LeetCode 242: Valid Anagram Solution

Master LeetCode problem 242 (Valid Anagram), a easy 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.

242. Valid Anagram

Easy

Hash Table String Sorting

Problem Explanation

Explanation

To solve this problem, we can first check if the lengths of the two strings are different. If they are not, we can create a frequency map for each character in both strings. Then, we compare the frequency maps to check if they are equal. If they are equal, it means the two strings are anagrams of each other.

Algorithm:

Check if the lengths of strings s and t are different. If they are, return false.
Create frequency maps for characters in strings s and t.
Compare the frequency maps. If they are equal, return true; otherwise, return false.

Time Complexity

The time complexity of this solution is O(n), where n is the length of the input strings s and t.

Space Complexity

The space complexity of this solution is O(1) because we are using a fixed-size array to store the frequency of characters.

Solution Code

class Solution {
    public boolean isAnagram(String s, String t) {
        if (s.length() != t.length()) {
            return false;
        }

        int[] count = new int[26];
        for (char c : s.toCharArray()) {
            count[c - 'a']++;
        }
        for (char c : t.toCharArray()) {
            count[c - 'a']--;
        }

        for (int i = 0; i < 26; i++) {
            if (count[i] != 0) {
                return false;
            }
        }

        return true;
    }
}

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The time complexity for LeetCode 242 (Valid Anagram) varies by solution. Check the detailed explanation section for specific complexities in Java, C++, and Python implementations.

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