Anagram

JavaView on GFG

Time: O(n)

Space: O(1)

Problem Overview

Two strings are anagrams if they have the same character frequencies.

See our study guide for structured GFG and LeetCode practice.

Intuition

Two strings are anagrams if they have the same character frequencies. Use a single int[256] array: increment for each character in s1, decrement for each in s2. If all entries are 0, they are anagrams.

Algorithm

1If lengths differ, return false.
2int[256] count. For each i: count[s1[i]]++; count[s2[i]]--.
3Return count.All(x => x == 0).

Example Walkthrough

Input: s1 = "listen", s2 = "silent"

1. After processing: all counts are 0. Return true.

Output: YES

Common Pitfalls

• Use a char frequency array, not sort — sort is O(n log n) vs O(n).

Anagram.java

Java

// Approach: Count character frequencies of both strings using a frequency array of size 26.
// If all counts match, they are anagrams.
// Time: O(n) Space: O(1)
import java.util.*;

class Solution {
    // Function is to check whether two strings are anagram of each other or not.
    public static boolean areAnagrams(String s1, String s2) {
        if (s1.length() != s2.length()) {
            return false;
        } else {
            HashMap<Character, Integer> h1 = new HashMap<>();
            HashMap<Character, Integer> h2 = new HashMap<>();
            for (int i = 0; i < s1.length(); i++) {
                if (h1.containsKey(s1.charAt(i)))
                    h1.put(s1.charAt(i), h1.get(s1.charAt(i)) + 1);
                else
                    h1.put(s1.charAt(i), 1);
            }
            for (int i = 0; i < s2.length(); i++) {
                if (h2.containsKey(s2.charAt(i)))
                    h2.put(s2.charAt(i), h2.get(s2.charAt(i)) + 1);
                else
                    h2.put(s2.charAt(i), 1);
            }

            return h1.equals(h2);
        }
    }
}

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