LeetCode Daily Challenge 🔥| DAY 94

🔥 LeetCode 2006 – Count Number of Pairs With Absolute Difference K | Java Solution | HashMap Explained

Learn how to count valid pairs efficiently using absolute difference logic 🚀

In this video, we solve LeetCode 2006: Count Number of Pairs With Absolute Difference K using an optimized counting approach in Java.

💡 What You’ll Learn:

✅ Absolute difference concept
✅ Pair counting optimization
✅ HashMap / frequency array approach
✅ Brute force vs optimized solution
✅ Time & Space Complexity analysis
✅ Interview-ready Java solution

📌 Example

nums = [1,2,2,1]
k = 1

Valid pairs:

(0,1) → |1 - 2| = 1 ✔️
(0,2) → |1 - 2| = 1 ✔️
(1,3) → |2 - 1| = 1 ✔️
(2,3) → |2 - 1| = 1 ✔️

Answer = 4 🎯

🚀 Approach Used

1. Traverse the array
2. For each number:
check previously seen values matching:
- num - k
- num + k
3. Add matching frequencies to answer
4. Update current number frequency

Core Logic:

count += freq[num - k];
count += freq[num + k];

📈 Complexity

Optimal Approach:
Time: O(n)
Space: O(1) (frequency array) / O(n) (HashMap)

Perfect for coding interviews, DSA prep & LeetCode practice.

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Видео LeetCode Daily Challenge 🔥| DAY 94 канала THE BACKEND BOI