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by Sumit Chourasia | Nov 20, 2020 | Category :coding | Tags : algorithm array data-structure leetcode medium

K-diff Pairs in an Array - Array - Medium - LeetCode

K-diff Pairs in an Array - Array - Medium - LeetCode

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).
Example 4:

Input: nums = [1,2,4,4,3,3,0,9,2,3], k = 3
Output: 2
Example 5:

Input: nums = [-1,-2,-3], k = 1
Output: 2
 

Constraints:

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

 

public class Solution {
    public int FindPairs(int[] nums, int k) {
        if(nums.Length ==0){
            return 0;
        }
        
        var map = new Dictionary<int,int>();
        for(int i=0;i<nums.Length;i++){
            int val=0;
            map.TryGetValue(nums[i],out val);
            map[nums[i]]= val+1;            
        }
        
        int count =0;
        foreach(var item in map){
            if(k==0){
                if(item.Value>1){
                    count++;
                }
            }
            else{
                if(map.ContainsKey(item.Key-k)){
                    count++;
                }
            }
        }
        
        return count;
    }
}

Time Complexity: O(n)

Space Complexity: O(n)