# Permutation in String LeetCode Solution

Here, We see Permutation in String LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

# List of all LeetCode Solution

## Problem Statement

Given two strings s1 and s2, return true if s2 contains a permutation of s1, or false otherwise.

In other words, return true if one of s1’s permutations is the substring of s2.

Example 1:
Input: s1 = “ab”, s2 = “eidbaooo”
Output: true
Explanation: s2 contains one permutation of s1 (“ba”).

Example 2:
Input: s1 = “ab”, s2 = “eidboaoo”
Output: false

## Permutation in String Leetcode Solution C++

class Solution {
public:
bool checkInclusion(string s1, string s2) {
vector<int> cur(26), goal(26);
for(char c : s1) goal[c - 'a']++;
for(int i = 0; i < s2.size(); i++) {
cur[s2[i] - 'a']++;
if(i >= s1.size()) cur[s2[i - s1.size()] - 'a']--;
if(goal == cur) return true;
}
return false;
}
};Code language: PHP (php)

## Permutation in String Leetcode Solution Java

class Solution {
public boolean checkInclusion(String s1, String s2) {
if(s1.length() > s2.length()) return false;

int[] arr1 = new int[26];
int[] arr2 = new int[26];

for(int i = 0; i < s1.length(); i++){
arr1[s1.charAt(i) - 'a']++;
arr2[s2.charAt(i) - 'a']++;
}

if(Arrays.equals(arr1, arr2)) return true;

int front = 0;
int back = s1.length();
while(back < s2.length()){
arr2[s2.charAt(front) - 'a']--;
arr2[s2.charAt(back) - 'a']++;

if(Arrays.equals(arr1, arr2)) return true;
front++;
back++;
}
return false;
}
}Code language: JavaScript (javascript)

## Permutation in String Solution JavaScript

var checkInclusion = function (s1, s2) {
const len1 = s1.length, len2 = s2.length;
if (len1 > len2) return false;
const count = Array(26).fill(0);
for (let i = 0; i < len1; i++) {
count[s1.charCodeAt(i)-97]++;
count[s2.charCodeAt(i)-97]--;
}
if (!count.some(e => e !== 0)) return true;
for (let i = len1; i < len2; i++) {
count[s2.charCodeAt(i)-97]--;
count[s2.charCodeAt(i-len1)-97]++;
if (!count.some(e => e !== 0)) return true;
}
return false;
};Code language: JavaScript (javascript)

## Permutation in String Solution Python

class Solution(object):
def checkInclusion(self, s1, s2):
window = len(s1)
s1_c = Counter(s1)
for i in range(len(s2)-window+1):
s2_c = Counter(s2[i:i+window])
if s2_c == s1_c:
return True
return False
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