2256.Minimum Average Difference

tags: `Medium`,`Array`,`Prefix Sum`

題目描述

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.

範例

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

解答

Python


















class Solution:
    def minimumAverageDifference(self, nums: List[int]) -> int:
        if len(nums) <= 1:
            return 0
        current_ans = [-1,1000000000]
        whole_array_sum = sum(nums)
        firstksum = 0
        #print(nums)
        for idx, v in enumerate(nums):
            firstksum += v
            if idx == len(nums)-1:
                cur = abs( whole_array_sum/len(nums) )
            else:
                cur = abs( floor(firstksum/(idx+1)) - floor( (whole_array_sum - firstksum)/(len(nums)-idx-1)) )
            #print(cur, firstksum, idx+1, whole_array_sum - firstksum, len(nums)-idx-1)
            if cur < current_ans[1]:
                current_ans = [idx, cur]
        return current_ans[0]

玉山 Dec 5, 2022

Javascript





















function minimumAverageDifference(nums) {
  let leftSum = 0;
  let rightSum = 0;
  for (let i = 0; i < nums.length; i++) {
    rightSum += nums[i];
  }

  let minimum = Infinity;
  let index = 0;
  for (let i = 0; i < nums.length; i++) {
    leftSum += nums[i];
    rightSum -= nums[i];
    const miniAvgDiff = Math.abs(~~(leftSum / (i + 1)) - ~~(rightSum / Math.max(1, nums.length - 1 - i)));
    if (miniAvgDiff < minimum) {
      minimum = miniAvgDiff;
      index = i;
    }
  }

  return index;
}

Marsgoat Dec 5, 2022

Reference

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