438. Find All Anagrams in a String

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438. Find All Anagrams in a String

Description

Difficulty: Medium

Related Topics: Hash Table, String, Sliding Window

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:

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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:

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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 * 104
  • s and p consist of lowercase English letters.

Hints/Notes

  • Sliding window

Solution

Language: C++

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class Solution {
public:
    vector<int> findAnagrams(string s, string p) {
        unordered_map<char, int> need, window;
        for (char c : p) need[c]++;
        int left = 0, right = 0, valid = 0, len = p.size();
        vector<int> res;
        while (right < s.size()) {
            char c = s[right++];
            if (need.count(c)) {
                window[c]++;
                if (window[c] == need[c]) {
                    valid++;
                }
            }
            if (right - left == len) {
                if (valid == need.size()) {
                    res.push_back(left);
                }
                c = s[left++];
                if (need.count(c)) {
                    if (window[c] == need[c]) {
                        valid--;
                    }
                    window[c]--;
                }
            }
        }
        return res;
    }
};