242. Valid Anagram 🎢

Difficulty: Easy - Tags: Hash Table, String, Sorting

LeetCode Problem Link

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Problem Statement 📜

Given two strings s and t, determine if t is an anagram of s. An anagram is a word or phrase formed by rearranging the letters of a different word or phrase, using all the original letters exactly once.

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Examples 🌟

🔹 Example 1:

Input:

s = "anagram", t = "nagaram"

Output:

true

🔹 Example 2:

Input:

s = "rat", t = "car"

Output:

false

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Constraints ⚙️

• 1 <= s.length, t.length <= 5 * 10^4

• s and t consist of lowercase English letters.

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Solution 💡

The simplest way to check if two strings are anagrams is to compare their sorted versions. Alternatively, we can use a frequency count for more efficiency.

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Java Solution (Sorting Approach)

import java.util.Arrays;

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

        char[] sArray = s.toCharArray();
        char[] tArray = t.toCharArray();

        Arrays.sort(sArray);
        Arrays.sort(tArray);

        return Arrays.equals(sArray, tArray);
    }
}

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Java Solution (HashMap Approach)

import java.util.HashMap;
import java.util.Map;

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

        Map<Character, Integer> charCount = new HashMap<>();

        for (char c : s.toCharArray()) {
            charCount.put(c, charCount.getOrDefault(c, 0) + 1);
        }

        for (char c : t.toCharArray()) {
            if (!charCount.containsKey(c) || charCount.get(c) == 0) {
                return false;
            }
            charCount.put(c, charCount.get(c) - 1);
        }

        return true;
    }
}

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Explanation of the Solution

1. Sorting Approach:

   • Convert both strings to character arrays.

   • Sort the arrays.

   • Compare the sorted arrays for equality.

2. HashMap Approach:

   • Count the frequency of each character in s using a HashMap.

   • For each character in t, decrement its count in the map.

   • If a character in t is missing or its count goes below zero, return false.

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Time Complexity ⏳

• Sorting Approach:

  • O(n log n) for sorting, where n is the length of the strings.

• HashMap Approach:

  • O(n) for iterating through the strings.

Space Complexity 💾

• Sorting Approach:

  • O(n) for storing the sorted arrays.

• HashMap Approach:

  • O(1) space for the map (since there are only 26 lowercase English letters).

You can find the full solution here.