Python 3:

from collections import Counter

counter = Counter(arr) num_pairs = 0

for i in arr: if i - k in counter: num_pairs += 1 # Note: integers in arr are distinct

return num_pairs

Simple C++ sol using set | O(n)

`c++

int pairs(int k, vector arr) { unordered_set st; int n = arr.size(), c = 0; for (int i=0; i

`

Nice and easy this time! First time I post a solution here, because I'm stunned it worked at the very first time, wihtout any corrections (rare case;)

Sort + caterpillar method.

Java solution:

    public static int pairs(int k, List<Integer> arr) {
        Collections.sort(arr);
        int a = 0;
        int b = 1;
        int count = 0;
        while (a < b && b < arr.size()) {            
            int left = arr.get(a);
            int right = arr.get(b);
            int diff = right - left;
            if (diff == k) {
                count++;
                b++;
            } else if (diff > k) {
                a++;
                if (a==b) b++;
            } else {
                b++;
            }            
        }
				}
        return count;
    }

Python3 Optimal Solution:

def pairs(k, arr):

from collections import defaultdict
arr_elements=set(arr)
hmap=defaultdict(int)

for i in range(len(arr)):
    if arr[i]+k in arr_elements and (arr[i], arr[i]+k) not in hmap:
        hmap[(arr[i], arr[i]+k)]+=1

    if arr[i]-k in arr_elements and (arr[i]-k, arr[i]) not in hmap:
        hmap[(arr[i]-k, arr[i])]+=1

return sum(hmap.values())

My idea is loop the array and plus earch number at index with K then push the result to Collections array. Next step I will check the element in Collections exist in arr and counting that.

With c#

public static int pairs(int k, List<int> arr)
    {
        var collections = arr.Select(number => k + number);
        var numberExisted = collections.Where(number => arr.Contains(number)).ToList();
        
        return numberExisted.Count();
    }