[📝LeetCode #242] Valid Anagram

#leetcode #java #100daysofcode #string

🎀 The Problem

Given two strings s and t, return true if t is an anagram of s, and false otherwise.

Example:

Input: s = "anagram", t = "nagaram"
Output: true

👩‍💻 My Answer

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

        HashMap map = new HashMap();

        char counter = 0;
        int scount = 0, tcount = 0;

        while (counter < s.length()) {
            char letter = s.charAt(counter);

            if (!map.containsKey(letter)) {
                for (int i = 0; i < s.length(); i++) {
                    if (s.charAt(i) == letter)
                        scount++;
                    if (t.charAt(i) == letter)
                        tcount++;
                }
            }

            if (scount != tcount)
                return false;
            else
                map.put(letter, scount);

            counter++;
        }

        return true;
    }
}

Pro & Con

✖️ Runtime & Memory

💋 Ideal Answer

Approach - "Array"

I watched the YouTube video and found that it's faster if I use int instead of a HashMap. Here is my array method.

New Code

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

        int[] scount = new int[26];
        int[] tcount = new int[26];

        for (int i = 0; i < s.length(); i++) {
            scount[s.charAt(i)-'a']++;
            tcount[t.charAt(i)-'a']++;
        }

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

        return true;
    }
}

This beats only 57%, so I was wondering what the ideal answer is. Then, I found the faster answer in LeetCode problem.

public class Solution {
    public boolean isAnagram(String s, String t) {
        int[] alphabet = new int[26];
        for (int i = 0; i < s.length(); i++) alphabet[s.charAt(i) - 'a']++;
        for (int i = 0; i < t.length(); i++) alphabet[t.charAt(i) - 'a']--;
        for (int i : alphabet) if (i != 0) return false;
        return true;
    }
}

💡 What I Learned

Use array instead of HashMap to make it faster.
int[] alphabet = new int[26];
I used to use two arrays for s and t. But use ONE instead.
If you find the letter in s, increment;t while the letter in t decreases the count in the array.

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