# Solving Leetcode Interviews in Seconds with AI: Minimum Average Difference


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2256" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# 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  

	# Explanation
	Here's a breakdown of the solution:

*   **Calculate Prefix Sum and Total Sum:** Efficiently compute the prefix sum of `nums` and the total sum of all elements. This avoids repeated calculations of sums within the loop.
*   **Iterate and Calculate Average Differences:** Iterate through the `nums` array. In each iteration, calculate the average of the prefix (first `i+1` elements) and the suffix (last `n-i-1` elements) using the precomputed sums. Compute the absolute difference between the averages.
*   **Track Minimum Average Difference:** Maintain variables to keep track of the minimum average difference encountered so far and the corresponding index. Update these variables whenever a smaller average difference is found.

*   **Runtime Complexity:** O(n), **Storage Complexity:** O(1)

	
	# Code
	```python
	def minimumAverageDifference(nums):
    n = len(nums)
    total_sum = sum(nums)
    prefix_sum = 0
    min_avg_diff = float('inf')
    min_index = 0

    for i in range(n):
        prefix_sum += nums[i]
        prefix_avg = prefix_sum // (i + 1)

        if i == n - 1:
            suffix_avg = 0
        else:
            suffix_avg = (total_sum - prefix_sum) // (n - i - 1)

        avg_diff = abs(prefix_avg - suffix_avg)

        if avg_diff < min_avg_diff:
            min_avg_diff = avg_diff
            min_index = i

    return min_index
	```