Skip to content

Latest commit

 

History

History
150 lines (121 loc) · 4.03 KB

File metadata and controls

150 lines (121 loc) · 4.03 KB

中文文档

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

Python3

class Solution:
    def reductionOperations(self, nums: List[int]) -> int:
        nums.sort()
        cnt, res, n = 0, 0, len(nums)
        for i in range(1, n):
            if nums[i] != nums[i - 1]:
                cnt += 1
            res += cnt
        return res

Java

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

TypeScript

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

C++

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

Go

func reductionOperations(nums []int) int {
	sort.Ints(nums)
	cnt, res, n := 0, 0, len(nums)
	for i := 1; i < n; i++ {
		if nums[i] != nums[i-1] {
			cnt++
		}
		res += cnt
	}
	return res
}

...