###### tags: `leetcode` `medium` `Array` `Prefix Sum`
# [2256. Minimum Average Difference](https://leetcode.com/problems/minimum-average-difference/)

## Description

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.

**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`.

## Examples
### Example 1:

**Input**: nums = [2,5,3,9,5,3]
**Output**: 3
**Explanation**:
- 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**: 0
**Explanation**:
The only index is 0 so return 0.
The average difference of index 0 is: |0 / 1 - 0| = |0 - 0| = 0.

## Constraints:

- $1 \leq nums.length \leq 10^5$
- $0 \leq nums[i] \leq 10^5$

## Code

int minimumAverageDifference(int* nums, int numsSize) {
    if (numsSize == 1)
        return 0;
    long* sum = calloc(numsSize, sizeof(long));
    int index = 0, min = INT_MAX, tmp = 0;
    // Collect all summation
    sum[0] = nums[0];
    for (int i = 1; i < numsSize; i++)
        sum[i] = sum[i - 1] + nums[i];
    for (int i = 0; i < numsSize; i++) {
        // Calculate all average difference
        tmp = abs(sum[i] / (i + 1) -
            ((numsSize - i - 1) ? ((sum[numsSize - 1] - sum[i]) / (numsSize - i - 1)) : 0));
        // Update minimum and index
        if (tmp < min) {
            min = tmp;
            index = i;
        }
    }
    return index;
}

## Complexity

|Space |Time |
|- |- |
|$O(N)$|$O(N)$|

## Result

- Runtime: 171 ms, faster than 81.82%
- Memory Usage: 18.4 MB, less than 18.18%