Last updated on October 5th, 2024 at 09:06 pm

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

Permutation in String 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

1. 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;        
    }
};

2. 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;        
    }
}

3. 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;
};

4. 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