What is the time complexity of Count Number of Pairs With Absolute Difference K?

The optimal solution has a time complexity of O(n^2).

What data structures are used to solve Count Number of Pairs With Absolute Difference K?

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

What are the key insights for solving Count Number of Pairs With Absolute Difference K?

Insight 1: Brute-force approach is feasible due to the relatively small constraint on array size (<=200). Insight 2: Nested loops are a straightforward way to examine all possible pairs of indices.

Which companies ask LeetCode 2006. Count Number of Pairs With Absolute Difference K in interviews?

This problem is frequently asked at Expedia, Goldman Sachs, Oracle.

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Count Number of Pairs With Absolute Difference K - LeetCode 2006 Solution

Q: How to solve Count Number of Pairs With Absolute Difference K?

The provided solution uses a brute-force approach. It iterates through all possible pairs of indices (i, j) in the input array `nums` where i < j. For each pair, it calculates the absolute difference between the elements at those indices using `abs(nums[i] - nums[j])`. If this difference is equal to `k`, the counter `count` is incremented. Finally, the function returns the value of `count`.

Q: Which companies ask LeetCode 2006. Count Number of Pairs With Absolute Difference K in interviews?

This problem is frequently asked at Expedia, Goldman Sachs, Oracle.

Count Number of Pairs With Absolute Difference K - Complete Solution Guide

Count Number of Pairs With Absolute Difference K is LeetCode problem 2006, 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 integer array nums and an integer k , return the number of pairs (i, j) where i < j such that |nums[i] - nums[j]| == k . The value of |x| is defined as: x if x >= 0 . -x if x < 0 . Example 1: Input: nums = [1,2,2,1], k = 1 Output: 4 Explanation: The pairs with an absolute difference of 1 are: - [ 1 , 2 ,2,1] - [ 1 ,2, 2 ,1] - [1, 2 ,2, 1 ] - [1,2, 2 , 1 ] Example 2: Input: nums = [1,3], k = 3 Output: 0 Explanation: There are no pairs with an absolute difference of 3. Example 3: Input: n

Detailed Explanation

The problem asks you to count the number of pairs of indices (i, j) in an integer array `nums` where i < j and the absolute difference between the elements at those indices, |nums[i] - nums[j]|, is equal to a given integer `k`. The input consists of an integer array `nums` and an integer `k`. The output is a single integer representing the count of such pairs. Constraints limit the size of the array and the range of values within the array.

Solution Approach

The provided solution uses a brute-force approach. It iterates through all possible pairs of indices (i, j) in the input array `nums` where i < j. For each pair, it calculates the absolute difference between the elements at those indices using `abs(nums[i] - nums[j])`. If this difference is equal to `k`, the counter `count` is incremented. Finally, the function returns the value of `count`.

Step-by-Step Algorithm

Step 1: Initialize a counter variable `count` to 0.
Step 2: Iterate through the array `nums` using nested loops. The outer loop iterates from i = 0 to nums.length - 1, and the inner loop iterates from j = i + 1 to nums.length.
Step 3: For each pair (i, j), calculate the absolute difference between `nums[i]` and `nums[j]` using `abs(nums[i] - nums[j])`.
Step 4: If the absolute difference is equal to `k`, increment the `count`.
Step 5: After iterating through all pairs, return the final value of `count`.

Key Insights

Insight 1: Brute-force approach is feasible due to the relatively small constraint on array size (<=200).
Insight 2: Nested loops are a straightforward way to examine all possible pairs of indices.
Insight 3: The `abs()` function is crucial for handling positive and negative differences.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(1)

Topics

This problem involves: Array, Hash Table, Counting.

Companies

Asked at: Expedia, Goldman Sachs, Oracle.

Count Number of Pairs With Absolute Difference K | LeetCode 2006

Easy

Array Hash Table Counting

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

Solution Code

Available in:

class Solution {
    public int countKDifference(int[] nums, int k) {
        int count = 0;
        for (int i = 0; i < nums.length; i++) {
            for (int j = i + 1; j < nums.length; j++) {
                if (Math.abs(nums[i] - nums[j]) == k) {
                    count++;
                }
            }
        }
        return count;
    }
}

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Problem Info

Description

Problem 2006: Count Number of Pairs With Absolute Difference K - Detailed explanation and solution approaches.

Topics

Array Hash Table Counting

Companies

ExpediaGoldman SachsOracle

Stats

Difficulty:Easy

Languages:4

Companies:3

Problem #:2006

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

5 steps

1
Step 1: Initialize a counter variable `count` to 0.
2
Step 2: Iterate through the array `nums` using nested loops. The outer loop iterates from i = 0 to nums.length - 1, and the inner loop iterates from j = i + 1 to nums.length.
3
Step 3: For each pair (i, j), calculate the absolute difference between `nums[i]` and `nums[j]` using `abs(nums[i] - nums[j])`.
4
Step 4: If the absolute difference is equal to `k`, increment the `count`.
5
Step 5: After iterating through all pairs, return the final value of `count`.

💡

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^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The nested loops iterate through approximately n(n-1)/2 pairs, resulting in a time complexity of O(n^2). The algorithm uses a constant amount of extra space to store the counter variable, resulting in a space complexity of O(1).

📖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

Hash Table Approach

Use a hash table (dictionary in Python) to store the frequency of each number in `nums`. Then, iterate through the unique numbers. For each number `num`, check if `num + k` or `num - k` exists in the hash table. The count can be calculated efficiently by multiplying the frequencies.

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

Trade-offs: This approach is more efficient in terms of time complexity (O(n) vs O(n^2)) but uses more space to store the hash table. The space tradeoff is worth it for larger input arrays.

✨

Key Insights

1Insight 1: Brute-force approach is feasible due to the relatively small constraint on array size (<=200).
2Insight 2: Nested loops are a straightforward way to examine all possible pairs of indices.
3Insight 3: The `abs()` function is crucial for handling positive and negative differences.

📚

Key Terms Explained

For Students

Array

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

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Edge Cases

Edge case 1: Empty array (`nums` is empty). The solution correctly handles this by not entering either loop, resulting in a count of 0.
Edge case 2: `k` is 0. The solution correctly identifies pairs with identical values.
Edge case 3: No pairs satisfy the condition. The solution correctly returns 0 in this case.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting the `abs()` function, leading to incorrect results for negative differences.

✗Common mistake 2: Incorrectly handling the loop indices (e.g., starting the inner loop from i instead of i + 1), resulting in duplicate pairs and incorrect counts.

✗Common mistake 3: Off-by-one errors in the loop bounds.

✅

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: Finding pairs of data points with a specific difference in datasets, such as finding pairs of stock prices with a specific price gap.

🏦Real-world scenario 2: Identifying similar items based on a certain metric. For example, detecting near-duplicate products in an e-commerce catalog based on their price difference.

💡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

Don't rush. Explain your thought process clearly. Easy problems are often about communication.

⚠️ 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: June 27, 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

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

@AlgoMasterJul 13

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

@InterviewProJul 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.

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

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 "Count Number of Pairs With Absolute Difference K"?

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 hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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