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

The optimal solution has a time complexity of O(n log n).

What data structures are used to solve Minimum Absolute Difference Between Elements With Constraint?

This problem uses Array, Binary Search, Ordered Set. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Minimum Absolute Difference Between Elements With Constraint?

Insight 1: The brute-force approach of checking all pairs (i, j) with |i - j| >= x is inefficient (O(n^2)), especially with the constraint of n <= 10^5. Insight 2: The key is to use a sorted data structure (like SortedList, TreeSet, or multiset) to efficiently find the closest number to `nums[j]` among the numbers `nums[j-x], nums[j-x-1], ... nums[0]`.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Minimum Absolute Difference Between Elements With Constraint - LeetCode 2817 Solution

Q: Which companies ask LeetCode 2817. Minimum Absolute Difference Between Elements With Constraint in interviews?

This problem is frequently asked at Capital One, Databricks, Roblox.

Minimum Absolute Difference Between Elements With Constraint - Complete Solution Guide

Minimum Absolute Difference Between Elements With Constraint is LeetCode problem 2817, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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

Detailed Explanation

The problem asks us to find the minimum absolute difference between any two numbers in a given array `nums` that are at least `x` indices apart. In other words, we need to iterate through all possible pairs of numbers `nums[i]` and `nums[j]` such that `abs(i - j) >= x`, and find the smallest value of `abs(nums[i] - nums[j])`. The input consists of an array `nums` of integers and an integer `x` representing the minimum index difference. The output is an integer representing the minimum absolute difference found.

Solution Approach

The solution uses a sliding window approach with a sorted data structure to efficiently find the minimum absolute difference. The algorithm iterates through the array starting from index `x`. For each element `nums[j]` (where j >= x), it adds `nums[j-x]` to the sorted set. Then, it finds the elements in the sorted set that are closest to `nums[j]` using binary search or similar efficient methods provided by the sorted data structure. The absolute difference between `nums[j]` and its closest neighbors in the sorted set are computed, and the minimum absolute difference is updated accordingly.

Step-by-Step Algorithm

Step 1: Initialize `min_diff` to infinity and create an empty sorted set (e.g., SortedList in Python, TreeSet in Java, or multiset in C++).
Step 2: Iterate through the array `nums` from index `x` to `n-1` (inclusive).
Step 3: In each iteration (at index `j`), add the element `nums[j-x]` to the sorted set. This ensures that the sorted set contains elements at least `x` indices apart from `nums[j]` in the original array.
Step 4: Find the element in the sorted set that is closest to `nums[j]`. This can be done using binary search or similar methods provided by the sorted set.
Step 5: Calculate the absolute difference between `nums[j]` and its closest neighbors (floor and ceiling in TreeSet terminology, or lower_bound and its predecessor in multiset terminology) in the sorted set. Update `min_diff` if a smaller absolute difference is found.
Step 6: If `min_diff` becomes 0, return 0 immediately, as this is the smallest possible absolute difference.
Step 7: After iterating through all the elements, return the final value of `min_diff`.

Key Insights

Insight 1: The brute-force approach of checking all pairs (i, j) with |i - j| >= x is inefficient (O(n^2)), especially with the constraint of n <= 10^5.
Insight 2: The key is to use a sorted data structure (like SortedList, TreeSet, or multiset) to efficiently find the closest number to `nums[j]` among the numbers `nums[j-x], nums[j-x-1], ... nums[0]`.
Insight 3: By maintaining a sorted set of elements within the `x` distance, we can use binary search to efficiently find the elements closest to the current element `nums[j]` and minimize the absolute difference calculation.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(n)

Topics

This problem involves: Array, Binary Search, Ordered Set.

Companies

Asked at: Capital One, Databricks, Roblox.

Minimum Absolute Difference Between Elements With Constraint | LeetCode 2817

Medium

Array Binary Search Ordered Set

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

Solution Code

Available in:

import java.util.List;
import java.util.TreeSet;

class Solution {
    public int minAbsoluteDifference(List<Integer> nums, int x) {
        if (x == 0) {
            return 0;
        }

        int n = nums.size();
        int min_diff = Integer.MAX_VALUE;

        TreeSet<Integer> sorted_past_elements = new TreeSet<>();

        for (int j = x; j < n; j++) {
            sorted_past_elements.add(nums.get(j - x));

            int current_element = nums.get(j);

            Integer ceiling = sorted_past_elements.ceiling(current_element);
            Integer floor = sorted_past_elements.floor(current_element);

            if (ceiling != null) {
                min_diff = Math.min(min_diff, ceiling - current_element);
            }

            if (floor != null) {
                min_diff = Math.min(min_diff, current_element - floor);
            }

            if (min_diff == 0) {
                return 0;
            }
        }

        return min_diff;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 2817: Minimum Absolute Difference Between Elements With Constraint - Detailed explanation and solution approaches.

