Problem Description

You have an integer array nums with indices starting from 0. For each index i in the array, you need to calculate the average difference.

The average difference at index i is calculated as follows:

• Take the average of the first i + 1 elements (elements from index 0 to i)
• Take the average of the last n - i - 1 elements (elements from index i+1 to the end)
• Find the absolute difference between these two averages
• Both averages should be rounded down to the nearest integer (integer division)

Special cases to note:

• If there are no elements in the second group (when i is the last index), the average of 0 elements is considered to be 0
• The absolute difference means you take the positive value of the difference

Your task is to find the index that has the minimum average difference. If multiple indices have the same minimum average difference, return the smallest index.

For example, if nums = [2,5,3,9,5,3]:

• At index 0: average of [2] is 2, average of [5,3,9,5,3] is 5, difference = |2-5| = 3
• At index 1: average of [2,5] is 3, average of [3,9,5,3] is 5, difference = |3-5| = 2
• At index 2: average of [2,5,3] is 3, average of [9,5,3] is 5, difference = |3-5| = 2
• And so on...

The solution iterates through each index, maintains running sums for the left part (pre) and right part (suf), calculates the averages using integer division, and keeps track of the index with the minimum difference seen so far.

Intuition

The key insight is that we need to calculate averages for every possible split point in the array. A naive approach would be to recalculate the sum of elements for each split, but this would be inefficient with O(n²) time complexity.

Instead, we can observe that as we move from one index to the next, we're essentially moving one element from the "right group" to the "left group". This means:

• The sum of the left part increases by the current element
• The sum of the right part decreases by the same element

This observation leads us to use a sliding window technique with prefix and suffix sums. We can maintain:

• pre: the running sum of elements we've included in the left part
• suf: the running sum of elements remaining in the right part

Initially, pre = 0 (no elements in the left part) and suf = sum(nums) (all elements in the right part).

As we iterate through each index i:

1. Add nums[i] to pre (include current element in left part)
2. Subtract nums[i] from suf (remove current element from right part)
3. Calculate the average of left part: pre // (i + 1)
4. Calculate the average of right part: suf // (n - i - 1) if there are elements left, otherwise 0
5. Find the absolute difference between these averages

By maintaining these running sums, we avoid recalculating sums from scratch at each index, reducing our time complexity to O(n). We simply keep track of the minimum difference found and its corresponding index, updating whenever we find a smaller difference.

Learn more about Prefix Sum patterns.

Solution Approach

We implement the solution using a single-pass traversal with running sums:

Initialization:

• pre = 0: Tracks the sum of elements in the left part (initially empty)
• suf = sum(nums): Tracks the sum of elements in the right part (initially contains all elements)
• n = len(nums): Store the array length
• ans = 0: Stores the index with minimum average difference
• mi = inf: Tracks the minimum average difference found so far

Main Loop: We iterate through the array using enumerate to get both index i and value x:

1. Update running sums:

   • pre += x: Add current element to left sum
   • suf -= x: Remove current element from right sum

2. Calculate averages:

   • Left average: a = pre // (i + 1)
     • We have i + 1 elements in the left part (indices 0 to i)
   • Right average: b = 0 if n - i - 1 == 0 else suf // (n - i - 1)
     • We have n - i - 1 elements in the right part
     • Special case: when i = n - 1 (last index), there are no elements in the right part, so average is 0

3. Update minimum:

   • Calculate absolute difference: t = abs(a - b)
   • If t < mi, update both:
     • ans = i (new minimum index)
     • mi = t (new minimum difference)

Return Result: After checking all indices, return ans which holds the index with the minimum average difference.

The algorithm achieves O(n) time complexity with a single pass through the array and O(1) extra space (not counting the input array).

Example Walkthrough

Let's trace through the algorithm with nums = [4, 2, 0]:

Initial Setup:

• n = 3 (length of array)
• pre = 0 (left sum starts empty)
• suf = 4 + 2 + 0 = 6 (right sum has all elements)
• ans = 0, mi = infinity

Iteration 1 (i = 0, x = 4):

• Update sums: pre = 0 + 4 = 4, suf = 6 - 4 = 2
• Calculate averages:
  • Left: a = 4 // 1 = 4 (average of [4])
  • Right: b = 2 // 2 = 1 (average of [2, 0])
• Difference: t = |4 - 1| = 3
• Since 3 < infinity: Update ans = 0, mi = 3

Iteration 2 (i = 1, x = 2):

• Update sums: pre = 4 + 2 = 6, suf = 2 - 2 = 0
• Calculate averages:
  • Left: a = 6 // 2 = 3 (average of [4, 2])
  • Right: b = 0 // 1 = 0 (average of [0])
• Difference: t = |3 - 0| = 3
• Since 3 = 3 (not less than mi): No update

Iteration 3 (i = 2, x = 0):

• Update sums: pre = 6 + 0 = 6, suf = 0 - 0 = 0
• Calculate averages:
  • Left: a = 6 // 3 = 2 (average of [4, 2, 0])
  • Right: b = 0 (no elements remain, so average is 0)
• Difference: t = |2 - 0| = 2
• Since 2 < 3: Update ans = 2, mi = 2

Result: Return ans = 2

The index 2 has the minimum average difference of 2. Notice how we efficiently maintained running sums instead of recalculating from scratch at each step, and handled the edge case where the right part becomes empty.

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the length of the array nums. This is because:

• The initial sum(nums) operation takes O(n) time
• The main loop iterates through each element exactly once, performing O(1) operations in each iteration (addition, subtraction, division, and comparison operations)
• Therefore, the overall time complexity is O(n) + O(n) = O(n)

The space complexity is O(1). This is because:

• The algorithm only uses a fixed number of variables (pre, suf, n, ans, mi, i, x, a, b, t) regardless of the input size
• No additional data structures that grow with input size are created
• All operations are performed in-place using constant extra space

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Integer Overflow with Large Arrays

Pitfall: When calculating the initial suffix_sum = sum(nums), if the array contains many large numbers, the sum might exceed the integer limits in some programming languages (though Python handles big integers automatically).

Solution: In languages with fixed integer sizes, use appropriate data types (e.g., long long in C++) or consider calculating sums incrementally if overflow is a concern.

2. Division by Zero Error

Pitfall: Forgetting to handle the special case when calculating the right average for the last index. When i = n - 1, there are no elements in the right part, leading to division by zero if not handled properly.

Solution: Always check if n - i - 1 == 0 before performing the division:

if n - i - 1 == 0:
    right_average = 0
else:
    right_average = suffix_sum // (n - i - 1)

3. Using Float Division Instead of Integer Division

Pitfall: Using regular division (/) instead of integer division (//) will produce floating-point results, which doesn't match the problem's requirement to round down to the nearest integer.

Wrong:

left_average = prefix_sum / (i + 1)  # Returns float

Correct:

left_average = prefix_sum // (i + 1)  # Returns integer (rounded down)

4. Off-by-One Errors in Element Counting

Pitfall: Incorrectly calculating the number of elements in each part. The left part has i + 1 elements (not i), and the right part has n - i - 1 elements (not n - i).

Solution: Remember that:

• Left part includes indices [0, i] → count = i + 1
• Right part includes indices [i+1, n-1] → count = n - i - 1

5. Not Returning the Smallest Index for Ties

Pitfall: Using <= instead of < when comparing differences, which would return the last index with minimum difference rather than the first.

Wrong:

if current_difference <= min_difference:  # Returns last index with minimum
    result_index = i
    min_difference = current_difference

Correct:

if current_difference < min_difference:  # Returns first index with minimum
    result_index = i
    min_difference = current_difference