9</code> inclusive, return all possible letter combinations that the number could represent. Return the answer in <strong>any order</strong>.</p>
<p>A mapping of digits to letters (just like on the telephone buttons) is given below. Note that 1 does not map to any lette... | Hash Table; String; Backtracking | Go | func letterCombinations(digits string) []string {
ans := []string{}
if len(digits) == 0 {
return ans
}
ans = append(ans, "")
d := []string{"abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"}
for _, i := range digits {
s := d[i-'2']
t := []string{}
for _, a := range ans {
for _, b := range s {
... |
17 | Letter Combinations of a Phone Number | Medium | <p>Given a string containing digits from <code>2-9</code> inclusive, return all possible letter combinations that the number could represent. Return the answer in <strong>any order</strong>.</p>
<p>A mapping of digits to letters (just like on the telephone buttons) is given below. Note that 1 does not map to any lette... | Hash Table; String; Backtracking | Java | class Solution {
public List<String> letterCombinations(String digits) {
List<String> ans = new ArrayList<>();
if (digits.length() == 0) {
return ans;
}
ans.add("");
String[] d = new String[] {"abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
for ... |
17 | Letter Combinations of a Phone Number | Medium | <p>Given a string containing digits from <code>2-9</code> inclusive, return all possible letter combinations that the number could represent. Return the answer in <strong>any order</strong>.</p>
<p>A mapping of digits to letters (just like on the telephone buttons) is given below. Note that 1 does not map to any lette... | Hash Table; String; Backtracking | JavaScript | /**
* @param {string} digits
* @return {string[]}
*/
var letterCombinations = function (digits) {
if (digits.length === 0) {
return [];
}
const ans = [''];
const d = ['abc', 'def', 'ghi', 'jkl', 'mno', 'pqrs', 'tuv', 'wxyz'];
for (const i of digits) {
const s = d[+i - 2];
... |
17 | Letter Combinations of a Phone Number | Medium | <p>Given a string containing digits from <code>2-9</code> inclusive, return all possible letter combinations that the number could represent. Return the answer in <strong>any order</strong>.</p>
<p>A mapping of digits to letters (just like on the telephone buttons) is given below. Note that 1 does not map to any lette... | Hash Table; String; Backtracking | PHP | class Solution {
/**
* @param string $digits
* @return string[]
*/
function letterCombinations($digits) {
$digitMap = [
'2' => ['a', 'b', 'c'],
'3' => ['d', 'e', 'f'],
'4' => ['g', 'h', 'i'],
'5' => ['j', 'k', 'l'],
'6' => ['m',... |
17 | Letter Combinations of a Phone Number | Medium | <p>Given a string containing digits from <code>2-9</code> inclusive, return all possible letter combinations that the number could represent. Return the answer in <strong>any order</strong>.</p>
<p>A mapping of digits to letters (just like on the telephone buttons) is given below. Note that 1 does not map to any lette... | Hash Table; String; Backtracking | Python | class Solution:
def letterCombinations(self, digits: str) -> List[str]:
if not digits:
return []
d = ["abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"]
ans = [""]
for i in digits:
s = d[int(i) - 2]
ans = [a + b for a in ans for b in s]
... |
17 | Letter Combinations of a Phone Number | Medium | <p>Given a string containing digits from <code>2-9</code> inclusive, return all possible letter combinations that the number could represent. Return the answer in <strong>any order</strong>.</p>
<p>A mapping of digits to letters (just like on the telephone buttons) is given below. Note that 1 does not map to any lette... | Hash Table; String; Backtracking | Rust | impl Solution {
pub fn letter_combinations(digits: String) -> Vec<String> {
let mut ans: Vec<String> = Vec::new();
if digits.is_empty() {
return ans;
}
ans.push("".to_string());
let d = ["abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"];
for i in d... |
17 | Letter Combinations of a Phone Number | Medium | <p>Given a string containing digits from <code>2-9</code> inclusive, return all possible letter combinations that the number could represent. Return the answer in <strong>any order</strong>.</p>
