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.
1def reduction_operations(nums):
2 nums.sort()
3 operations = 0
4 distinct_count = 0
5 for i in range(1, len(nums)):
6 if nums[i] != nums[i - 1]:
7 distinct_count += 1
8 operations += distinct_count
9 return operations
10
11print(reduction_operations([5, 1, 3]))In Python, we use the list sort method to order elements, then traverse through the sorted list while counting distinct numbers. This determines how many operations are needed by incrementing the operations count based on 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.
#include <vector>
int reductionOperations(std::vector<int>& nums) {
std::vector<int> count(50001, 0);
for (int num : nums) {
count[num]++;
}
int operations = 0, total = 0;
for (int i = 50000; i > 0; i--) {
if (count[i]) {
operations += total;
total += count[i];
}
}
return operations;
}
int main() {
std::vector<int> nums = {5, 1, 3};
std::cout << reductionOperations(nums) << std::endl;
return 0;
}This C++ solution tracks counts of each integer and processes from the highest possible integer down to compute operations. This enables efficient summation of the necessary steps by processing through accumulated frequencies.