Problem Description

This problem deals with an array of integers where the task is to find the index that minimizes the average difference between the two segments of the array divided by that index. The array is split at every index from 0 to n - 1, creating two non-overlapping segments. The left segment includes all elements from the beginning of the array to the current index, and the right segment includes all elements after the current index to the end of the array.

The average difference at an index i is the absolute difference between the average of the numbers to the left of (and including) i and the average of the numbers to the right of i. The averages are calculated as the sum of the elements in the segment divided by the number of elements in the segment, with the result rounded down to the nearest integer. If the right segment is empty (which happens when i is the last index), its average is considered to be 0.

The problem asks to determine the index that has the minimum average difference and to return the smallest index if there are multiple such indices.

Intuition

To solve this problem, one intuitive approach is to process each index and calculate the two averages (left and right segment averages) and their absolute difference. However, computing the sum of elements repeatedly for each index would result in a higher time complexity.

A more efficient method involves prefix sums, which can significantly optimize the repeated sum calculations. By maintaining a running total sum from the start of the array (pre) and a running total sum from the end of the array (suf), it becomes easier to compute the averages quickly at each index. As the index moves from left to right, elements are transferred from the right segment to the left segment. This is done by adding the current element to pre and subtracting it from suf.

The solution tracks the smallest average difference found (mi) and its corresponding index (ans). For each index i, it calculates the average of the left segment, the average of the right segment (except when the right segment is empty), and their absolute difference (t). If t is smaller than the smallest difference seen so far, mi is updated with t and ans with i. After traversing all indices, the index corresponding to the minimal average difference is returned.

This approach efficiently narrows down the index with the minimum average difference without calculating the sum for each segment more than once, resulting in an overall time complexity of O(n), where n is the length of the array.

Learn more about Prefix Sum patterns.

Solution Approach

The implementation of the solution uses a simple loop through the array and exploits the concept of prefix sums to efficiently calculate the averages required for the problem statement. The algorithm doesn't require any complex data structures or patterns beyond arrays and basic arithmetic operations.

Here’s the step-by-step implementation of the solution:

1. Initialize two variables pre and suf. Set pre to 0 because, initially, there are no elements in the left segment. Set suf to the sum of all elements in nums - this represents the sum of the right segment before any iteration.

2. Set up variables ans for storing the index of the minimum average difference found, and mi for storing the minimum average difference itself. mi is initialized to inf, which is a placeholder for infinity, to ensure that any actual difference calculated will be less than this initial value.

3. Iterate over the elements in nums. For each index i and its corresponding value x in nums, do the following:

   • Add x to pre, as x is now part of the left segment.
   • Subtract x from suf, as x is no longer part of the right segment.
   • Calculate the average of the left segment as a = pre // (i + 1).
   • If the right segment is non-empty (determined by n - i - 1 > 0), calculate its average as b = suf // (n - i - 1). If the right segment is empty, set b to 0.
   • Calculate the absolute difference between the two averages t = abs(a - b).

4. If the absolute difference t is less than the current minimum mi, update mi with t and ans with the current index i.

5. After the loop completes, the variable ans holds the index of the minimum average difference. Return ans as the result.

The algorithm solely uses integer division and basic arithmetic operations, which creates an efficient way of obtaining the minimum average difference without recalculating the sums for each segment for every index. The time complexity is O(n) because the algorithm requires a single pass through the array, and the space complexity is O(1) as it only uses a fixed number of extra variables.

Example Walkthrough

Let's consider the array nums = [3, 1, 2, 4, 3] to illustrate the solution approach.

1. Start by initializing pre to 0 and suf to the sum of all elements in nums, which is 3 + 1 + 2 + 4 + 3 = 13. Also, initialize mi to infinity, and ans to -1 (which will later hold the index of the minimum average difference).

2. Now, iteratively process each element of the array:

   • For index i = 0:

     • Left Segment [3] and Right Segment [1, 2, 4, 3]
     • pre = 0 + 3 = 3
     • suf = 13 - 3 = 10
     • Left average a = 3 // 1 = 3
     • Right average b = 10 // 4 = 2
     • Absolute difference t = abs(3 - 2) = 1
     • Since t < mi (1 < infinity), update mi to 1 and ans to 0.

   • For index i = 1:

     • Left Segment [3, 1] and Right Segment [2, 4, 3]
     • pre = 3 + 1 = 4
     • suf = 10 - 1 = 9
     • Left average a = 4 // 2 = 2
     • Right average b = 9 // 3 = 3
     • Absolute difference t = abs(2 - 3) = 1
     • Since t = mi (1 == 1), no update to mi or ans as we pick the smallest index.

   • For index i = 2:

     • Left Segment [3, 1, 2] and Right Segment [4, 3]
     • pre = 4 + 2 = 6
     • suf = 9 - 2 = 7
     • Left average a = 6 // 3 = 2
     • Right average b = 7 // 2 = 3
     • Absolute difference t = abs(2 - 3) = 1
     • Since t = mi (1 == 1), no update as ans is already at the smallest index 0.

   • For index i = 3:

     • Left Segment [3, 1, 2, 4] and Right Segment [3]
     • pre = 6 + 4 = 10
     • suf = 7 - 4 = 3
     • Left average a = 10 // 4 = 2
     • Right average b = 3 // 1 = 3
     • Absolute difference t = abs(2 - 3) = 1
     • Because t = mi (1 == 1), there's no update needed.

   • For index i = 4:

     • Left Segment [3, 1, 2, 4, 3] and no Right Segment
     • pre = 10 + 3 = 13
     • suf = 3 - 3 = 0
     • Left average a = 13 // 5 = 2
     • Right average b = 0 (since there are no elements on the right)
     • Absolute difference t = abs(2 - 0) = 2
     • No update required as t > mi (2 > 1).

3. After processing all indices, the smallest average difference mi is 1, and the ans associated with this difference is at index 0.

4. The result returned is ans = 0, which is the index of the minimum average difference.

In this example, the process was performed in a single pass (O(n) time complexity), and we did not require extra space aside from a few variables (O(1) space complexity). This walkthrough demonstrates the efficiency and correctness of the proposed solution.

Solution Implementation

Time and Space Complexity

The time complexity of the provided code is O(n), where n is the length of the array nums. The calculation of the sum of the array sum(nums) is done in O(n). Inside the loop, only constant time operations are done, and since the loop runs n times, the loop operations also amount to O(n). Thus, the entire function runs in linear time with respect to the size of the array.

The space complexity of the code is O(1). Independent of the length of the input array, a fixed number of variables are used (pre, suf, n, ans, mi, a, b, and t). These require a constant amount of space, and hence the space complexity does not scale with the input size.

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