<p>A mapping of digits to letters (just like on the telephone buttons) is given below. Note that 1 does not map to any lette... | Hash Table; String; Backtracking | TypeScript | function letterCombinations(digits: string): string[] {
if (digits.length === 0) {
return [];
}
const ans: string[] = [''];
const d = ['abc', 'def', 'ghi', 'jkl', 'mno', 'pqrs', 'tuv', 'wxyz'];
for (const i of digits) {
const s = d[+i - 2];
const t: string[] = [];
for... |
18 | 4Sum | Medium | <p>Given an array <code>nums</code> of <code>n</code> integers, return <em>an array of all the <strong>unique</strong> quadruplets</em> <code>[nums[a], nums[b], nums[c], nums[d]]</code> such that:</p>
<ul>
<li><code>0 <= a, b, c, d < n</code></li>
<li><code>a</code>, <code>b</code>, <code>c</code>, and <c... | Array; Two Pointers; Sorting | C++ | class Solution {
public:
vector<vector<int>> fourSum(vector<int>& nums, int target) {
int n = nums.size();
vector<vector<int>> ans;
if (n < 4) {
return ans;
}
sort(nums.begin(), nums.end());
for (int i = 0; i < n - 3; ++i) {
if (i && nums[i] ==... |
18 | 4Sum | Medium | <p>Given an array <code>nums</code> of <code>n</code> integers, return <em>an array of all the <strong>unique</strong> quadruplets</em> <code>[nums[a], nums[b], nums[c], nums[d]]</code> such that:</p>
<ul>
<li><code>0 <= a, b, c, d < n</code></li>
<li><code>a</code>, <code>b</code>, <code>c</code>, and <c... | Array; Two Pointers; Sorting | C# | public class Solution {
public IList<IList<int>> FourSum(int[] nums, int target) {
int n = nums.Length;
var ans = new List<IList<int>>();
if (n < 4) {
return ans;
}
Array.Sort(nums);
for (int i = 0; i < n - 3; ++i) {
if (i > 0 && nums[i] == num... |
18 | 4Sum | Medium | <p>Given an array <code>nums</code> of <code>n</code> integers, return <em>an array of all the <strong>unique</strong> quadruplets</em> <code>[nums[a], nums[b], nums[c], nums[d]]</code> such that:</p>
<ul>
<li><code>0 <= a, b, c, d < n</code></li>
<li><code>a</code>, <code>b</code>, <code>c</code>, and <c... | Array; Two Pointers; Sorting | Go | func fourSum(nums []int, target int) (ans [][]int) {
n := len(nums)
if n < 4 {
return
}
sort.Ints(nums)
for i := 0; i < n-3; i++ {
if i > 0 && nums[i] == nums[i-1] {
continue
}
for j := i + 1; j < n-2; j++ {
if j > i+1 && nums[j] == nums[j-1] {
continue
}
k, l := j+1, n-1
for k < l {
... |
18 | 4Sum | Medium | <p>Given an array <code>nums</code> of <code>n</code> integers, return <em>an array of all the <strong>unique</strong> quadruplets</em> <code>[nums[a], nums[b], nums[c], nums[d]]</code> such that:</p>
<ul>
<li><code>0 <= a, b, c, d < n</code></li>
<li><code>a</code>, <code>b</code>, <code>c</code>, and <c... | Array; Two Pointers; Sorting | Java | class Solution {
public List<List<Integer>> fourSum(int[] nums, int target) {
int n = nums.length;
List<List<Integer>> ans = new ArrayList<>();
if (n < 4) {
return ans;
}
Arrays.sort(nums);
for (int i = 0; i < n - 3; ++i) {
if (i > 0 && nums[i]... |
18 | 4Sum | Medium | <p>Given an array <code>nums</code> of <code>n</code> integers, return <em>an array of all the <strong>unique</strong> quadruplets</em> <code>[nums[a], nums[b], nums[c], nums[d]]</code> such that:</p>
<ul>
<li><code>0 <= a, b, c, d < n</code></li>
<li><code>a</code>, <code>b</code>, <code>c</code>, and <c... | Array; Two Pointers; Sorting | JavaScript | /**
* @param {number[]} nums
* @param {number} target
* @return {number[][]}
*/
var fourSum = function (nums, target) {
const n = nums.length;
const ans = [];
if (n < 4) {
return ans;
}
nums.sort((a, b) => a - b);
for (let i = 0; i < n - 3; ++i) {
if (i > 0 && nums[i] === num... |
18 | 4Sum | Medium | <p>Given an array <code>nums</code> of <code>n</code> integers, return <em>an array of all the <strong>unique</strong> quadruplets</em> <code>[nums[a], nums[b], nums[c], nums[d]]</code> such that:</p>
<ul>
<li><code>0 <= a, b, c, d < n</code></li>
<li><code>a</code>, <code>b</code>, <code>c</code>, and <c... | Array; Two Pointers; Sorting | PHP | class Solution {
/**
* @param int[] $nums
* @param int $target
* @return int[][]
*/
function fourSum($nums, $target) {
$result = [];
$n = count($nums);
sort($nums);
for ($i = 0; $i < $n - 3; $i++) {
if ($i > 0 && $nums[$i] === $nums[$i - 1]) {
... |
