### 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.

**Link to GitHub**: Code

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).

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**

2. We need to check the no of pairs in hash table by adding k value to the Hash table entries

3 . This condition map.put(i, map.getOrDefault(i,0)+1) would make sure that if there is a duplicate then

it would replace however it would increment the coutn

[3, 1, 4, 1, 5] . Here it would (1,2)

3. If K=0 then we need to increment the counter if the value pair is greater than 2

Example for checking if the K pair exists in the hash-map

when K=2

keyValue + K

1+2 =3 . It exists

package Algorithms; import java.util.Map; import java.util.HashMap; public class KdiffPairsInArray { public static int get(int [] nums,int k) { if(nums== null || nums.length==0 || k<0 ) return 0; //An object that maps keys to values. A map cannot contain duplicate keys; //each key can map to at most one value. Map<Integer, Integer> map = new HashMap<>(); //Insert all the contents of the array to hash table. for(int i:nums) { //getOrDefault used to return a default value whenever the value was not found

` //using the key specified in the hashkey. If there is a duplicate it would replace `

` //it with new value`

// map.put(i, map.getOrDefault(i,0)+1); } int count=0; //Count the no of pairs in mapping table for(Map.Entry<Integer, Integer> entry :map.entrySet()) { //If K=0 /* * Input: [1, 3, 1, 5, 4], k = 0 Output: 1 Explanation: There is one 0-diff pair in the array, (1, 1) * */ if(k==0){ if(entry.getValue() >=2) { count++; } } else { //Check if the K pair exists in the hash-map /* * ex" when K=2 * keyValue + K * 1+2 =3 . It exists */ if(map.containsKey(entry.getKey()+k)) count++; } } return count; } }

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