LeetCode #2817 — MEDIUM

Minimum Absolute Difference Between Elements With Constraint

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode
The Problem

Problem Statement

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
Patterns Used
🔍Modified Binary Search→📐Segment Tree→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Binary Search · Segment Tree

Example 1

[4,3,2,4]
2

Example 2

[5,3,2,10,15]
1

Example 3

[1,2,3,4]
3

Related Problems

  • K-diff Pairs in an Array (k-diff-pairs-in-an-array)
  • Find All K-Distant Indices in an Array (find-all-k-distant-indices-in-an-array)
  • Find Indices With Index and Value Difference I (find-indices-with-index-and-value-difference-i)
  • Find Indices With Index and Value Difference II (find-indices-with-index-and-value-difference-ii)
Step 02

Core Insight

What unlocks the optimal approach

  • <div class="_1l1MA">Let's only consider the cases where <code>i < j</code>, as the problem is symmetric.</div>
  • <div class="_1l1MA">For an index <code>j</code>, we are interested in an index <code>i</code> in the range <code>[0, j - x]</code> that minimizes <code>abs(nums[i] - nums[j])</code>.</div>
  • <div class="_1l1MA">For every index <code>j</code>, while going from left to right, add <code>nums[j - x]</code> to a set (C++ set, Java TreeSet, and Python sorted set).</div>
  • <div class="_1l1MA">After inserting <code>nums[j - x]</code>, we can calculate the closest value to <code>nums[j]</code> 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.</div>
  • <div class="_1l1MA">Calculate the minimum absolute difference among all indices.</div>
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Upper-end input sizes
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2817: Minimum Absolute Difference Between Elements With Constraint
class Solution {
    public int minAbsoluteDifference(List<Integer> nums, int x) {
        TreeMap<Integer, Integer> tm = new TreeMap<>();
        int ans = 1 << 30;
        for (int i = x; i < nums.size(); ++i) {
            tm.merge(nums.get(i - x), 1, Integer::sum);
            Integer key = tm.ceilingKey(nums.get(i));
            if (key != null) {
                ans = Math.min(ans, key - nums.get(i));
            }
            key = tm.floorKey(nums.get(i));
            if (key != null) {
                ans = Math.min(ans, nums.get(i) - key);
            }
        }
        return ans;
    }
}
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n × log n)
Space
O(n)

Approach Breakdown

LINEAR SCAN
O(n) time
O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH
O(log n) time
O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.

Related Problems
#315Count of Smaller Numbers After Selfhard→#327Count of Range Sumhard→#363Max Sum of Rectangle No Larger Than Khard→#456132 Patternmedium→#493Reverse Pairshard→