18 | 4Sum | Medium | <p>Given an array <code>nums</code> of <code>n</code> integers, return <em>an array of all the <strong>unique</strong> quadruplets</em> <code>[nums[a], nums[b], nums[c], nums[d]]</code> such that:</p>
<ul>
<li><code>0 <= a, b, c, d < n</code></li>
<li><code>a</code>, <code>b</code>, <code>c</code>, and <c... | Array; Two Pointers; Sorting | Python | class Solution:
def fourSum(self, nums: List[int], target: int) -> List[List[int]]:
n = len(nums)
ans = []
if n < 4:
return ans
nums.sort()
for i in range(n - 3):
if i and nums[i] == nums[i - 1]:
continue
for j in range(i + ... |
18 | 4Sum | Medium | <p>Given an array <code>nums</code> of <code>n</code> integers, return <em>an array of all the <strong>unique</strong> quadruplets</em> <code>[nums[a], nums[b], nums[c], nums[d]]</code> such that:</p>
<ul>
<li><code>0 <= a, b, c, d < n</code></li>
<li><code>a</code>, <code>b</code>, <code>c</code>, and <c... | Array; Two Pointers; Sorting | TypeScript | function fourSum(nums: number[], target: number): number[][] {
const n = nums.length;
const ans: number[][] = [];
if (n < 4) {
return ans;
}
nums.sort((a, b) => a - b);
for (let i = 0; i < n - 3; ++i) {
if (i > 0 && nums[i] === nums[i - 1]) {
continue;
}
... |
19 | Remove Nth Node From End of List | Medium | <p>Given the <code>head</code> of a linked list, remove the <code>n<sup>th</sup></code> node from the end of the list and return its head.</p>
<p> </p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0000-0099/0019.Remove%20Nth%20Node... | Linked List; Two Pointers | C++ | /**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
ListNode* removeNthFr... |
19 | Remove Nth Node From End of List | Medium | <p>Given the <code>head</code> of a linked list, remove the <code>n<sup>th</sup></code> node from the end of the list and return its head.</p>
<p> </p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0000-0099/0019.Remove%20Nth%20Node... | Linked List; Two Pointers | C# | /**
* Definition for singly-linked list.
* public class ListNode {
* public int val;
* public ListNode next;
* public ListNode(int val=0, ListNode next=null) {
* this.val = val;
* this.next = next;
* }
* }
*/
public class Solution {
public ListNode RemoveNthFromEnd(ListNode... |
19 | Remove Nth Node From End of List | Medium | <p>Given the <code>head</code> of a linked list, remove the <code>n<sup>th</sup></code> node from the end of the list and return its head.</p>
<p> </p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0000-0099/0019.Remove%20Nth%20Node... | Linked List; Two Pointers | Go | /**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func removeNthFromEnd(head *ListNode, n int) *ListNode {
dummy := &ListNode{0, head}
fast, slow := dummy, dummy
for ; n > 0; n-- {
fast = fast.Next
}
for fast.Next != nil {
slow, fast = slow.Next, ... |
19 | Remove Nth Node From End of List | Medium | <p>Given the <code>head</code> of a linked list, remove the <code>n<sup>th</sup></code> node from the end of the list and return its head.</p>
<p> </p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0000-0099/0019.Remove%20Nth%20Node... | Linked List; Two Pointers | Java | /**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode removeNthFromEnd(ListNo... |
19 | Remove Nth Node From End of List | Medium | <p>Given the <code>head</code> of a linked list, remove the <code>n<sup>th</sup></code> node from the end of the list and return its head.</p>
<p> </p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0000-0099/0019.Remove%20Nth%20Node... | Linked List; Two Pointers | JavaScript | /**
* Definition for singly-linked list.
* function ListNode(val, next) {
* this.val = (val===undefined ? 0 : val)
* this.next = (next===undefined ? null : next)
* }
*/
/**
* @param {ListNode} head
* @param {number} n
* @return {ListNode}
*/
var removeNthFromEnd = function (head, n) {
const dummy ... |
19 | Remove Nth Node From End of List | Medium | <p>Given the <code>head</code> of a linked list, remove the <code>n<sup>th</sup></code> node from the end of the list and return its head.</p>
<p> </p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0000-0099/0019.Remove%20Nth%20Node... | Linked List; Two Pointers | PHP | /**
* Definition for a singly-linked list.
