What is the time complexity of K-diff Pairs in an Array?

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

What data structures are used to solve K-diff Pairs in an Array?

This problem uses Array, Hash Table, Two Pointers, Binary Search, Sorting. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving K-diff Pairs in an Array?

Insight 1: The absolute difference `|nums[i] - nums[j]| == k` implies that `nums[i] - nums[j] == k` or `nums[j] - nums[i] == k`. Therefore, we are looking for pairs where one number is `k` more or `k` less than the other. Insight 2: A hash table (or Counter in Python) is an efficient data structure to count the frequency of each number in the array. This helps in identifying and counting unique pairs.

Which companies ask LeetCode 532. K-diff Pairs in an Array in interviews?

This problem is frequently asked at Capital One, tcs.

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K-diff Pairs in an Array - LeetCode 532 Solution

Q: How to solve K-diff Pairs in an Array?

The solution uses a hash table (or Counter in Python) to store the counts of each number in the input array. Then, it iterates through the hash table. If k is 0, it counts the number of elements with a count greater than 1 (meaning there are pairs of the same element). If k is greater than 0, it checks if the number plus k exists in the hash table and if so, increments the count.

K-diff Pairs in an Array - Complete Solution Guide

K-diff Pairs in an Array is LeetCode problem 532, 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

Given an array of integers nums and an integer k , return the number of unique k-diff pairs in the array . A k-diff pair is an integer pair (nums[i], nums[j]) , where the following are true: 0 <= i, j < nums.length i != j |nums[i] - nums[j]| == k Notice that |val| denotes the absolute value of val . Example 1: Input: nums = [3,1,4,1,5], k = 2 Output: 2 Explanation: There are two 2-diff pairs in the array, (1, 3) and (3, 5). Although we have two 1s in the input, we should only return the number o

Detailed Explanation

The problem asks us to find the number of unique k-diff pairs in a given array `nums`. A k-diff pair `(nums[i], nums[j])` satisfies the following conditions: `0 <= i, j < nums.length`, `i != j`, and `|nums[i] - nums[j]| == k`. We need to count how many such pairs exist, ensuring that we only count unique pairs, even if numbers appear multiple times in the input array. The input is an array of integers `nums` and an integer `k` (the difference). The output is the number of unique k-diff pairs.

Solution Approach

The solution uses a hash table (or Counter in Python) to store the counts of each number in the input array. Then, it iterates through the hash table. If k is 0, it counts the number of elements with a count greater than 1 (meaning there are pairs of the same element). If k is greater than 0, it checks if the number plus k exists in the hash table and if so, increments the count.

Step-by-Step Algorithm

Step 1: Initialize a hash table (or Counter) to store the frequency of each number in the input array `nums`.
Step 2: Handle the edge case where `k` is negative. If `k < 0`, return 0 because the absolute difference cannot be negative.
Step 3: Check if `k` is equal to 0. If it is, iterate through the hash table's values (counts). For each value, if the count is greater than 1, increment the `pair_count`.
Step 4: If `k` is not equal to 0, iterate through the keys (numbers) in the hash table. For each number, check if `number + k` exists as a key in the hash table. If it does, increment the `pair_count`.
Step 5: Return the final `pair_count`.

Key Insights

Insight 1: The absolute difference `|nums[i] - nums[j]| == k` implies that `nums[i] - nums[j] == k` or `nums[j] - nums[i] == k`. Therefore, we are looking for pairs where one number is `k` more or `k` less than the other.
Insight 2: A hash table (or Counter in Python) is an efficient data structure to count the frequency of each number in the array. This helps in identifying and counting unique pairs.
Insight 3: When k=0, we are looking for pairs of identical numbers. In this special case, we only need to count numbers that appear more than once in the array. When k>0, we need to check if `num + k` exists in the array.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Hash Table, Two Pointers, Binary Search, Sorting.

Companies

Asked at: Capital One, tcs.

K-diff Pairs in an Array | LeetCode 532

Medium

Array Hash Table Two Pointers Binary Search Sorting

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

Solution Code

Available in:

import java.util.HashMap;
import java.util.Map;

class Solution {
    public int findPairs(int[] nums, int k) {
        if (k < 0) {
            return 0;
        }

        Map<Integer, Integer> counts = new HashMap<>();
        for (int num : nums) {
            counts.put(num, counts.getOrDefault(num, 0) + 1);
        }

        int pair_count = 0;

        if (k == 0) {
            for (int count : counts.values()) {
                if (count > 1) {
                    pair_count += 1;
                }
            }
        } else {
            for (int num : counts.keySet()) {
                if (counts.containsKey(num + k)) {
                    pair_count += 1;
                }
            }
        }

        return pair_count;
    }
}

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Problem Info

Description

Problem 532: K-diff Pairs in an Array - Detailed explanation and solution approaches.

