Last updated on July 18th, 2024 at 10:13 pm

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

Topics

Array, Two Pointers

Companies

Amazon

Level of Question

Medium

K-diff Pairs in an Array 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.

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).

1. K-diff Pairs in an Array LeetCode Solution C++

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;
    }
};

2. K-diff Pairs in an Array LeetCode Solution Java

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;
 	}
}

3. K-diff Pairs in an Array Solution JavaScript

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
};

4. K-diff Pairs in an Array Solution Python

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