# K-diff Pairs in an Array LeetCode Solution

Here, We see K-diff Pairs in an Array LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

# List of all LeetCode Solution

## Problem Statement

Given an array of integers `nums` and an integer `k`, return the number of unique k-diff pairs in the array.

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).

## K-diff Pairs in an Array LeetCode SolutionC++

``````class Solution {
public:
int findPairs(vector<int>& nums, int k) {
unordered_map<int,int> a;
for(int i:nums)
a[i]++;
int ans=0;
for(auto x:a){
if(k==0){
if(x.second>1)
ans++;
}
else if(a.find(x.first+k)!=a.end())
ans++;
}
return ans;
}
};```Code language: PHP (php)```

## K-diff Pairs in an Array LeetCode SolutionJava

``````class Solution {
public int findPairs(int[] nums, int k) {
Map<Integer, Integer> map = new HashMap();
for (int num : nums)
map.put(num, map.getOrDefault(num, 0) + 1);
int result = 0;
for (int i : map.keySet())
if (k > 0 && map.containsKey(i + k) || k == 0 && map.get(i) > 1)
result++;
return result;
}
}```Code language: JavaScript (javascript)```

## K-diff Pairs in an Array SolutionJavaScript

``````var findPairs = function(nums, k) {
if(nums.length === 0 || k < 0) return 0
let myMap = new Map(),
count = 0
for(num of nums){
myMap.set(num,(myMap.get(num)+1) || 1)
}
myMap.forEach((value,key) =>{
if(k === 0){
if(value > 1) count++
}
else{
if(myMap.has(key+k)) count++
}
})
return count
};```Code language: JavaScript (javascript)```

## K-diff Pairs in an Array SolutionPython

``````class Solution(object):
def findPairs(self, nums, k):
cnt=0
c=Counter(nums)
if k==0:
for key,v in c.items():
if v>1:
cnt+=1
else:
for key,v in c.items():
if key+k in c:
cnt+=1
return cnt``````
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