What is the time complexity of Minimum Absolute Difference Between Elements With Constraint?

The optimal solution for Minimum Absolute Difference Between Elements With Constraint runs in O(n log n) time and uses O(n) space. The complexity arises from maintaining a sorted list of elements, where each insertion and search operation takes logarithmic time.

What is the brute force solution for Minimum Absolute Difference Between Elements With Constraint?

The brute force approach for Minimum Absolute Difference Between Elements With Constraint is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every possible pair of elements in the array to find the minimum absolute difference. This is straightforward but inefficient for large arrays. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Minimum Absolute Difference Between Elements With Constraint?

Minimum Absolute Difference Between Elements With Constraint uses the following concepts: Array, Binary Search, Ordered Set. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Minimum Absolute Difference Between Elements With Constraint?

Always clarify the constraints before jumping into coding. Think about how data structures can help optimize your solution. Dry run your algorithm with sample inputs to ensure it works as expected.

What are common mistakes when solving Minimum Absolute Difference Between Elements With Constraint?

Not considering the index constraint properly, leading to incorrect pairs being checked. Overlooking the need for efficient searching when the array size is large.

#2817

Minimum Absolute Difference Between Elements With Constraint

Medium

ArrayBinary SearchOrdered SetHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

💡

Intuition

Time O(n log n)Space O(n)

Using a sorted data structure allows us to efficiently find the closest values to each element while maintaining the index constraint. This reduces the time complexity significantly.

⚙️

Algorithm

4 steps

1Step 1: Initialize a sorted list to keep track of elements seen so far.
2Step 2: Iterate through the array, and for each element nums[j], check the sorted list for the closest elements.
3Step 3: For each nums[j], calculate the absolute difference with the closest elements in the sorted list, ensuring they are at least x indices apart.
4Step 4: Add nums[j] to the sorted list for future comparisons.

solution.py16 lines

1# Full working Python code
2from sortedcontainers import SortedList
3
4def minAbsDifference(nums, x):
5    sorted_list = SortedList()
6    min_diff = float('inf')
7    for j in range(len(nums)):
8        if j >= x:
9            sorted_list.add(nums[j - x])
10        if sorted_list:
11            pos = sorted_list.bisect_left(nums[j])
12            if pos < len(sorted_list):
13                min_diff = min(min_diff, abs(sorted_list[pos] - nums[j]))
14            if pos > 0:
15                min_diff = min(min_diff, abs(sorted_list[pos - 1] - nums[j]))
16    return min_diff

ℹ

Complexity note: The complexity arises from maintaining a sorted list of elements, where each insertion and search operation takes logarithmic time.

1The problem requires careful consideration of index constraints.
2Using sorted data structures can significantly reduce the time complexity.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.