Leetcode #2256: Minimum Average Difference
In this guide, we solve Leetcode #2256 Minimum Average Difference in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Array, Prefix Sum
Intuition
Range queries become simple once we precompute cumulative sums.
We can transform subarray conditions into prefix comparisons.
Approach
Compute prefix sums and use a map to find matching prefixes.
This avoids nested loops while keeping the logic clear.
Steps:
- Compute prefix sums.
- Use a map to find valid ranges.
- Update the answer.
Example
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.
Python Solution
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
Complexity
The time complexity is , where is the length of the array . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.