LeetCode 438: Find All Anagrams in a String Solution

Master LeetCode problem 438 (Find All Anagrams in a String), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

438. Find All Anagrams in a String

Medium

Hash Table String Sliding Window

Problem Explanation

Explanation

To solve this problem, we can use a sliding window approach. We will maintain a frequency map for the characters in string p, and then slide a window of length p.length() over string s, updating the frequency map accordingly. If the frequency map matches the frequency map of string p, we have found an anagram at the current window position. We can keep track of the start indices of such anagrams.

Initialize two frequency arrays, one for string p and one for the sliding window in string s.
Slide the window over string s, updating the frequency arrays.
If the frequency arrays match, add the start index of the window to the result.
Return the list of start indices where anagrams of p are found in s.

Time Complexity: O(n), where n is the length of string s. Space Complexity: O(1) since the frequency map will have at most 26 characters.

Solution Code

import java.util.*;

class Solution {
    public List<Integer> findAnagrams(String s, String p) {
        List<Integer> result = new ArrayList<>();
        int[] pFreq = new int[26];
        int[] sFreq = new int[26];
        
        for (char c : p.toCharArray()) {
            pFreq[c - 'a']++;
        }
        
        int pLength = p.length();
        for (int i = 0; i < s.length(); i++) {
            sFreq[s.charAt(i) - 'a']++;
            if (i >= pLength) {
                sFreq[s.charAt(i - pLength) - 'a']--;
            }
            if (Arrays.equals(pFreq, sFreq)) {
                result.add(i - pLength + 1);
            }
        }
        
        return result;
    }
}

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