You are given two 0-indexed arrays nums and cost consisting each of n positive integers.
You can do the following operation any number of times:
\n\nnums by 1.The cost of doing one operation on the ith element is cost[i].
Return the minimum total cost such that all the elements of the array nums become equal.
\n
Example 1:
\n\n\nInput: nums = [1,3,5,2], cost = [2,3,1,14]\nOutput: 8\nExplanation: We can make all the elements equal to 2 in the following way:\n- Increase the 0th element one time. The cost is 2.\n- Decrease the 1st element one time. The cost is 3.\n- Decrease the 2nd element three times. The cost is 1 + 1 + 1 = 3.\nThe total cost is 2 + 3 + 3 = 8.\nIt can be shown that we cannot make the array equal with a smaller cost.\n\n\nExample 2:
\n\n\nInput: nums = [2,2,2,2,2], cost = [4,2,8,1,3]\nOutput: 0\nExplanation: All the elements are already equal, so no operations are needed.\n\n\n
\n
Constraints:
\n\nn == nums.length == cost.length1 <= n <= 1051 <= nums[i], cost[i] <= 106给你两个下标从 0 开始的数组 nums 和 cost ，分别包含 n 个 正 整数。
你可以执行下面操作 任意 次：
\n\nnums 中 任意 元素增加或者减小 1 。对第 i 个元素执行一次操作的开销是 cost[i] 。
请你返回使 nums 中所有元素 相等 的 最少 总开销。
\n\n
示例 1：
\n\n\n输入：nums = [1,3,5,2], cost = [2,3,1,14]\n输出：8\n解释：我们可以执行以下操作使所有元素变为 2 ：\n- 增加第 0 个元素 1 次，开销为 2 。\n- 减小第 1 个元素 1 次，开销为 3 。\n- 减小第 2 个元素 3 次，开销为 1 + 1 + 1 = 3 。\n总开销为 2 + 3 + 3 = 8 。\n这是最小开销。\n\n\n
示例 2：
\n\n\n输入：nums = [2,2,2,2,2], cost = [4,2,8,1,3]\n输出：0\n解释：数组中所有元素已经全部相等，不需要执行额外的操作。\n\n\n
\n\n
提示：
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