Sponsored
Sponsored
To solve this problem, we can first sort the array. After sorting, for each distinct number except the smallest one, count how many times smaller numbers need to be stepped up to equal the larger ones. Iterate over the sorted array and accumulate these steps until all numbers are equal.
Time Complexity: O(n log n) due to sorting, Space Complexity: O(1) as sorting is done in place.
1import java.util.Arrays;
2
3public class Solution {
4 public int reductionOperations(int[] nums) {
5 Arrays.sort(nums);
6 int operations = 0, distinctCount = 0;
7 for (int i = 1; i < nums.length; i++) {
8 if (nums[i] != nums[i - 1]) {
9 distinctCount++;
10 }
11 operations += distinctCount;
12 }
13 return operations;
14 }
15
16 public static void main(String[] args) {
17 Solution solution = new Solution();
18 int[] nums = {5, 1, 3};
19 System.out.println(solution.reductionOperations(nums));
20 }
21}In Java, we sort the array using Arrays.sort and then follow the same approach as in C++: iterate through the array calculating the number of operations by maintaining a counter for distinct elements.
This approach involves counting the duplicates of each number without explicitly sorting. By iterating from the maximum value to the minimum, we calculate how many numbers need to be converted at each step by leveraging the array structure and gaps between numbers.
Time Complexity: O(n + k) where k is the maximum possible value in nums, Space Complexity: O(k) for the count array.
JavaScript implementation similar to prior solutions, involves an array to count occurrences. Reducing from the maximum possible value gives a straightforward approach to calculate necessary operations efficiently.