What is the time complexity of Minimum Absolute Difference?

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

What data structures are used to solve Minimum Absolute Difference?

This problem uses Array, Sorting. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Minimum Absolute Difference?

**Sorting as a Prerequisite for Localized Search:** Sorting the input array `arr` is the foundational step. It transforms the problem from a potentially quadratic search space (any `arr[i]` vs. any `arr[j]`) into a linear one, where we only need to examine adjacent elements. This pre-processing makes the subsequent comparison phase incredibly efficient. **Minimum Difference Resides Between Neighbors:** A key mathematical insight here is that for distinct numbers in a sorted array, the minimum absolute difference between any two elements will always be found by comparing an element `arr[i]` with its immediate predecessor `arr[i-1]`. This allows us to sweep through the array once, checking only `N-1` differences, rather than `N*(N-1)/2` differences, after sorting.

Which companies ask LeetCode 1200. Minimum Absolute Difference in interviews?

This problem is frequently asked at Agoda, Audible, IBM, J.P. Morgan, PayPal and 1 more companies.

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Minimum Absolute Difference - LeetCode 1200 Solution

Q: How to solve Minimum Absolute Difference?

The provided solution leverages a fundamental property of sorted data to dramatically simplify the problem. Its first, crucial step is to sort the input array `arr`. This is a game-changer because, in a sorted sequence, the smallest difference between any two distinct numbers *must* always occur between adjacent elements. Consider three distinct numbers `x < y < z`; it's clear that `z - x` will always be greater than both `y - x` and `z - y`. This insight eliminates the need for an expensive `O(N^2)` brute-force comparison of every possible pair.

Minimum Absolute Difference - Complete Solution Guide

Minimum Absolute Difference is LeetCode problem 1200, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given an array of distinct integers arr , find all pairs of elements with the minimum absolute difference of any two elements. Return a list of pairs in ascending order(with respect to pairs), each pair [a, b] follows a, b are from arr a < b b - a equals to the minimum absolute difference of any two elements in arr Example 1: Input: arr = [4,2,1,3] Output: [[1,2],[2,3],[3,4]] Explanation: The minimum absolute difference is 1. List all pairs with difference equal to 1 in ascending order. Example

Detailed Explanation

This problem challenges us to pinpoint the absolute smallest difference possible between any two distinct numbers within a given array, `arr`. Once that global minimum difference is identified, our task isn't just to report *a* pair exhibiting it, but to meticulously collect *all* such pairs. Each pair `[a, b]` must satisfy `a < b`, and their difference `b - a` must precisely equal this minimum. The final output is expected as a list of these pairs, ordered ascendingly based on the pairs themselves (e.g., `[1,2]` before `[2,3]`).

Solution Approach

The provided solution leverages a fundamental property of sorted data to dramatically simplify the problem. Its first, crucial step is to sort the input array `arr`. This is a game-changer because, in a sorted sequence, the smallest difference between any two distinct numbers *must* always occur between adjacent elements. Consider three distinct numbers `x < y < z`; it's clear that `z - x` will always be greater than both `y - x` and `z - y`. This insight eliminates the need for an expensive `O(N^2)` brute-force comparison of every possible pair.

Step-by-Step Algorithm

Sort the input array `arr` in ascending order.
Initialize `min_diff` to a large value (infinity in Python, Integer.MAX_VALUE in Java, INT_MAX in C++). Initialize an empty result list `result`.
Iterate through the sorted array, starting from the second element. For each element, calculate the difference with the preceding element: `diff = arr[i] - arr[i-1]`.
If `diff` is less than `min_diff`, update `min_diff` to `diff`, and reset `result` to contain only the current pair `[arr[i-1], arr[i]]`.
If `diff` is equal to `min_diff`, append the current pair `[arr[i-1], arr[i]]` to `result`.
After iterating through the entire array, return the `result` list.

Key Insights

**Sorting as a Prerequisite for Localized Search:** Sorting the input array `arr` is the foundational step. It transforms the problem from a potentially quadratic search space (any `arr[i]` vs. any `arr[j]`) into a linear one, where we only need to examine adjacent elements. This pre-processing makes the subsequent comparison phase incredibly efficient.
**Minimum Difference Resides Between Neighbors:** A key mathematical insight here is that for distinct numbers in a sorted array, the minimum absolute difference between any two elements will always be found by comparing an element `arr[i]` with its immediate predecessor `arr[i-1]`. This allows us to sweep through the array once, checking only `N-1` differences, rather than `N*(N-1)/2` differences, after sorting.
**Single-Pass Dynamic Collection:** The solution efficiently combines finding the minimum difference with collecting all corresponding pairs in a single pass. It doesn't require a separate second pass after determining the `min_diff`. Instead, it maintains `min_diff` dynamically; if a smaller difference is found, the `result` list is reset. If an equal difference is found, the pair is simply appended. This strategy is elegant, avoids redundant work, and naturally produces the pairs in the required ascending order due to the initial sort.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(n)

Topics

This problem involves: Array, Sorting.

