567. Permutation in String

Problem

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

  Constraints:

• 1 <= s1.length, s2.length <= 10^4

• s1 and s2 consist of lowercase English letters.

Solution (Java)

class Solution {
    public boolean checkInclusion(String s1, String s2) {
        int n = s1.length();
        int m = s2.length();
        if (n > m) {
            return false;
        }
        int[] cntS1 = new int[26];
        int[] cntS2 = new int[26];
        for (int i = 0; i < n; i++) {
            cntS1[s1.charAt(i) - 'a']++;
        }
        for (int i = 0; i < n; i++) {
            cntS2[s2.charAt(i) - 'a']++;
        }
        if (check(cntS1, cntS2)) {
            return true;
        }
        for (int i = n; i < m; i++) {
            cntS2[s2.charAt(i - n) - 'a']--;
            cntS2[s2.charAt(i) - 'a']++;
            if (check(cntS1, cntS2)) {
                return true;
            }
        }
        return false;
    }

    private boolean check(int[] cnt1, int[] cnt2) {
        for (int i = 0; i < 26; i++) {
            if (cnt1[i] != cnt2[i]) {
                return false;
            }
        }
        return true;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).