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Pairs

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1243 Discussions

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  • agentkaumal88
     + 0 comments

    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;
}

  • mhuang005
     + 0 comments

    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

  • sid_grv
     + 0 comments

    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

    `

  • krokiet
     + 0 comments

    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;
    }

  • sarthakhare15
     + 1 comment

    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())