题目如下：

You have an array arr of length n where arr[i] = (2 * i) + 1 for all valid values of i (i.e. 0 <= i < n).

In one operation, you can select two indices x and y where 0 <= x, y < n and subtract 1 from arr[x] and add 1 to arr[y] (i.e. perform arr[x] -=1 and arr[y] += 1). The goal is to make all the elements of the array equal. It is guaranteed that all the elements of the array can be made equal using some operations.

Given an integer n, the length of the array. Return the minimum number of operations needed to make all the elements of arr equal.

Example 1:

Input: n = 3
Output: 2
Explanation: arr = [1, 3, 5]
First operation choose x = 2 and y = 0, this leads arr to be [2, 3, 4]
In the second operation choose x = 2 and y = 0 again, thus arr = [3, 3, 3].

Example 2:

Input: n = 6
Output: 9 

Constraints:

• 1 <= n <= 10^4

解题思路：每次操作都是加一减一，即数组的和是不变的，由此可知最后必定是所有元素的值等于数组的平均值。因此结果是所有元素与平均值的差的绝对值的和除以二。

代码如下：

class Solution(object):
    def minOperations(self, n):
        """
        :type n: int
        :rtype: int
        """
        res = 0
        if n % 2 == 1:
            target =  2*(n/2) + 1
        else:
            target =  ((2*(n/2) + 1) + 2*(n/2-1) + 1)/2

        for i in range(n/2):
            res += (target - 2*i -1)
        return res