Companies

Asked at: Agoda, Audible, IBM, J.P. Morgan, PayPal, Paycom.

Minimum Absolute Difference | LeetCode 1200

Easy

Array Sorting

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

Solution Code

Available in:

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

class Solution {
    public List<List<Integer>> minimumAbsDifference(int[] arr) {
        Arrays.sort(arr);
        int minDiff = Integer.MAX_VALUE;
        for (int i = 1; i < arr.length; i++) {
            minDiff = Math.min(minDiff, arr[i] - arr[i - 1]);
        }

        List<List<Integer>> result = new ArrayList<>();
        for (int i = 1; i < arr.length; i++) {
            if (arr[i] - arr[i - 1] == minDiff) {
                List<Integer> pair = new ArrayList<>();
                pair.add(arr[i - 1]);
                pair.add(arr[i]);
                result.add(pair);
            }
        }

        return result;
    }
}

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Problem Info

Description

Problem 1200: Minimum Absolute Difference - Detailed explanation and solution approaches.

Topics

Array Sorting

Companies

AgodaAudibleIBMJ.P. MorganPayPalPaycom

Stats

Difficulty:Easy

Languages:4

Companies:6

Problem #:1200

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

6 steps

1
Sort the input array `arr` in ascending order.
2
Initialize `min_diff` to a large value (infinity in Python, Integer.MAX_VALUE in Java, INT_MAX in C++). Initialize an empty result list `result`.
3
Iterate through the sorted array, starting from the second element. For each element, calculate the difference with the preceding element: `diff = arr[i] - arr[i-1]`.
4
If `diff` is less than `min_diff`, update `min_diff` to `diff`, and reset `result` to contain only the current pair `[arr[i-1], arr[i]]`.
5
If `diff` is equal to `min_diff`, append the current pair `[arr[i-1], arr[i]]` to `result`.
6
After iterating through the entire array, return the `result` list.

💡

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 dominant operation is the sorting of the array, which takes O(n log n) time using efficient algorithms like merge sort or quicksort. The subsequent iteration through the sorted array takes O(n) time. The space complexity is O(n) due to the potential need to store all pairs in the result list in the worst case (when all pairs have the minimum difference).

📖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

Using a HashMap (for smaller ranges)

If the range of numbers is relatively small, you could use a HashMap to count the occurrences of each number. Then iterate through the possible differences and check if you have at least two numbers with that difference. This might be slightly faster for certain input distributions but would have higher space complexity.

Time: O(n + k) where k is the range of numbers.Space: O(k)

Trade-offs: This approach might be more efficient if the range of numbers is smaller than the number of elements. However, it's less efficient for larger ranges of numbers.

✨

Key Insights

1**Sorting as a Prerequisite for Localized Search:** Sorting the input array `arr` is the foundational step. It transforms the problem from a potentially quadratic search space (any `arr[i]` vs. any `arr[j]`) into a linear one, where we only need to examine adjacent elements. This pre-processing makes the subsequent comparison phase incredibly efficient.
2**Minimum Difference Resides Between Neighbors:** A key mathematical insight here is that for distinct numbers in a sorted array, the minimum absolute difference between any two elements will always be found by comparing an element `arr[i]` with its immediate predecessor `arr[i-1]`. This allows us to sweep through the array once, checking only `N-1` differences, rather than `N*(N-1)/2` differences, after sorting.
3**Single-Pass Dynamic Collection:** The solution efficiently combines finding the minimum difference with collecting all corresponding pairs in a single pass. It doesn't require a separate second pass after determining the `min_diff`. Instead, it maintains `min_diff` dynamically; if a smaller difference is found, the `result` list is reset. If an equal difference is found, the pair is simply appended. This strategy is elegant, avoids redundant work, and naturally produces the pairs in the required ascending order due to the initial sort.

📚

Key Terms Explained

For Students

Array

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

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Empty array: The provided solutions handle this implicitly; an empty array would result in an empty output list.
All elements are the same: This case would result in an empty output list as the minimum difference would be 0, and no pairs (a < b) would be found.

⚠️

Common Mistakes to Avoid

✗Forgetting to sort the array: Without sorting, finding the minimum absolute difference efficiently is not possible.

✗Not handling the case where multiple pairs have the minimum difference: The solution needs to correctly store all pairs that satisfy the minimum difference condition.

✗Inefficiently calculating absolute differences: Repeatedly calculating absolute differences within the loop should be avoided; it's already implicit when working with a sorted array.

✅

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?

💼Finding closest data points: This algorithm can be adapted to find the closest pairs of data points in a dataset (e.g., finding similar customer profiles based on features).

🏦Signal processing: Identifying closely spaced frequencies or signals in a spectrum analysis.

💡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

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ Watch Out

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

👨‍💻 From Experience

This is one of the first problems I solved when learning Array. It's a great foundation for more complex problems.

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 29, 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

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 13

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Minimum Absolute Difference"?

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

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