README_EN.md

438. Find All Anagrams in a String

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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

Python3

class Solution:
    def findAnagrams(self, s: str, p: str) -> List[int]:
        counter = Counter(p)
        ans = []
        left = right = 0
        t = Counter()
        while right < len(s):
            t[s[right]] += 1
            while t[s[right]] > counter[s[right]]:
                t[s[left]] -= 1
                left += 1
            if right - left + 1 == len(p):
                ans.append(left)
            right += 1
        return ans

Java

class Solution {
    public List<Integer> findAnagrams(String s, String p) {
        int[] counter = new int[26];
        for (char c : p.toCharArray()) {
            ++counter[c - 'a'];
        }
        List<Integer> ans = new ArrayList<>();
        int left = 0, right = 0;
        int[] t = new int[26];
        while (right < s.length()) {
            int i = s.charAt(right) - 'a';
            ++t[i];
            while (t[i] > counter[i]) {
                --t[s.charAt(left) - 'a'];
                ++left;
            }
            if (right - left + 1 == p.length()) {
                ans.add(left);
            }
            ++right;
        }
        return ans;
    }
}

TypeScript

function findAnagrams(s: string, p: string): number[] {
    let n = s.length,
        m = p.length;
    let cnt = new Array(26).fill(0);
    let ans = [];
    for (let i = 0; i < m; i++) {
        cnt[p.charCodeAt(i) - 97]--;
        cnt[s.charCodeAt(i) - 97]++;
    }
    if (cnt.every(v => v == 0)) {
        ans.push(0);
    }
    for (let i = m; i < n; i++) {
        cnt[s.charCodeAt(i) - 97]++;
        cnt[s.charCodeAt(i - m) - 97]--;
        if (cnt.every(v => v == 0)) {
            ans.push(i - m + 1);
        }
    }
    return ans;
}

C++

class Solution {
public:
    vector<int> findAnagrams(string s, string p) {
        vector<int> counter(26);
        for (char c : p) ++counter[c - 'a'];
        vector<int> ans;
        int left = 0, right = 0;
        vector<int> t(26);
        while (right < s.size()) {
            int i = s[right] - 'a';
            ++t[i];
            while (t[i] > counter[i]) {
                --t[s[left] - 'a'];
                ++left;
            }
            if (right - left + 1 == p.size()) ans.push_back(left);
            ++right;
        }
        return ans;
    }
};

Go

func findAnagrams(s string, p string) []int {
	counter := make([]int, 26)
	for _, c := range p {
		counter[c-'a']++
	}
	var ans []int
	left, right := 0, 0
	t := make([]int, 26)
	for right < len(s) {
		i := s[right] - 'a'
		t[i]++
		for t[i] > counter[i] {
			t[s[left]-'a']--
			left++
		}
		if right-left+1 == len(p) {
			ans = append(ans, left)
		}
		right++
	}
	return ans
}

Rust

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 res = vec![];
        if n > m {
            return res;
        }

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

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