* class ListNode {
* public $val = 0;
* public $next = null;
* function __construct($val = 0, $next = null) {
* $this->val = $val;
* $this->next = $next;
* }
* }
*/
class Solution {
/**
* @param ListNode $head
* @param In... |
19 | Remove Nth Node From End of List | Medium | <p>Given the <code>head</code> of a linked list, remove the <code>n<sup>th</sup></code> node from the end of the list and return its head.</p>
<p> </p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0000-0099/0019.Remove%20Nth%20Node... | Linked List; Two Pointers | Python | # Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def removeNthFromEnd(self, head: Optional[ListNode], n: int) -> Optional[ListNode]:
dummy = ListNode(next=head)
fast = slow = dummy
... |
19 | Remove Nth Node From End of List | Medium | <p>Given the <code>head</code> of a linked list, remove the <code>n<sup>th</sup></code> node from the end of the list and return its head.</p>
<p> </p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0000-0099/0019.Remove%20Nth%20Node... | Linked List; Two Pointers | Ruby | # Definition for singly-linked list.
# class ListNode
# attr_accessor :val, :next
# def initialize(val = 0, _next = nil)
# @val = val
# @next = _next
# end
# end
# @param {ListNode} head
# @param {Integer} n
# @return {ListNode}
def remove_nth_from_end(head, n)
dummy = ListNode.new(0, he... |
19 | Remove Nth Node From End of List | Medium | <p>Given the <code>head</code> of a linked list, remove the <code>n<sup>th</sup></code> node from the end of the list and return its head.</p>
<p> </p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0000-0099/0019.Remove%20Nth%20Node... | Linked List; Two Pointers | Rust | // Definition for singly-linked list.
// #[derive(PartialEq, Eq, Clone, Debug)]
// pub struct ListNode {
// pub val: i32,
// pub next: Option<Box<ListNode>>
// }
//
// impl ListNode {
// #[inline]
// fn new(val: i32) -> Self {
// ListNode {
// next: None,
// val
// }
// }
// }
impl Solutio... |
19 | Remove Nth Node From End of List | Medium | <p>Given the <code>head</code> of a linked list, remove the <code>n<sup>th</sup></code> node from the end of the list and return its head.</p>
<p> </p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0000-0099/0019.Remove%20Nth%20Node... | Linked List; Two Pointers | Swift | /**
* Definition for singly-linked list.
* public class ListNode {
* public var val: Int
* public var next: ListNode?
* public init() { self.val = 0; self.next = nil; }
* public init(_ val: Int) { self.val = val; self.next = nil; }
* public init(_ val: Int, _ next: ListNode?) { self.va... |
19 | Remove Nth Node From End of List | Medium | <p>Given the <code>head</code> of a linked list, remove the <code>n<sup>th</sup></code> node from the end of the list and return its head.</p>
<p> </p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0000-0099/0019.Remove%20Nth%20Node... | Linked List; Two Pointers | TypeScript | /**
* Definition for singly-linked list.
* class ListNode {
* val: number
* next: ListNode | null
* constructor(val?: number, next?: ListNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.next = (next===undefined ? null : next)
* }
* }
*/
function removeNthFromEnd(... |
20 | Valid Parentheses | Easy | <p>Given a string <code>s</code> containing just the characters <code>'('</code>, <code>')'</code>, <code>'{'</code>, <code>'}'</code>, <code>'['</code> and <code>']'</code>, determine if the input string is valid.</p>
<p>An input string is valid if:</p>
<ol>
<li>Open ... | Stack; String | C++ | class Solution {
public:
bool isValid(string s) {
string stk;
for (char c : s) {
if (c == '(' || c == '{' || c == '[')
stk.push_back(c);
else if (stk.empty() || !match(stk.back(), c))
return false;
else
stk.pop_back(... |
20 | Valid Parentheses | Easy | <p>Given a string <code>s</code> containing just the characters <code>'('</code>, <code>')'</code>, <code>'{'</code>, <code>'}'</code>, <code>'['</code> and <code>']'</code>, determine if the input string is valid.</p>
<p>An input string is valid if:</p>
<ol>
<li>Open ... | Stack; String | C# | public class Solution {
public bool IsValid(string s) {
Stack<char> stk = new Stack<char>();
foreach (var c in s.ToCharArray()) {
if (c == '(') {
stk.Push(')');
} else if (c == '[') {
stk.Push(']');
} else if (c == '{') {
... |
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