LeetCode每日一题(2256. Minimum Average Difference)

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

---

比较简单的问题， 只要想到 prefix sum， 答案就出来了， 注意除以 0 的情况就好了

---

impl Solution {
    pub fn minimum_average_difference(nums: Vec<i32>) -> i32 {
        let prefix_sum: Vec<i64> = nums
            .into_iter()
            .scan(0, |s, v| {
                *s += v as i64;
                Some(*s)
            })
            .collect();
        let mut min = i64::MAX;
        let mut min_index = 0;
        for i in 0..prefix_sum.len() {
            let first = prefix_sum[i] / (i + 1) as i64;
            let second = if i != prefix_sum.len() - 1 {
                (*prefix_sum.last().unwrap() - prefix_sum[i]) / (prefix_sum.len() - i - 1) as i64
            } else {
                0
            }
            .abs();
            let diff = (first - second).abs();
            if diff < min {
                min = diff;
                min_index = i;
            }
        }
        min_index as i32
    }
}