Valid Anagrams - Leet Code Solution

Problem Statement

Given two strings s and t , write a function to determine if t is an anagram of s.

Example

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

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

Note: You may assume the string contains only lowercase alphabets.

What is Anagram

First try to understand what an Anagram is. Its NOT about checking order of characters in a string.

Its about checking that:

Each character in both strings has equal number of occurrence.

Solution - 1

A simple solution can be to sort the strings first, then compare.

Code

public boolean isAnagram_sort(String s, String t) {
   if (s.length() != t.length()) {
      return false;
   }
   char[] s1 = s.toCharArray();
   char[] s2 = t.toCharArray();
   Arrays.sort(s1);
   Arrays.sort(s2);

   return Arrays.equals(s1, s2);
}

Complexity

It is equal to complexity taken by sorting.
Its O(nlogn)

Solution using counting number of characters - HashMap

Another simple solution is that we can use a HashMap<Character, Integer>.

In one pass of first array, we can populate HashMap, which will have count of each character
In iteration of second array, we can simply decrement count of found characters
At any time, if the count becomes zero before we decrementing it. Which means, character count do not match.

Code

public boolean isAnagram(String s, String t) {
   if (s.length() != t.length()) {
      return false;
   }
   
   Map<Character, Integer> map = new HashMap<Character, Integer>();
   for (int i=0; i<s.length(); i++) {
      int count = map.getOrDefault(s.charAt(i), 0);
      count ++;
      
      map.put(s.charAt(i), count);
   }
   
   for (int i=0; i<t.length(); i++) {
      int count = map.getOrDefault(t.charAt(i), 0);
      if (count == 0) {
         return false;
      }
      
      count --;
      map.put(t.charAt(i), count);
   }
   
   return true;
}

Complexity

Its O(n)

Solution by using array

Since we know that there are only lowercase characters. We know the unique number of characters will be 26.

We can take an Integer array of count 26
We can assume that first index corresponds to a, second to b and so on.
In first pass of an array, we can increment count according to location mentioned above
While iterating second array, we can simply start decrementing count.
And, at any point if we found the count to be negative. We return false.

Code

public boolean isAnagram_array(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'] ++;
   }
   
   for (int i=0; i<t.length(); i++) {
      if (count[t.charAt(i) - 'a'] <= 0) {
         return false;
      }
      count[t.charAt(i) - 'a'] --;
   }
   
   return true;
}