1887. Reduction Operations to Make the Array Elements Equal

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Description

Given an integer array nums, your goal is to make all elements in nums equal. To complete one operation, follow these steps:

1. Find the largest value in nums. Let its index be i (0-indexed) and its value be largest. If there are multiple elements with the largest value, pick the smallest i.
2. Find the next largest value in nums strictly smaller than largest. Let its value be nextLargest.
3. Reduce nums[i] to nextLargest.

Return the number of operations to make all elements in nums equal.

 

Example 1:

Input: nums = [5,1,3]
Output: 3
Explanation: It takes 3 operations to make all elements in nums equal:
1. largest = 5 at index 0. nextLargest = 3. Reduce nums[0] to 3. nums = [3,1,3].
2. largest = 3 at index 0. nextLargest = 1. Reduce nums[0] to 1. nums = [1,1,3].
3. largest = 3 at index 2. nextLargest = 1. Reduce nums[2] to 1. nums = [1,1,1].

Example 2:

Input: nums = [1,1,1]
Output: 0
Explanation: All elements in nums are already equal.

Example 3:

Input: nums = [1,1,2,2,3]
Output: 4
Explanation: It takes 4 operations to make all elements in nums equal:
1. largest = 3 at index 4. nextLargest = 2. Reduce nums[4] to 2. nums = [1,1,2,2,2].
2. largest = 2 at index 2. nextLargest = 1. Reduce nums[2] to 1. nums = [1,1,1,2,2].
3. largest = 2 at index 3. nextLargest = 1. Reduce nums[3] to 1. nums = [1,1,1,1,2].
4. largest = 2 at index 4. nextLargest = 1. Reduce nums[4] to 1. nums = [1,1,1,1,1].

 

Constraints:

• 1 <= nums.length <= 5 * 104
• 1 <= nums[i] <= 5 * 104

Solutions

Solution 1: Sorting

We first sort the array $\textit{nums}$, then iterate from the second element of the array. If the current element is not equal to the previous element, we increment $\textit{cnt}$, indicating the number of operations needed to reduce the current element to the minimum value. Then we add $\textit{cnt}$ to $\textit{ans}$ and continue to the next element.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(\log n)$. Here, $n$ is the length of the array $\textit{nums}$.

Python3

class Solution:
    def reductionOperations(self, nums: List[int]) -> int:
        nums.sort()
        ans = cnt = 0
        for a, b in pairwise(nums):
            if a != b:
                cnt += 1
            ans += cnt
        return ans

Java

class Solution {
    public int reductionOperations(int[] nums) {
        Arrays.sort(nums);
        int ans = 0, cnt = 0;
        for (int i = 1; i < nums.length; ++i) {
            if (nums[i] != nums[i - 1]) {
                ++cnt;
            }
            ans += cnt;
        }
        return ans;
    }
}

C++

class Solution {
public:
    int reductionOperations(vector<int>& nums) {
        ranges::sort(nums);
        int ans = 0, cnt = 0;
        for (int i = 1; i < nums.size(); ++i) {
            cnt += nums[i] != nums[i - 1];
            ans += cnt;
        }
        return ans;
    }
};

Go

func reductionOperations(nums []int) (ans int) {
	sort.Ints(nums)
	cnt := 0
	for i, x := range nums[1:] {
		if x != nums[i] {
			cnt++
		}
		ans += cnt
	}
	return
}

TypeScript

function reductionOperations(nums: number[]): number {
    nums.sort((a, b) => a - b);
    let [ans, cnt] = [0, 0];
    for (let i = 1; i < nums.length; ++i) {
        if (nums[i] !== nums[i - 1]) {
            ++cnt;
        }
        ans += cnt;
    }
    return ans;
}

JavaScript

/**
 * @param {number[]} nums
 * @return {number}
 */
var reductionOperations = function (nums) {
    nums.sort((a, b) => a - b);
    let [ans, cnt] = [0, 0];
    for (let i = 1; i < nums.length; ++i) {
        if (nums[i] !== nums[i - 1]) {
            ++cnt;
        }
        ans += cnt;
    }
    return ans;
};

C#

public class Solution {
    public int ReductionOperations(int[] nums) {
        Array.Sort(nums);
        int ans = 0, cnt = 0;
        for (int i = 1; i < nums.Length; i++) {
            if (nums[i] != nums[i - 1]) {
                ++cnt;
            }
            ans += cnt;
        }
        return ans;
    }
}