Minimum Absolute Difference Between Elements With Constraint Interview Question for Google

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.

Example 1:

Input: nums = [4,3,2,4], x = 2
Output: 0
Explanation: We can select nums[0] = 4 and nums[3] = 4.
They are at least 2 indices apart, and their absolute difference is the minimum, 0.
It can be shown that 0 is the optimal answer.

Example 2:

Input: nums = [5,3,2,10,15], x = 1
Output: 1
Explanation: We can select nums[1] = 3 and nums[2] = 2.
They are at least 1 index apart, and their absolute difference is the minimum, 1.
It can be shown that 1 is the optimal answer.

Example 3:

Input: nums = [1,2,3,4], x = 3
Output: 3
Explanation: We can select nums[0] = 1 and nums[3] = 4.
They are at least 3 indices apart, and their absolute difference is the minimum, 3.
It can be shown that 3 is the optimal answer.

Constraints:

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

Brute Force 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, we calculate the absolute difference between the elements and update the minimum difference found so far.

Code (Python)

def min_abs_difference_brute_force(nums, x):
    n = len(nums)
    min_diff = float('inf')

    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

Time Complexity

The time complexity of this brute-force approach is O(n^2), where n is the length of the nums array. This is because we have nested loops that iterate through all possible pairs of elements.

Space Complexity

The space complexity is O(1) because we are only using a constant amount of extra space to store the minimum difference.

Optimal Solution Using Binary Search and Sorted List

To improve the time complexity, we can use a sorted list to keep track of the elements within the range [i+x, n-1]. For each element nums[i], we want to find the element in the sorted list that is closest to nums[i]. We can efficiently find this closest element using binary search.

Algorithm

Initialize min_diff to infinity.
Iterate through the array from i = 0 to n - x - 1.
For each i, maintain a sorted list of nums[j] for j >= i + x.
Use binary search in the sorted list to find the element closest to nums[i].
Update min_diff with the minimum absolute difference found.

Code (Python)

import bisect

def min_abs_difference_optimal(nums, x):
    n = len(nums)
    min_diff = float('inf')
    sorted_list = []

    for i in range(n - x):
        # Ensure sorted_list is populated for binary search
        if i == 0:
            for j in range(x, n):
                bisect.insort(sorted_list, nums[j])

        # Find closest element to nums[i] in sorted_list
        idx = bisect.bisect_left(sorted_list, nums[i])

        if idx > 0:
            min_diff = min(min_diff, abs(nums[i] - sorted_list[idx - 1]))
        if idx < len(sorted_list):
            min_diff = min(min_diff, abs(nums[i] - sorted_list[idx]))

        # Remove nums[i+x] as it is outside the valid range for the next iteration
        sorted_list.remove(nums[i + x])

    return min_diff

Time Complexity

The time complexity of this approach is O(n log n), where n is the length of the nums array. The bisect.insort and bisect.bisect_left operations each take O(log n) time, and we perform these operations n times in the loop. Removing an element from list also take O(n).

Space Complexity

The space complexity is O(n), where n is the length of the nums array. This is because we are storing elements in a sorted list, which can contain at most n - x elements, but in the worst case it is O(n).

Edge Cases

If x is 0, any two elements can be compared.
If x is greater than or equal to the length of the array, there are no valid pairs.
If all the numbers are the same, the minimum difference will be 0.