2. Valid Anagram

Description

Valid Anagram

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

Example 1:
Input: s = "anagram", t = "nagaram"
Output: true

Example 2:
Input: s = "rat", t = "car"
Output: false

Constraints:

• 1 <= s.length, t.length <= 5 * 104
• s and t consist of lowercase English letters.

Follow up: What if the inputs contain Unicode characters? How would you adapt your solution to such a case?

Approach 1

• Two unordered count map and a final loop to check if counts are same in both
• Time: O(n) Space: O(n)

class Solution {
public:
    bool isAnagram(string s, string t) {
        if(s.size() != t.size()) {
            return false;
        }
 
        unordered_map<char,int> countS;
        unordered_map<char,int> countT;
 
        for(int i = 0; i < s.size(); i++) {
            countS[s[i]]++;
            countT[t[i]]++;
        }
 
        for(const auto& pair: countS) {
            if(countS[pair.first] != countT[pair.first]) {
                return false;
            }
        }
        return true;
    }
};

Approach 2

• Create an integer array of size 26
• Loop through string and add 1 to array if it is in first string and -1 for second
• Loop through the array and if it is not zero for any key then it is not anagram

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

• The number 256 represents the total number of possible character codes in the **Extended ASCII character set**.

class Solution {
    public boolean isAnagram(String s, String t) {
        if(s.length() != t.length()) {
            return false;
        }
        int[] count = new int[256];
        for(int i = 0; i < s.length(); i++) {
            count[s.charAt(i)]++;
            count[t.charAt(i)]--;
        }
 
        for(int n: count) {
            if(n != 0) {
                return false;
            }
        }
 
        return true;
    }
}

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Approach 1: Sorting ($O(N\log N)$ Time, $O(N)$ Space)

Intuition

If two strings are anagrams, sorting their characters alphabetically will produce identical arrays. Convert both strings to character arrays, sort them, and check if they are equal.

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);
    }
}
 

Complexity

• Time Complexity: $O(N\log N)$ — Sorting two character arrays of length $N$.
• Space Complexity: $O(N)$ — Auxiliary space needed to convert strings to character arrays.

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Approach 2: Frequency Counting (Most Optimized — $O(N)$ Time, $O(1)$ Space)

Intuition

Because both strings consist of lowercase English letters ('a' to 'z'), track character frequency using a fixed-size integer array of length 26.

1. Return false immediately if the lengths of s and t differ.
2. Increment the count for each character in s while decrementing for each character in t.
3. If all array values net out to zero, t is a valid anagram of s.

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

Complexity

• Time Complexity: $O(N)$ — Single linear pass over strings of length $N$.
• Space Complexity: $O(1)$ — Fixed auxiliary space for an array of size 26.

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Easy Memory Rule

“Lengths must match $\to$ increment for s, decrement for t $\to$ ensure every frequency nets to 0!”