Minimum Absolute Difference Between Elements With Constraint Python/Java solution

In Python, a sorted list with the bisect module can maintain ordered elements and perform binary search for neighbors. In Java, TreeSet provides built-in methods like ceiling() and floor() to quickly locate the closest values. Both implementations run in O(n log n) time.

What is the best approach for Minimum Absolute Difference Between Elements With Constraint?

The most efficient approach uses a sliding window with an ordered set or sorted list. While iterating through the array, insert elements that are at least x indices behind the current index. Use binary search to find the closest values to the current number in O(log n). This produces an overall time complexity of O(n log n) with O(n) space.

Is Minimum Absolute Difference Between Elements With Constraint asked at Google/Amazon/Meta?

Problems combining ordered sets, binary search, and sliding window techniques commonly appear in interviews at companies like Google, Amazon, and Meta. This question tests the ability to maintain a dynamically sorted structure while enforcing index constraints.

What data structure is used in Minimum Absolute Difference Between Elements With Constraint?

The optimal solution uses an ordered set or balanced binary search tree. Examples include TreeSet in Java, multiset in C++, or a sorted list with bisect in Python. These structures allow efficient insertion and nearest-value lookup using binary search.

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

The brute force method runs in O(n²) time because it compares all valid pairs that satisfy the index constraint. The optimized ordered-set approach reduces the complexity to O(n log n) by maintaining a sorted structure and using binary search to find the closest values efficiently.

Description Hints 5 Solution Video 6:43

2817. Minimum Absolute Difference Between Elements With Constraint

Q: Is Minimum Absolute Difference Between Elements With Constraint easy or hard?

The problem is rated Medium on LeetCode because the brute force idea is simple but inefficient. The challenge is recognizing that a dynamically maintained sorted structure enables fast nearest-value queries while respecting the index distance constraint.

Q: How to solve Minimum Absolute Difference Between Elements With Constraint in O(n)?

A strict O(n) solution is generally not feasible because you must maintain ordering to find the closest numeric value efficiently. Most optimal implementations rely on a balanced tree, TreeSet, or sorted container with binary search, resulting in O(n log n) time.

Medium38.0% acceptance8 companies

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

Asked at 8 companies

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Topics

Hints open one at a time

Hint 1

Let's only consider the cases where i < j, as the problem is symmetric.

Hint 2

For an index j, we are interested in an index i in the range [0, j - x] that minimizes abs(nums[i] - nums[j]).

Hint 3

For every index j, while going from left to right, add nums[j - x] to a set (C++ set, Java TreeSet, and Python sorted set).

Hint 4

After inserting nums[j - x], we can calculate the closest value to nums[j] in the set using binary search and store the absolute difference. In C++, we can achieve this by using lower_bound and/or upper_bound.

Hint 5

Calculate the minimum absolute difference among all indices.

6:43

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