Topics

Array Hash Table Two Pointers Binary Search Sorting

Companies

Capital Onetcs

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:532

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 hash table (or Counter) to store the frequency of each number in the input array `nums`.
2
Step 2: Handle the edge case where `k` is negative. If `k < 0`, return 0 because the absolute difference cannot be negative.
3
Step 3: Check if `k` is equal to 0. If it is, iterate through the hash table's values (counts). For each value, if the count is greater than 1, increment the `pair_count`.
4
Step 4: If `k` is not equal to 0, iterate through the keys (numbers) in the hash table. For each number, check if `number + k` exists as a key in the hash table. If it does, increment the `pair_count`.
5
Step 5: Return the final `pair_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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is determined by the initial population of the hashmap and then the linear scan of the keys. Space complexity is determined by the maximum size of the hashmap, which is bounded by the number of distinct elements in the input array.

📖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

Sorting and Two Pointers

Sort the input array and then use two pointers to find pairs that satisfy the condition `|nums[i] - nums[j]| == k`. To avoid counting duplicate pairs, skip duplicate numbers in the sorted array.

Time: O(n log n) due to sorting. The two pointer scan is O(n).Space: O(1) if sorting is done in-place; otherwise, O(n) for the sorted array.

Trade-offs: This approach has a higher time complexity due to sorting but may have a lower space complexity if in-place sorting is used. If the array is already sorted or sorting is undesirable due to space constraints and the hashmap would be quite large, this can be a useful solution. However, sorting modifies the input array, which may not be desirable.

Brute Force

Iterate through all possible pairs of numbers in the array and check if their absolute difference equals k. Keep track of the unique pairs to avoid counting duplicates.

Time: O(n^2), where n is the length of the array.Space: O(1), excluding the space for the input array.

Trade-offs: This approach is simple to understand but inefficient for large arrays. It should be avoided for larger inputs where the hash map approach or sorting + two pointers are much more efficient.

✨

Key Insights

1Insight 1: The absolute difference `|nums[i] - nums[j]| == k` implies that `nums[i] - nums[j] == k` or `nums[j] - nums[i] == k`. Therefore, we are looking for pairs where one number is `k` more or `k` less than the other.
2Insight 2: A hash table (or Counter in Python) is an efficient data structure to count the frequency of each number in the array. This helps in identifying and counting unique pairs.
3Insight 3: When k=0, we are looking for pairs of identical numbers. In this special case, we only need to count numbers that appear more than once in the array. When k>0, we need to check if `num + k` exists in the array.

📚

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!

Two Pointers

Using two indices to traverse data from different positions simultaneously.

Binary Search

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

Sorting

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

Edge Cases

Edge case 1: `k < 0`. The problem defines k as the *absolute* difference, so if k is negative, there can be no such pairs. The solution handles this by returning 0 if k is negative.
Edge case 2: `k == 0`. This means we are looking for pairs of the same number. The solution correctly handles this by checking if the count of a number is greater than 1.
Edge case 3: Duplicate numbers in `nums`. The solution uses a hash table to keep track of counts, ensuring that duplicate numbers are handled correctly and only unique pairs are counted.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case where `k = 0` correctly. It's important to check for duplicate numbers when `k = 0` and only count them once.

✗Common mistake 2: Counting the same pair twice (e.g., (a, b) and (b, a)). The hash table approach avoids this issue because it checks for the existence of `num + k` only once for each number. The sorting + two pointers must explicitly handle this by advancing the left pointer if the values are the same to avoid redundant counts

✗Common mistake 3: Forgetting to handle the edge case where `k` is negative. Returning 0 is the correct approach.

✗Common mistake 4: Not accounting for duplicates in the input array properly which causes overcounting of unique pairs.

✅

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: In data analysis, finding data points that are within a certain range (k-diff) of each other can be useful for identifying clusters or anomalies.

🏦Real-world scenario 2: In financial analysis, finding stock prices that are correlated within a certain price difference (k-diff) can be used for identifying trading opportunities.

🎮Real-world scenario 3: Recommendation systems, where you might want to recommend items that are 'similar' based on a metric where k represents a threshold for similarity.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ 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: November 20, 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

❓

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.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

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 "K-diff Pairs in an Array"?

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

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