# 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".


Solution:

Use a vector to represent a string, sliding window

class Solution {
public:
vector<int> findAnagrams(string s, string p) {
if(s.size()<p.size()) return vector<int>();
vector<int> pcnt(26,0),scnt(26,0), ans;
for(int i = 0; i<p.size()-1; i++){
pcnt[p[i]-'a']++;
scnt[s[i]-'a']++;
}
pcnt[p[p.size()-1]-'a']++;
for(int i = p.size()-1; i<s.size();i++){
scnt[s[i]-'a']++;
if(pcnt==scnt){
ans.push_back(i-p.size()+1);
}
scnt[s[i-p.size()+1]-'a']--;
}
return ans;
}
};