leetcode - minimum number of array elements so that movement of equal

Thinking (assuming that each array have been sorted)

• Or less so that each maximum number n-1 plus 1, times out direct violence Solution
• Improved First: In order to make the smallest element is equal to the largest element, at least array maximum - minimum time, so in order to solve the violence (see below first python code) based again
• Improved II: whole solution process is the smallest ever catch maximum, until the two are equal so far. \ (max_ {0} \) is the maximum value of the original array, the updated \ (max_ {1} \) of the original array of the n-1 plus the last \ ((max-min) \ ) , \ (max_ {2} \) of the original array of n-2-th plus the last \ ((min-max) \) ... and can be found in every min is the last max,
• The final number of movements is \ (moves = \ sum_ {i = 1} ^ {n-1} (a [i] -a [0]) \) where n is the length of the array

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# 暴力求解 -> 超时
class Solution:
    def minMoves(self, nums) -> int:
        nums = sorted(nums)
        if len(nums) == 2:
            return nums[1]-nums[0]
        if len(nums) == 1 or len(nums) == 0:
            return 0
        if nums[0] == nums[-2]:
            return nums[-1] - nums[0]
        # 达到目的必然会移动不少于max-min步, 因为最后的结果的max一定是大于原有max的, 所以min至少要先移动到原有max那里去
        count = nums[-1]-nums[0]
        for i in range(len(nums) - 1):
            nums[i] += count
        nums = sorted(nums)
        while nums[0] != nums[-1]:
            for i in range(len(nums)-1):
                nums[i] += 1
            count += 1
            nums = sorted(nums)
        return count

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# 学习官方求解 修改后的 -> 通过
# 其实优化过程, 就是个找规律的过程
class Solution:
    def minMoves(self, nums):
        sums = sum(nums)
        mins = min(nums)
        mul = mins*len(nums)
        return sums - mul

to sum up

If I could find the law, it is not optimized thing