How do you solve Count Number of Pairs With Absolute Difference K on LeetCode?

Count Number of Pairs With Absolute Difference K (LeetCode #2006) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a HashMap can significantly reduce the number of comparisons needed.

What is the brute force solution for Count Number of Pairs With Absolute Difference K?

The brute force approach for Count Number of Pairs With Absolute Difference K is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every possible pair of elements in the array to see if their absolute difference equals k. This is straightforward but can be slow for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count Number of Pairs With Absolute Difference K?

Count Number of Pairs With Absolute Difference K uses the following concepts: Array, Hash Table, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Count Number of Pairs With Absolute Difference K?

Always clarify the problem constraints and edge cases with the interviewer. Explain your thought process while coding to show your understanding. Consider both time and space complexity when discussing your solution.

What are common mistakes when solving Count Number of Pairs With Absolute Difference K?

Not considering pairs (i, j) where i < j correctly, leading to double counting. Overlooking edge cases where k is larger than the maximum difference in the array.

#2006

Count Number of Pairs With Absolute Difference K

Easy

ArrayHash TableCountingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Using a HashMap allows us to count occurrences of each number, making it easy to find how many numbers can form valid pairs with a given number. This significantly reduces the number of comparisons needed.

⚙️

Algorithm

5 steps

1Step 1: Create a HashMap to count occurrences of each number in the array.
2Step 2: Initialize a counter to 0.
3Step 3: For each unique number in the HashMap, check if the number + k and number - k exist in the map.
4Step 4: If they exist, multiply their counts and add to the counter.
5Step 5: Return the counter.

solution.py8 lines

1def countKDifference(nums, k):
2    from collections import Counter
3    count = 0
4    num_count = Counter(nums)
5    for num in num_count:
6        count += num_count[num] * num_count.get(num + k, 0)
7        count += num_count[num] * num_count.get(num - k, 0)
8    return count // 2

ℹ

Complexity note: The time complexity is O(n) because we traverse the array once to count occurrences and then iterate through the unique numbers. The space complexity is O(n) due to the HashMap storing counts of each number.

1Using a HashMap can significantly reduce the number of comparisons needed.
2Understanding absolute differences can help simplify the problem.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.