LeetCode-in-Java

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:

Example 1:

Input: nums = [2,5,3,9,5,3]

Output: 3

Explanation:

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:

Solution

public class Solution {
    public int minimumAverageDifference(int[] nums) {
        long numsSum = 0;
        for (int num : nums) {
            numsSum += num;
        }
        long minAverageDiff = Long.MAX_VALUE;
        long sumFromFront = 0;
        int index = 0;
        for (int i = 0; i < nums.length; i++) {
            sumFromFront += nums[i];
            int numbersRight = i == nums.length - 1 ? 1 : nums.length - i - 1;
            long averageDiff =
                    Math.abs(sumFromFront / (i + 1) - (numsSum - sumFromFront) / numbersRight);
            if (minAverageDiff > averageDiff) {
                minAverageDiff = averageDiff;
                index = i;
            }
            if (averageDiff == 0) {
                break;
            }
        }
        return index;
    }
}