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LeetCode 2531: Make Number of Distinct Characters Equal Solution

Master LeetCode problem 2531 (Make Number of Distinct Characters Equal), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2531. Make Number of Distinct Characters Equal

Medium
Hash TableStringCounting

Problem Explanation

Explanation:

To determine if it is possible to make the number of distinct characters in word1 and word2 equal with exactly one move, we need to consider the following cases:

  1. If the total number of distinct characters in both strings is already equal, return true.
  2. If the difference between the number of distinct characters in word1 and word2 is more than 1, return false.
  3. If the difference is exactly 1, we can achieve the goal with one swap.

Algorithm:

  1. Count the occurrences of each character in word1 and word2.
  2. Calculate the number of distinct characters in word1 and word2.
  3. If the number of distinct characters in both strings is the same, return true.
  4. If the difference is more than 1, return false.
  5. If the difference is exactly 1, check if one of the strings contains a character that is not present in the other string. If found, return true; otherwise, return false.

Time Complexity: O(n) Space Complexity: O(1)

:

Solution Code

class Solution {
    public boolean makeEqual(String word1, String word2) {
        int[] count1 = new int[26];
        int[] count2 = new int[26];
        
        for (char c : word1.toCharArray()) {
            count1[c - 'a']++;
        }
        
        for (char c : word2.toCharArray()) {
            count2[c - 'a']++;
        }
        
        int distinct1 = 0, distinct2 = 0;
        
        for (int count : count1) {
            if (count > 0) distinct1++;
        }
        
        for (int count : count2) {
            if (count > 0) distinct2++;
        }
        
        if (distinct1 == distinct2) {
            return true;
        }
        
        if (Math.abs(distinct1 - distinct2) > 1) {
            return false;
        }
        
        if (distinct1 > distinct2) {
            return checkDistinct(count1, count2);
        } else {
            return checkDistinct(count2, count1);
        }
    }
    
    private boolean checkDistinct(int[] count1, int[] count2) {
        for (int i = 0; i < 26; i++) {
            if (count1[i] > 0 && count2[i] == 0) {
                return true;
            }
        }
        return false;
    }
}

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