2256. Minimum Average Difference

Medium

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.

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 <= nums.length <= 105

• 0 <= nums[i] <= 105

解題

func minimumAverageDifference(nums []int) int {
    l := len(nums)
    presum := make([]int, l)
    presum[0] = nums[0]

    for i:=1; i<l; i++ {
        presum[i] = presum[i - 1] + nums[i]
    }

    ans := 0
    diff := math.MaxInt64

    for i:=0; i<l; i++ {
        curdiff:= presum[i] / (i + 1)
        if i != l - 1{
            curdiff -= (presum[l - 1] - presum[i]) / (l - i - 1)
        }
        
        if  abs(curdiff) < diff {
            diff = abs(curdiff)
            ans = i
        }
    }

    return ans
}

func abs(a int) int {
    if a < 0 { return -a }
    return a
}