MAY-17 2020 Challenge

17. Find All Anagrams in the String

Given a string s and a non-empty string p, find all the start indices of p's anagrams in s.

Strings consists of lowercase English letters only and the length of both strings s and p will not be larger than 20,100.

The order of output does not matter.

Example 1:

Input:
s: "cbaebabacd" p: "abc"

Output:
[0, 6]

Explanation:
The substring with start index = 0 is "cba", which is an anagram of "abc".
The substring with start index = 6 is "bac", which is an anagram of "abc".

Example 2:

Input:
s: "abab" p: "ab"

Output:
[0, 1, 2]

Explanation:
The substring with start index = 0 is "ab", which is an anagram of "ab".
The substring with start index = 1 is "ba", which is an anagram of "ab".
The substring with start index = 2 is "ab", which is an anagram of "ab".

Solution in Java : 

class Solution {
    public List<Integer> findAnagrams(String s, String p) {
        List<Integer> rst = new ArrayList<>();
        if (s == null || s.length() == 0 || s.length() < p.length()) {
            return rst;
        }

        int[] map_p = new int[26];
        int[] map_s = new int[26];
        for (int i = 0; i < p.length(); i++) {
            map_p[p.charAt(i) - 'a']++;
        }
        for (int i = 0; i < p.length(); i++) {
            map_s[s.charAt(i) - 'a']++;
        }

        for (int i = 0; i < s.length() - p.length(); i++) {
            if (isMatch(map_p, map_s)) {
                rst.add(i);
            }
                          map_s[s.charAt(i+p.length()) - 'a']++;
            map_s[s.charAt(i) - 'a']--;
        }
        if (isMatch(map_p, map_s)) {
            rst.add(s.length() - p.length());
        }
        return rst;
    }

    public boolean isMatch(int[] arr1, int[] arr2) {
        for (int i = 0; i < arr1.length; i++) {
            if (arr1[i] != arr2[i]) {
                return false;
            }
        }
        return true;
    }
}