# 438. Find All Anagrams in a String

## Description

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:

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".


Constraints:

• 1 <= s.length, p.length <= 3 * 104
• s and p consist of lowercase English letters.

## Solutions

• class Solution {
public List<Integer> findAnagrams(String s, String p) {
int m = s.length(), n = p.length();
List<Integer> ans = new ArrayList<>();
if (m < n) {
return ans;
}
int[] cnt1 = new int[26];
for (int i = 0; i < n; ++i) {
++cnt1[p.charAt(i) - 'a'];
}
int[] cnt2 = new int[26];
for (int i = 0; i < n - 1; ++i) {
++cnt2[s.charAt(i) - 'a'];
}
for (int i = n - 1; i < m; ++i) {
++cnt2[s.charAt(i) - 'a'];
if (Arrays.equals(cnt1, cnt2)) {
}
--cnt2[s.charAt(i - n + 1) - 'a'];
}
return ans;
}
}

• class Solution {
public:
vector<int> findAnagrams(string s, string p) {
int m = s.size(), n = p.size();
vector<int> ans;
if (m < n) {
return ans;
}
vector<int> cnt1(26);
for (char& c : p) {
++cnt1[c - 'a'];
}
vector<int> cnt2(26);
for (int i = 0; i < n - 1; ++i) {
++cnt2[s[i] - 'a'];
}
for (int i = n - 1; i < m; ++i) {
++cnt2[s[i] - 'a'];
if (cnt1 == cnt2) {
ans.push_back(i - n + 1);
}
--cnt2[s[i - n + 1] - 'a'];
}
return ans;
}
};

• class Solution:
def findAnagrams(self, s: str, p: str) -> List[int]:
m, n = len(s), len(p)
ans = []
if m < n:
return ans
cnt1 = Counter(p)
cnt2 = Counter(s[: n - 1])
for i in range(n - 1, m):
cnt2[s[i]] += 1
if cnt1 == cnt2:
ans.append(i - n + 1)
cnt2[s[i - n + 1]] -= 1
return ans


• func findAnagrams(s string, p string) (ans []int) {
m, n := len(s), len(p)
if m < n {
return
}
cnt1 := [26]int{}
cnt2 := [26]int{}
for _, c := range p {
cnt1[c-'a']++
}
for _, c := range s[:n-1] {
cnt2[c-'a']++
}
for i := n - 1; i < m; i++ {
cnt2[s[i]-'a']++
if cnt1 == cnt2 {
ans = append(ans, i-n+1)
}
cnt2[s[i-n+1]-'a']--
}
return
}

• function findAnagrams(s: string, p: string): number[] {
const m = s.length;
const n = p.length;
const ans: number[] = [];
if (m < n) {
return ans;
}
const cnt1: number[] = new Array(26).fill(0);
const cnt2: number[] = new Array(26).fill(0);
const idx = (c: string) => c.charCodeAt(0) - 'a'.charCodeAt(0);
for (const c of p) {
++cnt1[idx(c)];
}
for (const c of s.slice(0, n - 1)) {
++cnt2[idx(c)];
}
for (let i = n - 1; i < m; ++i) {
++cnt2[idx(s[i])];
if (cnt1.toString() === cnt2.toString()) {
ans.push(i - n + 1);
}
--cnt2[idx(s[i - n + 1])];
}
return ans;
}


• public class Solution {
public IList<int> FindAnagrams(string s, string p) {
int m = s.Length, n = p.Length;
IList<int> ans = new List<int>();
if (m < n) {
return ans;
}
int[] cnt1 = new int[26];
int[] cnt2 = new int[26];
for (int i = 0; i < n; ++i) {
++cnt1[p[i] - 'a'];
}
for (int i = 0, j = 0; i < m; ++i) {
int k = s[i] - 'a';
++cnt2[k];
while (cnt2[k] > cnt1[k]) {
--cnt2[s[j++] - 'a'];
}
if (i - j + 1 == n) {
}
}
return ans;
}
}

• impl Solution {
pub fn find_anagrams(s: String, p: String) -> Vec<i32> {
let (s, p) = (s.as_bytes(), p.as_bytes());
let (m, n) = (s.len(), p.len());
let mut ans = vec![];
if m < n {
return ans;
}

let mut cnt = [0; 26];
for i in 0..n {
cnt[(p[i] - b'a') as usize] += 1;
cnt[(s[i] - b'a') as usize] -= 1;
}
for i in n..m {
if cnt.iter().all(|&v| v == 0) {
ans.push((i - n) as i32);
}
cnt[(s[i] - b'a') as usize] -= 1;
cnt[(s[i - n] - b'a') as usize] += 1;
}
if cnt.iter().all(|&v| v == 0) {
ans.push((m - n) as i32);
}
ans
}
}