**Minimum Average Difference solution leetcode** – 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.

## [Solution] Minimum Average Difference solution leetcode

Return* the index with the minimum average difference*. If there are multiple such indices, return the

**smallest**one.

**Note:**

- The
**absolute difference**of two numbers is the absolute value of their difference. - The
**average**of`n`

elements is the**sum**of the`n`

elements divided (**integer division**) by`n`

. - The average of
`0`

elements is considered to be`0`

.

## [Solution] Minimum Average Difference solution leetcode

Input:nums = [2,5,3,9,5,3]Output:3Explanation:- The average difference of index 0 is: |2 / 1 - (5 + 3 + 9 + 5 + 3) / 5| = |2 / 1 - 25 / 5| = |2 - 5| = 3. - The average difference of index 1 is: |(2 + 5) / 2 - (3 + 9 + 5 + 3) / 4| = |7 / 2 - 20 / 4| = |3 - 5| = 2. - The average difference of index 2 is: |(2 + 5 + 3) / 3 - (9 + 5 + 3) / 3| = |10 / 3 - 17 / 3| = |3 - 5| = 2. - The average difference of index 3 is: |(2 + 5 + 3 + 9) / 4 - (5 + 3) / 2| = |19 / 4 - 8 / 2| = |4 - 4| = 0. - The average difference of index 4 is: |(2 + 5 + 3 + 9 + 5) / 5 - 3 / 1| = |24 / 5 - 3 / 1| = |4 - 3| = 1. - The average difference of index 5 is: |(2 + 5 + 3 + 9 + 5 + 3) / 6 - 0| = |27 / 6 - 0| = |4 - 0| = 4. The average difference of index 3 is the minimum average difference so return 3.

**Example 2:**

Input:nums = [0]Output:0Explanation:The only index is 0 so return 0. The average difference of index 0 is: |0 / 1 - 0| = |0 - 0| = 0.

## Minimum Average Difference solution leetcode

`1 <= nums.length <= 10`

^{5}`0 <= nums[i] <= 10`

^{5}