//438

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".

#include <iostream>
#include <vector>

using namespace std;

class Solution {
public:
    vector<int> findAnagrams(string s, string p) {
        //freq key-value ,like 'b' --2;
        vector<int> pv(26,0),sv(26,0),res;
        if(s.size()<p.size()){
            return res;
        }

        for(int i=0;i<p.size();i++){
            pv[p[i]-'a']++;
            sv[s[i]-'a']++;
        }

        if(pv==sv){
            res.push_back(0);
        }
        //slide window length==p.size()
        for(int i=p.size();i<s.size();i++){
            sv[s[i]-'a']++;
            sv[s[i-p.size()]-'a']--;
            if(pv==sv){
                res.push_back(i-p.size()+1);
            }
        }

        return res;
    }
};

int main(){
    string s="abab",p="ab";
    vector<int> ret=Solution().findAnagrams(s,p);
    for(int i=0;i<ret.size();i++){
        cout<<ret[i]<<endl;
    }
    return 0;
}