567. Permutation In String¶

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567. Permutation in String

Medium

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 <= 104
• s1 and s2 consist of lowercase English letters.

Solution¶

class Solution {
    public boolean checkInclusion(String s1, String s2) {
        int n1 = s1.length(), n2 = s2.length();
        if (n1 > n2)  return false;
        int[] arr1 = new int[26];
        int[] arr2 = new int[26];
        for (char ch : s1.toCharArray())  arr1[ch - 'a']++;
        for (int i = 0; i < n2; i++) {
            arr2[s2.charAt(i) - 'a']++;
            if (i >= n1)  arr2[s2.charAt(i - n1) - 'a']--;
            if (Arrays.equals(arr1, arr2)) return true;
        }
        return false;
    }
}

Complexity Analysis¶

• Time Complexity: O(?)
• Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here