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

Minimum Absolute Difference Between Elements With Constraint

Medium

Google

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

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 bounds on the length of the `nums` array and the value of `x`?
Can the elements in the `nums` array be negative, zero, or floating-point numbers?
If no two elements satisfy the constraint `|i - j| >= x`, what value should I return?
Are duplicate values allowed in the `nums` array, and if so, how should they be handled?
Is x guaranteed to be non-negative and less than the length of the input array?

Brute Force Solution

Approach

The goal is to find two numbers in a list that are at least a certain distance apart, and have the smallest possible difference. The brute force approach involves checking every single pair of numbers that meets the distance requirement.

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

Consider the first number in the list.
Compare it with every other number in the list that is far enough away based on the required distance.
Calculate the difference between that first number and all the numbers that meet the distance requirement.
Keep track of the smallest difference found so far.
Now, do the same thing with the second number in the list. Compare it with all other numbers that are far enough away.
Again, calculate the difference between the second number and those meeting the distance criteria and update the smallest difference if necessary.
Repeat this process for every number in the list.
After checking all possible pairs, the smallest difference that you kept track of is the answer.

Code Implementation

def min_absolute_difference(numbers, index_distance):
    minimum_difference = float('inf')

    for first_index in range(len(numbers)): 
        # Iterate through each number in the list

        for second_index in range(len(numbers)): 
            # Compare with every other number.

            if abs(first_index - second_index) >= index_distance: 
                # Ensure indices are far enough apart.

                current_difference = abs(numbers[first_index] - numbers[second_index])

                if current_difference < minimum_difference:
                    minimum_difference = current_difference

    return minimum_difference

Big(O) Analysis

Time Complexity

O(n²) – The provided solution iterates through the array of size n. For each element, it compares it with other elements satisfying the distance constraint. In the worst case, for each of the n elements, it may need to compare with almost all other elements (close to n). Therefore, the total number of operations is approximately n * n, which means the time complexity is O(n²).

Space Complexity

O(1) – The provided algorithm's space complexity is determined by the auxiliary space used beyond the input list. The description mentions tracking only the smallest difference found so far, which can be stored in a single variable. No additional data structures like lists, hash maps, or recursion are employed. Therefore, the space used remains constant, independent of the input list size N.

Optimal Solution

Approach

The goal is to find two numbers in a list that are at least a certain distance apart and have the smallest possible difference. The trick is to first organize the list and then efficiently look for the best pairs using this sorted order.

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

First, arrange the numbers in the list from smallest to largest.
Start with the smallest number in the list.
Move through the list, but only consider numbers that are far enough away (based on the given distance constraint).
Calculate the difference between the starting number and each valid number you find that is far enough away.
Keep track of the smallest difference you find.
Now, repeat this process starting with the next number in the list, but continue where you left off, remembering the distance constraint.
By moving in order, you avoid checking the same pairs twice and can quickly find the minimum difference without looking at every single possible combination.

Code Implementation

def min_absolute_difference_with_constraint(numbers, distance):
    numbers.sort()
    minimum_difference = float('inf')
    list_length = len(numbers)

    for i in range(list_length):
        for j in range(i + 1, list_length):
            # Ensure elements are at least 'distance' indices apart.
            if abs(i - j) >= distance:

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

                # Update minimum difference if a smaller difference is found.
                if current_difference < minimum_difference:

                    minimum_difference = current_difference

    #If min_diff remains unchanged, no valid pair was found
    if minimum_difference == float('inf'):
        return -1
    else:
        return minimum_difference

Big(O) Analysis

Time Complexity

O(n log n) – The dominant cost comes from sorting the input array of n elements, which takes O(n log n) time. The subsequent nested loop iterates through the sorted array. The outer loop runs up to n times, and the inner loop, while seeming nested, considers each element no more than once in the context of finding the minimum difference with the distance constraint. Thus, the inner loop's contribution is bounded by O(n). Consequently, the time complexity for iterating and finding the minimum difference is O(n). Therefore, the overall time complexity is dominated by the initial sorting step, resulting in O(n log n).

Space Complexity

O(1) – The algorithm sorts the input list in-place, which doesn't require extra memory proportional to the input size. The algorithm uses a constant number of variables like 'min_difference' and indices to keep track of the current elements being compared. Therefore, the auxiliary space used does not depend on the input size N, and remains constant.

Edge Cases

Case	How to Handle
nums is null or empty	Return positive infinity or a large sentinel value indicating no valid difference exists since no elements are present.
nums has only one element	Return positive infinity or a large sentinel value, as no pairs satisfying the distance constraint can be formed.
x is 0	Return the minimum absolute difference between any two elements in the array.
x is greater than or equal to the length of nums	Return positive infinity or a large sentinel value since no pairs can satisfy the distance constraint.
nums contains all identical values	Return 0, as the absolute difference between any two elements will be 0.
nums contains very large or very small (negative) numbers leading to potential integer overflow when calculating the absolute difference.	Use long data type to avoid integer overflow when computing the absolute difference.
nums contains duplicate values	The algorithm should work correctly as it iterates through all possible pairs satisfying the index constraint regardless of duplicated values.
Large input array size	Ensure the algorithm used has a time complexity that scales well, preferably O(n log n) or better, to avoid timeouts.