# 532. K-diff Pairs in an Array

## Description

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.

Example 1:

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


Constraints:

• 1 <= nums.length <= 104
• -107 <= nums[i] <= 107
• 0 <= k <= 107

## Solutions

• class Solution {
public int findPairs(int[] nums, int k) {
Set<Integer> vis = new HashSet<>();
Set<Integer> ans = new HashSet<>();
for (int v : nums) {
if (vis.contains(v - k)) {
}
if (vis.contains(v + k)) {
}
}
return ans.size();
}
}

• class Solution {
public:
int findPairs(vector<int>& nums, int k) {
unordered_set<int> vis;
unordered_set<int> ans;
for (int& v : nums) {
if (vis.count(v - k)) ans.insert(v - k);
if (vis.count(v + k)) ans.insert(v);
vis.insert(v);
}
return ans.size();
}
};

• class Solution:
def findPairs(self, nums: List[int], k: int) -> int:
vis, ans = set(), set()
for v in nums:
if v - k in vis:
if v + k in vis:
return len(ans)


• func findPairs(nums []int, k int) int {
vis := map[int]bool{}
ans := map[int]bool{}
for _, v := range nums {
if vis[v-k] {
ans[v-k] = true
}
if vis[v+k] {
ans[v] = true
}
vis[v] = true
}
return len(ans)
}

• impl Solution {
pub fn find_pairs(mut nums: Vec<i32>, k: i32) -> i32 {
nums.sort();
let n = nums.len();
let mut res = 0;
let mut left = 0;
let mut right = 1;
while right < n {
let num = i32::abs(nums[left] - nums[right]);
if num == k {
res += 1;
}
if num <= k {
right += 1;
while right < n && nums[right - 1] == nums[right] {
right += 1;
}
} else {
left += 1;
while left < right && nums[left - 1] == nums[left] {
left += 1;
}
if left == right {
right += 1;
}
}
}
res
}
}