K-diff Pairs in an Array

K-diff Pairs in an Array - Problem

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.

Input & Output

Example 1 — Basic Case

$ Input: nums = [3,1,4,1,5], k = 2

› Output: 2

💡 Note: There are two 2-diff pairs: (1,3) and (3,5). Note that we only count unique pairs, so (1,3) from different positions counts as one pair.

Example 2 — No Valid Pairs

$ Input: nums = [1,2,3,4,5], k = 1

› Output: 4

💡 Note: Valid pairs are (1,2), (2,3), (3,4), and (4,5). Each consecutive pair has difference 1.

Example 3 — Duplicate Elements

$ Input: nums = [1,3,1,5,4], k = 0

› Output: 1

💡 Note: For k=0, we need identical pairs. Only the number 1 appears more than once, giving us one unique pair (1,1).

Constraints

1 ≤ nums.length ≤ 10⁴
-10⁷ ≤ nums[i] ≤ 10⁷
0 ≤ k ≤ 10⁷

Visualization

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Asked in

a Amazon 25 G Google 18 M Microsoft 12 f Facebook 8

The key insight is to avoid checking all pairs by using a hash map or sorted two-pointer approach. Hash map approach provides O(n) time by checking if each number's complement (num+k) exists. Best approach is hash map with Time: O(n), Space: O(n).

Common Approaches

✓ Dp 2d

⏱️ Time: N/A Space: N/A

Brute Force - Check All Pairs

⏱️ Time: O(n²) Space: O(n)

Compare each element with every other element to find pairs with absolute difference k. Use a set to store unique pairs to avoid duplicates.

Hash Map - O(n) Lookup

⏱️ Time: O(n) Space: O(n)

Store all numbers in a hash map with their frequencies. For each unique number, check if num+k exists (or num-k). Handle the special case when k=0 by checking frequency > 1.

Two Pointers After Sorting

⏱️ Time: O(n log n) Space: O(1)

Sort the array first, then use two pointers. If difference is less than k, move right pointer. If greater than k, move left pointer. If equal to k, count the pair and handle duplicates.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

Time & Space Complexity

Time Complexity

⏱️

✓ Linear Growth

Space Complexity

⚡ Linearithmic Space

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Medium Frequency

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Smart Actions

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