You are given a 0-indexed integer array nums of length n.
The average difference of the index i is the absolute difference between the average of the first i + 1 elements of nums and the average of the last n - i - 1 elements. Both averages should be rounded down to the nearest integer.
Return the index with the minimum average difference. If there are multiple such indices, return the smallest one.
\n\nNote:
\n\nn elements is the sum of the n elements divided (integer division) by n.0 elements is considered to be 0.\n
Example 1:
\n\n\nInput: nums = [2,5,3,9,5,3]\nOutput: 3\nExplanation:\n- The average difference of index 0 is: |2 / 1 - (5 + 3 + 9 + 5 + 3) / 5| = |2 / 1 - 25 / 5| = |2 - 5| = 3.\n- The average difference of index 1 is: |(2 + 5) / 2 - (3 + 9 + 5 + 3) / 4| = |7 / 2 - 20 / 4| = |3 - 5| = 2.\n- The average difference of index 2 is: |(2 + 5 + 3) / 3 - (9 + 5 + 3) / 3| = |10 / 3 - 17 / 3| = |3 - 5| = 2.\n- The average difference of index 3 is: |(2 + 5 + 3 + 9) / 4 - (5 + 3) / 2| = |19 / 4 - 8 / 2| = |4 - 4| = 0.\n- The average difference of index 4 is: |(2 + 5 + 3 + 9 + 5) / 5 - 3 / 1| = |24 / 5 - 3 / 1| = |4 - 3| = 1.\n- The average difference of index 5 is: |(2 + 5 + 3 + 9 + 5 + 3) / 6 - 0| = |27 / 6 - 0| = |4 - 0| = 4.\nThe average difference of index 3 is the minimum average difference so return 3.\n\n\n
Example 2:
\n\n\nInput: nums = [0]\nOutput: 0\nExplanation:\nThe only index is 0 so return 0.\nThe average difference of index 0 is: |0 / 1 - 0| = |0 - 0| = 0.\n\n\n
\n
Constraints:
\n\n1 <= nums.length <= 1050 <= nums[i] <= 105给你一个下标从 0 开始长度为 n 的整数数组 nums 。
下标 i 处的 平均差 指的是 nums 中 前 i + 1 个元素平均值和 后 n - i - 1 个元素平均值的 绝对差 。两个平均值都需要 向下取整 到最近的整数。
请你返回产生 最小平均差 的下标。如果有多个下标最小平均差相等，请你返回 最小 的一个下标。
\n\n注意：
\n\nn 个元素的平均值是 n 个元素之 和 除以（整数除法） n 。0 个元素的平均值视为 0 。\n\n
示例 1：
\n\n输入：nums = [2,5,3,9,5,3]\n输出：3\n解释：\n- 下标 0 处的平均差为：|2 / 1 - (5 + 3 + 9 + 5 + 3) / 5| = |2 / 1 - 25 / 5| = |2 - 5| = 3 。\n- 下标 1 处的平均差为：|(2 + 5) / 2 - (3 + 9 + 5 + 3) / 4| = |7 / 2 - 20 / 4| = |3 - 5| = 2 。\n- 下标 2 处的平均差为：|(2 + 5 + 3) / 3 - (9 + 5 + 3) / 3| = |10 / 3 - 17 / 3| = |3 - 5| = 2 。\n- 下标 3 处的平均差为：|(2 + 5 + 3 + 9) / 4 - (5 + 3) / 2| = |19 / 4 - 8 / 2| = |4 - 4| = 0 。 \n- 下标 4 处的平均差为：|(2 + 5 + 3 + 9 + 5) / 5 - 3 / 1| = |24 / 5 - 3 / 1| = |4 - 3| = 1 。\n- 下标 5 处的平均差为：|(2 + 5 + 3 + 9 + 5 + 3) / 6 - 0| = |27 / 6 - 0| = |4 - 0| = 4 。\n下标 3 处的平均差为最小平均差，所以返回 3 。\n\n\n
示例 2：
\n\n输入：nums = [0]\n输出：0\n解释：\n唯一的下标是 0 ，所以我们返回 0 。\n下标 0 处的平均差为：|0 / 1 - 0| = |0 - 0| = 0 。\n\n\n
\n\n
提示：
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