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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.
1#include <stdio.h>
2#include <stdlib.h>
3
4int compare(const void *a, const void *b) {
5 return (*(int*)a - *(int*)b);
6}
7
8int reductionOperations(int* nums, int numsSize) {
9 qsort(nums, numsSize, sizeof(int), compare);
10 int operations = 0, distinctCount = 0;
11 for (int i = 1; i < numsSize; i++) {
12 if (nums[i] != nums[i - 1]) {
13 distinctCount++;
14 }
15 operations += distinctCount;
16 }
17 return operations;
18}
19
20int main() {
21 int nums[] = {5, 1, 3};
22 int size = sizeof(nums) / sizeof(nums[0]);
23 printf("%d\n", reductionOperations(nums, size));
24 return 0;
25}In this C implementation, we use the qsort function to sort the array. Then, we iterate through the sorted array, counting distinct elements. Each time a new distinct element is found, it represents a level of operations needed to equalize numbers, accumulated into 'operations'.
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.
We maintain a count array for each potential value of the elements in nums. From the largest possible value, we calculate the cumulative operations needed by summing through the stored counts for all numbers greater than the current one.