567. Permutation in String

Given two strings s1 and s2, write a function to return true if s2 contains the permutation of s1. In other words, one of the first string’s permutations is the substring of the second string.

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Example 1:

Input: s1 = "ab" s2 = "eidbaooo"
Output: True
Explanation: s2 contains one permutation of s1 ("ba").

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Example 2:

Input:s1= "ab" s2 = "eidboaoo"
Output: False

Constraints:

The input strings only contain lower case letters.
The length of both given strings is in range [1, 10,000].

Solution

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class Solution {
    public boolean checkInclusion(String s1, String s2) {
        int len1 = s1.length();
        int len2 = s2.length();
        
        int[] count1 = new int[256];
        int[] count2 = new int[256];
        for (int i = 0; i < len1; i++) {
            count1[s1.charAt(i)]++;
        }
        for (int i = 0; i < len2; i++) {
            count2[s2.charAt(i)]++;
            if (i >= len1) {
                count2[s2.charAt(i - len1)]--;
            }
            if (Arrays.equals(count1, count2)) {
                return true;
            }
        }
        return false;
    }

        public boolean checkInclusion_RollingHash(String s1, String s2) {
        int len1 = s1.length();
        int len2 = s2.length();
        if (len1 > len2) return false;
        int hash1 = 0;
        int hash2 = 0;
        for (int i = 0; i < len1; i++) {
            char c1 = s1.charAt(i);
            char c2 = s2.charAt(i);
            hash1 += pow(c1);
            hash2 += pow(c2);
        }
        if (hash1 == hash2) {
            return true;
        }
        for (int i = len1; i < len2; i++) {
            hash2 = hash2 - pow(s2.charAt(i - len1)) + pow(s2.charAt(i)); 
            if (hash1 == hash2) {
                return true;
            }
        }
        return false;
    }
    
    int pow(int v) {
        return (int) Math.pow(v, 3);
    }
}

Solution#

Solution