include

using namespace std; int test(int n, long long k, vector a) { int dem = 0; unordered_sets(a.begin(), a.end()); for(int i = 0; i < n; i++) { if(a[i] > k){ long long tam = a[i] - k; if(s.find(tam) != s.end()) { dem++; } } } return dem; } int main() { int n; long long k; cin >> n >> k; vector a(n); for(int i = 0; i < n; i++) { cin >> a[i]; } cout << test(n, k, a); return 0; }

C#

public static int pairs(int k, List<int> arr)
{
    var hashSet = new HashSet<int>(arr);
    return hashSet.Count(x => hashSet.Contains(x + k));
}


public static int pairs(int k, List<int> arr)
{
    var total = 0;
    var hashSet = new HashSet<int>(arr);

    foreach (var n in hashSet)
    {
        if (hashSet.Contains(n + k))
        {
            total++;
        }
    }

    return total;
}

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;
    }