Minimum Absolute Difference Between Elements With Constraint

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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 <= 10⁵
1 <= nums[i] <= 10⁹
0 <= x < nums.length

Try coding it in LeetCode

Minimum Absolute Difference Between Elements With Constraint

Medium

Databricks

6 views

Topics:

ArraysBinary Search

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 <= 10⁵
1 <= nums[i] <= 10⁹
0 <= x < nums.length

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the possible ranges for values in the `nums` array and for `x`?
Can the input array `nums` be empty or contain null values? What should I return in that case?
If no two elements are at least `x` positions apart, what value should I return?
Is the array `nums` sorted or unsorted?
Are there any memory constraints that I should be aware of?

Brute Force Solution

Approach

The brute force method involves examining every possible pairing of elements within a given collection, ensuring a specified distance separates them. We calculate the absolute difference for each valid pair. The minimum difference found across all these pairs is our solution.

Here's how the algorithm would work step-by-step:

Consider every possible combination of two numbers from the given collection.
For each pair of numbers, verify if they are far enough apart based on the constraint (i.e., if their positions satisfy the minimum distance requirement).
If the numbers are far enough apart, calculate the absolute difference between them.
Keep track of the smallest absolute difference encountered so far.
Once we've checked all possible pairs, the smallest absolute difference we tracked is the answer.

Code Implementation

def minimum_absolute_difference_brute_force(
    numbers, index_distance_constraint
):
    minimum_absolute_difference = float('inf')

    # Iterate through all possible pairs of numbers
    for first_index in range(len(numbers)):
        for second_index in range(len(numbers)):

            # Ensure we're not comparing the same element to itself
            if first_index != second_index:

                # Check if indices satisfy the distance constraint
                if (
                    abs(first_index - second_index)
                    >= index_distance_constraint
                ):

                    # Calculate the absolute difference
                    absolute_difference = abs(
                        numbers[first_index] - numbers[second_index]
                    )

                    # Update the minimum difference if needed
                    minimum_absolute_difference = min(
                        minimum_absolute_difference,
                        absolute_difference,
                    )

    # Return the minimum absolute difference found
    if minimum_absolute_difference == float('inf'):
        return -1
    else:
        return minimum_absolute_difference

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through all possible pairs of elements in the input collection of size n. For each element, the algorithm checks its absolute difference with every other element to see if the distance constraint is met. This process inherently involves checking approximately n * (n-1) / 2 pairs. Therefore, the number of operations grows proportionally to n squared, making the time complexity O(n²).

Space Complexity

O(1) – The provided brute force algorithm iterates through pairs of numbers from the input collection. It only uses a few variables: to store the indices of the pair being considered, and another to keep track of the minimum absolute difference found so far. Since the number of extra variables does not depend on the input size N (the number of elements in the collection), the auxiliary space complexity is constant.

Optimal Solution

Approach

The goal is to find two numbers within the list that are at least a certain distance apart, such that the difference between them is as small as possible. We can achieve this efficiently by first sorting the list and then looking at nearby elements, instead of comparing every pair.

Here's how the algorithm would work step-by-step:

First, arrange the numbers in ascending order from smallest to largest. This helps bring potential candidates closer together.
Then, consider each number in the ordered list, one at a time.
For each number, look ahead to the first number that is far enough away (meeting the distance requirement).
Calculate the difference between these two numbers.
Keep track of the smallest difference you've found so far.
Continue this process until you've checked all possible pairs that meet the distance requirement.
The smallest difference you kept track of is the answer.

Code Implementation

def find_min_absolute_difference(
    numbers, index_difference
):
    numbers.sort()

    minimum_difference = float('inf')

    for i in range(len(numbers)): 
        # Iterate to second to last valid index
        if i + index_difference >= len(numbers):
            break

        # Find the element at the minimum index difference
        j = i + index_difference

        current_difference = abs(numbers[i] - numbers[j])

        # Keep track of the smallest difference found so far.
        minimum_difference = min(minimum_difference, current_difference)

    return minimum_difference

Big(O) Analysis

Time Complexity

O(n log n) – The dominant operation in this algorithm is sorting the input list of size n, which takes O(n log n) time using an efficient sorting algorithm like merge sort or quicksort. After sorting, the algorithm iterates through the sorted list once, performing a linear scan to find the minimum absolute difference satisfying the distance constraint. This linear scan takes O(n) time. Since O(n log n) grows faster than O(n), the overall time complexity is determined by the sorting step, making the algorithm's time complexity O(n log n).

Space Complexity

O(N) – The dominant space complexity comes from sorting the input list of N numbers. Most sorting algorithms, like merge sort or quicksort (in worst case), require O(N) auxiliary space for temporary storage during the sorting process. The remaining operations, such as calculating differences and tracking the minimum, use a constant amount of extra space. Thus, the overall space complexity is determined by the sorting step, resulting in O(N).

Edge Cases

Case	How to Handle
Empty or null input array	Return infinity or a sufficiently large value, indicating no valid difference was found.
x is 0	Return 0 if the array contains any duplicate values, or the minimum absolute difference if all values are unique
x is greater than or equal to the length of the array	Return infinity or a sufficiently large value, as no two elements can be x positions apart.
Input array contains only one element	Return infinity or a sufficiently large value, as a difference requires at least two elements.
Array contains large positive and negative numbers	Use long or a data type with sufficient range to avoid integer overflow during difference calculation.
Array contains all identical values	If x > 0, return infinity; else if x == 0, return 0.
Large input array with many possible pairs	Optimize the search using efficient data structures (e.g., balanced trees) or sorting to reduce time complexity.
All numbers in the array are zero	If x > 0 return infinity else if x is zero, return 0.