Permutation in String - Leetcode 567

Permutation in String - Leetcode 567

Problem - Leetcode

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 <= s1.length, s2.length <= 10<sup>4</sup>

  • s1 and s2 consists of lowercase English letters.

Solution in Golang

func checkInclusion(s1 string, s2 string) bool {
    if len(s1) > len(s2) {
        return false
    s1Count, s2Count := [26]int{}, [26]int{}
    for i, _ := range s1 {
    matches := 0
    for i := 0; i < 26; i++ {
        if s1Count[i] == s2Count[i] {
            matches += 1
        } else {
            matches += 0
    l := 0
    for r := len(s1); r < len(s2); r++ {
        if matches == 26 {
            return true
        index := s2[r] - 'a'
        if s1Count[index] == s2Count[index] {
        } else if s1Count[index]+1 == s2Count[index] {
        index = s2[l] - 'a'
        if s1Count[index] == s2Count[index] {
        } else if s1Count[index]-1 == s2Count[index] {
    return matches == 26

This code defines a Go function called checkInclusion that checks if one string, s1, is a permutation of any substring in another string, s2. In other words, it determines if all the characters in s1 can be found in any order within s2. Here's a detailed explanation of how the code works:

  1. The function starts by comparing the lengths of s1 and s2. If s1 is longer than s2, it immediately returns false, because it's not possible for s1 to be a permutation of a longer string.

  2. Two arrays, s1Count and s2Count, of size 26 (representing the lowercase English alphabet) are initialized to store the frequency of characters in s1 and s2, respectively.

  3. The code iterates through the characters in s1 and increments the corresponding index in s1Count and s2Count arrays based on the ASCII value difference between the character and 'a'.

  4. The variable matches is initialized to track the number of characters that have matching frequencies between s1 and the current substring of s2.

  5. A loop from 0 to 25 iterates through each index representing a character in the arrays. If the character frequency in s1 and s2 match, matches is incremented.

  6. The code then enters a sliding window approach to compare the substrings of s2 with the length of s1. The l variable represents the left boundary of the sliding window, and the r variable represents the right boundary.

  7. At each step, the code checks if matches is equal to 26 (representing all characters of the alphabet). If yes, it means s1 is a permutation of the current substring of s2, and it immediately returns true.

  8. The character at index r of s2 is considered. Its frequency in s2Count is incremented. If this increment makes the frequency match that in s1Count, matches is incremented. If the frequency in s2Count exceeds s1Count by one, matches is decremented.

  9. Similarly, the character at index l of the previous substring is considered. Its frequency in s2Count is decremented. If this decrement makes the frequency match that in s1Count, matches is incremented. If the frequency in s2Count becomes one less than s1Count, matches is decremented.

  10. The l index is incremented to slide the window one step to the right.

  11. The loop continues until the end of s2 is reached. Finally, the function returns true if matches equals 26, indicating that all characters of s1 are permuted within some substring of s2.

This code essentially uses the sliding window technique along with character frequency tracking to efficiently check for the permutation condition between two strings.

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