LeetCode 1551. The minimum operand to make all elements in an array equal

The title

There exists an array arr of length n, where arr[I] = (2 * I) + 1 (0 <= I < n).

In one operation, you can pick two subscripts, call them x and y (0 <= x, y < n) and make arr[x] minus 1, arr[y] plus 1 (i.e. Arr [x] -=1 and arr[y] += 1). The ultimate goal is to make all elements in the array equal. The problem test case will ensure that, after several steps, all elements in the array will eventually be equal.

I give you an integer n, the length of the array. Return the minimum operand required to make all elements in array ARR equal.

Example 1: Input: n = 3 Output: 2 Arr = [1, 3, 5] The first operation selects x = 2 and y = 0, making the array [2, 3, 4] the second operation continues to select x = 2 and y = 0, which will make the array [3, 3, 3] example 2: Input: n = 6 Output: 9Copy the code

Tip:

1 <= n <= 10^4

Their thinking

class Solution:
    def minOperations(self, n: int) -> int:
        average = n
        maxNum = (n-1)*2+1
        res = 0
        # print(average, maxNum)
        while maxNum > average:
            res += maxNum - average
            maxNum -= 2
        return res


if __name__ == '__main__':
    n = 6
    ret = Solution().minOperations(n)
    print(ret)
Copy the code