LeetCode #2256 — MEDIUM

Minimum Average Difference

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

[2,5,3,9,5,3]

Example 2

[0]

Core Insight

What unlocks the optimal approach

How can we use precalculation to efficiently calculate the average difference at an index?
Create a prefix and/or suffix sum array.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2256: Minimum Average Difference
class Solution {
    public int minimumAverageDifference(int[] nums) {
        int n = nums.length;
        long pre = 0, suf = 0;
        for (int x : nums) {
            suf += x;
        }
        int ans = 0;
        long mi = Long.MAX_VALUE;
        for (int i = 0; i < n; ++i) {
            pre += nums[i];
            suf -= nums[i];
            long a = pre / (i + 1);
            long b = n - i - 1 == 0 ? 0 : suf / (n - i - 1);
            long t = Math.abs(a - b);
            if (t < mi) {
                ans = i;
                mi = t;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2256: Minimum Average Difference
func minimumAverageDifference(nums []int) (ans int) {
	n := len(nums)
	pre, suf := 0, 0
	for _, x := range nums {
		suf += x
	}
	mi := suf
	for i, x := range nums {
		pre += x
		suf -= x
		a := pre / (i + 1)
		b := 0
		if n-i-1 != 0 {
			b = suf / (n - i - 1)
		}
		if t := abs(a - b); t < mi {
			ans = i
			mi = t
		}
	}
	return
}

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

# Accepted solution for LeetCode #2256: Minimum Average Difference
class Solution:
    def minimumAverageDifference(self, nums: List[int]) -> int:
        pre, suf = 0, sum(nums)
        n = len(nums)
        ans, mi = 0, inf
        for i, x in enumerate(nums):
            pre += x
            suf -= x
            a = pre // (i + 1)
            b = 0 if n - i - 1 == 0 else suf // (n - i - 1)
            if (t := abs(a - b)) < mi:
                ans = i
                mi = t
        return ans

// Accepted solution for LeetCode #2256: Minimum Average Difference
/**
 * [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 <= 10^5
 * 	0 <= nums[i] <= 10^5
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/minimum-average-difference/
// discuss: https://leetcode.com/problems/minimum-average-difference/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    pub fn minimum_average_difference(nums: Vec<i32>) -> i32 {
        0
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    #[ignore]
    fn test_2256_example_1() {
        let nums = vec![2, 5, 3, 9, 5, 3];

        let result = 3;

        assert_eq!(Solution::minimum_average_difference(nums), result);
    }

    #[test]
    #[ignore]
    fn test_2256_example_2() {
        let nums = vec![2, 5, 3, 9, 5, 3];

        let result = 0;

        assert_eq!(Solution::minimum_average_difference(nums), result);
    }
}

// Accepted solution for LeetCode #2256: Minimum Average Difference
function minimumAverageDifference(nums: number[]): number {
    const n = nums.length;
    let pre = 0;
    let suf = nums.reduce((a, b) => a + b);
    let ans = 0;
    let mi = suf;
    for (let i = 0; i < n; ++i) {
        pre += nums[i];
        suf -= nums[i];
        const a = Math.floor(pre / (i + 1));
        const b = n - i - 1 === 0 ? 0 : Math.floor(suf / (n - i - 1));
        const t = Math.abs(a - b);
        if (t < mi) {
            ans = i;
            mi = t;
        }
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.