438. Find All Anagrams in a 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".

class Solution {
public:
    vector<int> findAnagrams(string s, string p) {
        vector<int> ans;
        if (s.empty() || p.empty()) {
            return ans;
        }
       
        map<char, int> mp;
        for (auto ch : p) {
            mp[ch]++;
        }
       
        int start = 0, end = 0;
        map<char, int>mp2;
        while (end < s.size()) {
            int len = end - start + 1;
            if (len > p.size()) {
                if (--mp2[s[start++]] == 0) {
                    mp2.erase(s[start - 1]);
                }
            }
            mp2[s[end]]++;
            if (mp2 == mp) {
                ans.push_back(start);
            }
            end++;
        }
        return ans;
    }
};