LeetCode

## Problem Statement

Given an array of integers and an integer k, you need to find the number of unique k-diff pairs in the array. Here a k-diff pair is defined as an integer pair (i, j), where i and j are both numbers in the array and their absolute difference is k.

Example 1:

``````Input: [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 of unique pairs.
``````

Example 2:

``````Input:[1, 2, 3, 4, 5], k = 1
Output: 4
Explanation: There are four 1-diff pairs in the array, (1, 2), (2, 3), (3, 4) and (4, 5).
``````

Example 3:

``````Input: [1, 3, 1, 5, 4], k = 0
Output: 1
Explanation: There is one 0-diff pair in the array, (1, 1).
``````

## Solution

``````func findPairs(nums []int, k int) int {
if k < 0 {
return 0
}

dict, count := make(map[int]int), 0

for _, v := range nums {
dict[v]++
}

for v, c := range dict {
if k == 0 {
if c >= 2 {
count++
}
} else {
if dict[v+k] > 0 {
count++
}
}
}

return count
}
``````