Given an integer array nums of size n, return the minimum number of moves required to make all array elements equal.
In one move, you can increment n - 1 elements of the array by 1.
Example 1:
Input: nums = [1,2,3]
Output: 3
Explanation: Only three moves are needed (remember each move increments two elements):
[1,2,3] => [2,3,3] => [3,4,3] => [4,4,4]
翻译：
给定一个大小为size的数组nums，每次你需要对它的“除最大值以外”的值进行+1操作，当数组中所有值都一样时结束。问：经过多少次可以达到平衡
我们先来观察一组数据：
| 数组 | 迭代次数 |
|---|---|
| [1, 3, 7, 11] | 0 |
| [2, 4, 8, 11] | 1 |
| [3, 5, 9, 11] | 2 |
| [4, 6, 10, 11] | 3 |
| [5, 7, 11, 11] | 4 |
| [6, 8, 11, 12] | 5 |
| [7, 9, 12, 12] | 6 |
| [8, 10, 12, 13] | 7 |
| [9, 11, 13, 13] | 8 |
| [10, 12, 13, 14] | 9 |
| [11, 13, 14, 14] | 10 |
| [12, 14, 14, 15] | 11 |
| [13, 15, 15, 15] | 12 |
| [14, 15, 16, 16] | 13 |
| [15, 16, 16, 17] | 14 |
| [16, 17, 17, 17] | 15 |
| [17, 17, 18, 18] | 16 |
| [18, 18, 18, 19] | 17 |
| [19, 19, 19, 19] | 18 |
通过观察这组数据，可以发现：
- 经过 ( 11 − 7 ) × 1 = 4 (11-7) \times 1=4 (11−7)×1=4 次后，最后两位数可以达到平衡
- 经过 ( 7 − 3 ) × 2 = 8 (7-3) \times 2 =8 (7−3)×2=8 次后，最后三位可以达到平衡，因为第二位每次加1，需要后两位交替2次
- 经过 ( 3 − 1 ) × 3 = 6 (3-1)\times 3=6 (3−1)×3=6 次后，最后四位（全部）可以达到平衡，最后第第一位每加1，需要后面三位交替3次
经过总结，可以得到如下公式：
设 nums有序，且 nums = [ x 1 , x 2 , x 3 , ⋯ , x n ] \text{设~~nums有序，且~~nums} = [ x_1,x_2,x_3, \cdots , x_n] 设 nums有序，且 nums=[x1,x2,x3,⋯,xn]
则，总次数为
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\begin{aligned} \\\\ T&=(x_n-x_{n-1})\times 1 +(x_{n-1} - x_{n-2})\times2 + \cdots +(x_2-x_1)\times (n-1) \\\\ &= x_{n}+x_{n-1}+\ldots \cdots+x_{2}-x_{1}(n-1) \\\\ & =x_{n}+x_{n-1}+\ldots \cdots+x_{2} + x_1 - x_{1}n \end{aligned}
T=(xn−xn−1)×1+(xn−1−xn−2)×2+⋯+(x2−x1)×(n−1)=xn+xn−1+…⋯+x2−x1(n−1)=xn+xn−1+…⋯+x2+x1−x1n
将测试数据代入该公式，为：
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18
T = 11+7+3 + 1 - 1\times 4 = 18
T=11+7+3+1−1×4=18
class Solution:
def minMoves(self, nums) -> int:
return sum(nums) - len(nums)*min(nums)
