tags: leetcode medium Array Prefix Sum

2256. 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%