Topics

Array Binary Search Ordered Set

Companies

Capital OneDatabricksRoblox

Stats

Difficulty:Medium

Languages:4

Companies:3

Problem #:2817

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

7 steps

1
Step 1: Initialize `min_diff` to infinity and create an empty sorted set (e.g., SortedList in Python, TreeSet in Java, or multiset in C++).
2
Step 2: Iterate through the array `nums` from index `x` to `n-1` (inclusive).
3
Step 3: In each iteration (at index `j`), add the element `nums[j-x]` to the sorted set. This ensures that the sorted set contains elements at least `x` indices apart from `nums[j]` in the original array.
4
Step 4: Find the element in the sorted set that is closest to `nums[j]`. This can be done using binary search or similar methods provided by the sorted set.
5
Step 5: Calculate the absolute difference between `nums[j]` and its closest neighbors (floor and ceiling in TreeSet terminology, or lower_bound and its predecessor in multiset terminology) in the sorted set. Update `min_diff` if a smaller absolute difference is found.
6
Step 6: If `min_diff` becomes 0, return 0 immediately, as this is the smallest possible absolute difference.
7
Step 7: After iterating through all the elements, return the final value of `min_diff`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n log n)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is O(n log n) because we iterate through the array once (O(n)), and for each element, we perform an insertion and a search operation (binary search) in the sorted set, both of which take O(log n) time. The space complexity is O(n) because, in the worst case, the sorted set can contain up to n-x elements.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Brute Force

Iterate through all possible pairs (i, j) with abs(i - j) >= x and calculate the absolute difference between nums[i] and nums[j]. Find the minimum among all such differences.

Time: O(n^2)Space: O(1)

Trade-offs: Simpler to implement but much slower than the sorted set approach for large input sizes. Not suitable for the given constraints.

✨

Key Insights

1Insight 1: The brute-force approach of checking all pairs (i, j) with |i - j| >= x is inefficient (O(n^2)), especially with the constraint of n <= 10^5.
2Insight 2: The key is to use a sorted data structure (like SortedList, TreeSet, or multiset) to efficiently find the closest number to `nums[j]` among the numbers `nums[j-x], nums[j-x-1], ... nums[0]`.
3Insight 3: By maintaining a sorted set of elements within the `x` distance, we can use binary search to efficiently find the elements closest to the current element `nums[j]` and minimize the absolute difference calculation.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Edge Cases

Edge case 1: x = 0. The problem states that if x=0, the function should return 0. The code handles this by immediately returning 0 if x is 0.
Edge case 2: Empty array or very small array. The given constraints ensure that nums.length >= 1 and x < nums.length. So, there is no need to handle empty array cases
Edge case 3: All elements in the array are the same. The algorithm correctly finds the minimum absolute difference as 0 in this scenario.
Edge case 4: Integer overflow. The problem constraints state 1 <= nums[i] <= 10^9. Therefore, the absolute difference between two numbers can be up to 10^9 which fits in an int. Thus, integer overflow isn't a significant concern in most languages. If the upper bound for nums[i] was higher, then we might need to use `long` or `long long` to prevent overflow.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not using a sorted data structure. Using an unsorted data structure and searching for the closest element linearly within the window would lead to O(n^2) time complexity.

✗Common mistake 2: Incorrectly implementing the binary search or the sorted set operations, leading to incorrect results or increased time complexity.

✗Common mistake 3: Not handling the edge case of x = 0 correctly.

✗Common mistake 4: Not considering elements both greater and smaller than the current number when finding the closest number. The ceiling *and* floor (or lower_bound and its predecessor) are needed to find the closest numbers.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Financial data analysis: Finding the minimum price difference between assets traded at least a certain time interval apart.

🏦Real-world scenario 2: Signal processing: Identifying the smallest variation in signal amplitude within a specified time window.

🎮Real-world scenario 3: Recommendation systems: Finding items with the closest feature values (e.g., price, rating) separated by a minimum interaction time in a user's purchase history.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: September 26, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

Can binary search be applied to non-sorted arrays?

Binary search requires a monotonic property, not necessarily sorted data. It can be applied to any search space where you can determine if the answer is in the left or right half. Examples include searching rotated arrays or finding minimum in optimization problems.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

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

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether binary search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?