LeetCode 2442: Count Number of Distinct Integers After Reverse Operations Solution

Master LeetCode problem 2442 (Count Number of Distinct Integers After Reverse Operations), 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.

2442. Count Number of Distinct Integers After Reverse Operations

Medium

Array Hash Table Math Counting

Problem Explanation

Explanation

To solve this problem, we can iterate through the given array nums and for each number, reverse its digits and add it to a new set. We can use a set data structure to store the distinct integers in the final array. By iterating through each number in nums and its reversed version, we can count the number of distinct integers in the final array.

Algorithm:
1. Initialize a set distinctIntegers to store distinct integers.
2. Iterate through each number in nums:
  - Reverse the digits of the number.
  - Add the original number and its reversed version to the set distinctIntegers.
3. Return the size of the set distinctIntegers.
Time Complexity: O(N * M), where N is the number of elements in nums and M is the maximum number of digits in any element.
Space Complexity: O(N), where N is the number of elements in nums.

Solution Code

class Solution {
    public int countDistinct(int[] nums) {
        Set<Integer> distinctIntegers = new HashSet<>();
        for (int num : nums) {
            int reverseNum = 0;
            while (num > 0) {
                reverseNum = reverseNum * 10 + num % 10;
                num /= 10;
            }
            distinctIntegers.add(num);
            distinctIntegers.add(reverseNum);
        }
        return distinctIntegers.size();
    }
}

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