To solve this problem, we need to count the number of pairs (i, j) in an array such that their absolute difference equals k. The best approach is to use array scanning combined with hash lookup for optimal performance, as this will allow us to efficiently track previously seen values and avoid checking every pair explicitly.
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.
To solve this, we can utilize an array scanning approach alongside a hash table to track previously seen values, avoiding the need to check every possible pair directly.
Examples
Example 1
Input: nums = [1,2,2,1], k = 1
Output: 4
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
There are no pairs with an absolute difference of 3.
Example 3
Input: nums = [3,2,1,5,4], k = 2
Output: 3
The pairs with an absolute difference of 2 are:
- [3,2,1,5,4]
- [3,2,1,5,4]
- [3,2,1,5,4]
Constraints
- 1 <= nums.length <= 200
- 1 <= nums[i] <= 100
- 1 <= k <= 99
Solution Approach
Array Scanning with Hash Table
Iterate through the array while maintaining a hash table that tracks the counts of each element seen so far. For each element nums[i], check if nums[i] - k or nums[i] + k exists in the hash table. If it does, increase the count accordingly.
Efficient Pair Counting
The hash table allows for efficient lookup in constant time, making this approach optimal. Instead of checking every possible pair, we check each element and its potential pair in the hash table, ensuring that no unnecessary calculations are done.
Edge Case Handling
Handle cases where the array contains duplicates or where no valid pairs exist. Ensure that you do not double-count pairs and that you correctly handle when no pairs with the required absolute difference are found.
Complexity Analysis
| Metric | Value |
|---|---|
| Time | Depends on the final approach |
| Space | Depends on the final approach |
The time complexity is O(n) where n is the length of the array, due to the linear scan and constant-time hash lookups. The space complexity is O(n) because we store each element in the hash table.
What Interviewers Usually Probe
- Candidate efficiently uses a hash table for constant-time lookups.
- Candidate demonstrates understanding of avoiding brute force pair checking.
- Candidate handles edge cases like duplicates and missing pairs effectively.
Common Pitfalls or Variants
Common pitfalls
- Overcomplicating the problem with nested loops.
- Failing to manage hash table updates correctly, leading to double-counting.
- Not handling edge cases like empty arrays or arrays without valid pairs.
Follow-up variants
- Consider using a set for tracking seen numbers instead of a hash table.
- Adapt the solution for cases where negative values are allowed in the array.
- Optimize further if the array is guaranteed to be sorted.
How GhostInterview Helps
- GhostInterview assists by offering optimized solutions to count pairs with the exact absolute difference.
- It helps avoid the brute-force O(n^2) approach by providing better methods using hash tables.
- Guidance through the problem-solving process ensures that you consider all edge cases and performance trade-offs.
Topic Pages
Related GhostInterview Pages
- LeetCode Interview Copilot - Use GhostInterview as a live solver when you want direct help with LeetCode-style coding questions.
- Coding Interview Assistant - See how GhostInterview supports array, string, linked list, graph, and tree interview workflows.
- How GhostInterview Works - Review the screenshot, reasoning, and answer flow before using the solver in a live interview.
FAQ
How does array scanning help in solving this problem?
Array scanning ensures that we only need to pass through the array once, making the solution efficient compared to checking all pairs directly.
Can I solve this without using a hash table?
While it's possible to use brute-force methods, they would be inefficient with time complexity O(n^2). A hash table significantly improves the time complexity to O(n).
How does GhostInterview assist with this problem?
GhostInterview offers a step-by-step breakdown of how to efficiently solve this problem, helping you understand the hash table-based approach and avoid common pitfalls.
What are the edge cases to consider in this problem?
Consider arrays with duplicates, arrays where no pairs meet the condition, and cases with a very small or large array size.
What if there are negative numbers in the array?
If negative numbers are allowed, the solution can be adapted by adjusting the conditions for pair matching to account for negative differences.
Need direct help with Count Number of Pairs With Absolute Difference K instead of spending more time grinding it?
Download GhostInterview when you want a LeetCode solver, not another long practice loop. Capture Count Number of Pairs With Absolute Difference K from a screenshot, get the answer path and complexity, and use supported stealth workflows that stay outside captured layers.
Capture the prompt fast instead of rewriting the problem by hand.
Get the solution path, trade-offs, and complexity summary in one pass.
Stay outside captured layers on supported screen-share workflows.
Stay in the same pattern family
Count all pairs of rectangles that share the exact width-to-height ratio using efficient array scanning and hash lookup.
Open problem page#2013 Detect SquaresDetect Squares requires tracking points on a 2D plane to quickly count all possible axis-aligned squares using efficient hash-based lookup.
Open problem page#1994 The Number of Good SubsetsFind the number of good subsets in an integer array, where each subset's product is the product of distinct primes.
Open problem page