You are given a 0-indexed integer array nums and an integer x.

Find the minimum absolute difference between two elements in the array that are at least x indices apart.

In other words, find two indices i and j such that abs(i - j) >= x and abs(nums[i] - nums[j]) is minimized.

Return* an integer denoting the minimum absolute difference between two elements that are at least* x indices apart.

For example:

nums = [4,3,2,4], x = 2 should return 0.

nums = [5,3,2,10,15], x = 1 should return 1.

nums = [1,2,3,4], x = 3 should return 3.

Constraints:

• 1 <= nums.length <= 10^5
• 1 <= nums[i] <= 10^9
• 0 <= x < nums.length

Naive Solution

The most straightforward approach is to iterate through all possible pairs of elements in the array and check if their indices are at least x apart. If they are, calculate the absolute difference between the elements and update the minimum difference found so far.

def min_abs_difference_naive(nums, x):
    min_diff = float('inf')
    n = len(nums)
    for i in range(n):
        for j in range(n):
            if abs(i - j) >= x:
                min_diff = min(min_diff, abs(nums[i] - nums[j]))
    return min_diff

Optimal Solution

To optimize, we can use a sliding window approach combined with a sorted data structure (like a sorted list or a binary search tree) to efficiently find the minimum absolute difference. We maintain a window of elements that are at least x indices behind the current element. For each element, we search for the closest elements in the sorted window to find the minimum absolute difference.

import bisect

def min_abs_difference_optimal(nums, x):
    min_diff = float('inf')
    n = len(nums)
    sorted_window = []
    
    for i in range(x, n):
        sorted_window.append(nums[i-x])
        sorted_window.sort()
        
        # Find the element closest to nums[i] in the sorted window
        
        left = 0
        right = len(sorted_window)-1
        while left <= right:
            mid = (left+right)//2
            min_diff = min(min_diff, abs(nums[i] - sorted_window[mid]))
            if nums[i] > sorted_window[mid]:
                left = mid+1
            else:
                right = mid-1
        
    return min_diff

Big(O) Run-time Analysis

Naive Solution:

• The naive solution has nested loops, each iterating through the array of size n. Therefore, the run-time complexity is O(n^2).

Optimal Solution:

• The optimal solution iterates through the array once (O(n)). Inside the loop, we insert into and sort the sorted_window. The sort() is O(k log k) where k is the size of the sorted_window. However, the max size of the sorted_window is x, so it's O(x log x). Therefore, the overall runtime is O(n * x log x)

Big(O) Space Usage Analysis

Naive Solution:

• The naive solution uses a constant amount of extra space. Therefore, the space complexity is O(1).

Optimal Solution:

• The optimal solution uses a sorted_window which can have up to x elements in it. Therefore, the space complexity is O(x).

Edge Cases

1. Empty Array: If the input array nums is empty, the function should return float('inf') or raise an exception, depending on the desired behavior.
2. x = 0: If x is 0, then any two elements can be compared, and we should return the minimum absolute difference among all pairs.
3. x >= len(nums): If x is greater than or equal to the length of nums, then no two elements can satisfy the distance condition, and the function should return float('inf') (or an error).
4. All elements are the same: In this case, the minimum absolute difference will be 0.
5. Large input array: The optimal solution is significantly faster for large arrays because of its O(n log n) time complexity, compared to the naive approach's O(n^2) time complexity. But if n is small, the difference is negligible.

Here's how we can handle these edge cases in the optimal solution:

import bisect

def min_abs_difference_optimal(nums, x):
    if not nums:
        return float('inf')

    n = len(nums)
    if x >= n:
        return float('inf')

    min_diff = float('inf')
    sorted_window = []
    
    for i in range(x, n):
        sorted_window.append(nums[i-x])
        sorted_window.sort()
        
        # Find the element closest to nums[i] in the sorted window
        
        left = 0
        right = len(sorted_window)-1
        while left <= right:
            mid = (left+right)//2
            min_diff = min(min_diff, abs(nums[i] - sorted_window[mid]))
            if nums[i] > sorted_window[mid]:
                left = mid+1
            else:
                right = mid-1
        
    return min_diff