438. Find All Anagrams in a String

438. Find All Anagrams in a String (Medium)

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.

 

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.

Companies:
Amazon, Microsoft, Bloomberg, Snapchat, Facebook, Paypal, Yandex, Apple

Related Topics:
Hash Table, String, Sliding Window

Similar Questions:

• Valid Anagram (Easy)
• Permutation in String (Medium)

Solution 1.

// OJ: https://leetcode.com/problems/find-all-anagrams-in-a-string/
// Author: github.com/lzl124631x
// Time: O(M + N)
// Space: O(1)
class Solution {
public:
    vector<int> findAnagrams(string s, string p) {
        int target[26] = {}, cnt[26] = {};
        for (char c : p) target[c - 'a']++;
        vector<int> ans;
        for (int i = 0; i < s.size(); ++i) {
            cnt[s[i] - 'a']++;
            if (i >= p.size()) cnt[s[i - p.size()] - 'a']--;
            int j = 0;
            for (; j < 26 && target[j] == cnt[j]; ++j);
            if (j == 26) ans.push_back(i - p.size() + 1);
        }
        return ans;
    }
};

Solution 2.

// OJ: https://leetcode.com/explore/challenge/card/may-leetcoding-challenge/536/week-3-may-15th-may-21st/3332/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    vector<int> findAnagrams(string s, string p) {
        int cnt[26] = {}, matched = 0;
        for (char c : p) cnt[c - 'a']++;
        vector<int> ans;
        for (int i = 0; i < s.size(); ++i) {
            if (--cnt[s[i] - 'a'] >= 0) ++matched;
            if (i >= p.size() && ++cnt[s[i - p.size()] - 'a'] > 0) --matched;
            if (matched == p.size()) ans.push_back(i - p.size() + 1);
        }
        return ans;
    }
};