Minimum average difference

Array Prefix Sum

43.8% acceptance

Feb 25, 2026

1569

182

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.

Solution

Rust

Time O(n)

Space O(1)

LeetCode

solution.rs

impl Solution {
  pub fn minimum_average_difference(nums: Vec<i32>) -> i32 {
    let n = nums.len() as i64;
    let total: i64 = nums.iter().map(|&x| x as i64).sum();
    let mut prefix = 0i64;
    let mut min_diff = i64::MAX;
    let mut ans = 0usize;
    for i in 0..nums.len() {
      prefix += nums[i] as i64;
      let left_avg = prefix / (i as i64 + 1);
      let right_cnt = n - i as i64 - 1;
      let right_avg = if right_cnt == 0 { 0 } else { (total - prefix) / right_cnt };
      let diff = (left_avg - right_avg).abs();
      if diff < min_diff {
        min_diff = diff;
        ans = i;
      }
    }
    ans as i32
  }
}