# Find All Anagrams In A String Problem

## Description

LeetCode Problem 438.

Given two strings s and p, return an array of all the start indices of p’s anagrams in s. You may return the answer in any order.

An Anagram is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.

Example 1:

``````1
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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:

``````1
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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".
``````

Constraints:

• 1 <= s.length, p.length <= 3 * 10^4
• s and p consist of lowercase English letters.

## Sample C++ Code

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class Solution {
public:
vector<int> findAnagrams(string s, string p) {
int len_s = s.size(), len_p = p.size();
vector<int> ans;
if (len_s < len_p)
return ans;

vector<int> cnt_s(26, 0);
vector<int> cnt_p(26, 0);

for (int i = 0; i < len_p; i ++) {
cnt_s[s[i]-'a'] ++;
cnt_p[p[i]-'a'] ++;
}

bool eq = true;
for (int i = 0; i < 26; i ++) {
if (cnt_s[i] != cnt_p[i]) {
eq = false;
break;
}
}
if (eq) ans.push_back(0);

for (int i = 0; i < len_s - len_p; i ++) {
cnt_s[s[i]-'a'] --;
cnt_s[s[i+len_p]-'a'] ++;
bool eq = true;
for (int j = 0; j < 26; j ++) {
if (cnt_s[j] != cnt_p[j]) {
eq = false;
break;
}
}
if (eq) ans.push_back(i+1);
}

return ans;
}
};
``````