K-diff Pairs in an Array in C++

Suppose we have an array and an integer k, we have to find the number of unique k-diff pairs in the array. Here the k-diff pair is like (i, j), where i and j are both are present in the array and their absolute difference is k.

So, if the input is like [3,1,4,1,5], k = 2, then the output will be 2, as there are two 2-diff pairs in the array-like (1,3) and (3,5).

To solve this, we will follow these steps −

• Define maps called seen and done

• Define one set s

• if k < 0, then −

  • return 0

• for initialize i := 0, when i < size of nums, update (increase i by 1), do −

  • (increase seen[nums[i]] by 1)

  • insert nums[i] into s

• ans := 0

• for each element it in s, do −

  • if k is same as 0, then −

    • if seen[it] > 1, then −

      • (increase ans by 1)

  • Otherwise

    • increase done[it] by 1

    • if (it + k) is in seen but not in done, then −

      • (increase ans by 1)

    • if (it - k) is in seen but not in done, then −

      • (increase ans by 1)

• return ans

Example 

Let us see the following implementation to get a better understanding −

 Live Demo

#include <bits/stdc++.h&g;
using namespace std;
class Solution {
public:
   int findPairs(vector<int>& nums, int k) {
      map<int, int> seen, done;
      set<int> s;
      if (k < 0)
         return 0;
      for (int i = 0; i < nums.size(); i++) {
         seen[nums[i]]++;
         s.insert(nums[i]);
      }
      int ans = 0;
      for (auto it = s.begin(); it != s.end(); it++) {
         if (k == 0) {
            if (seen[*it] > 1)
            ans++;
         }
         else {
            done[*it]++;
            if (seen.find(*it + k) != seen.end() && done.find(*it + k) == done.end())
               ans++;
            if (seen.find(*it - k) != seen.end() && done.find(*it - k) == done.end())
               ans++;
         }
      }
      return ans;
   }
};
main(){
   Solution ob;
   vector<int> v = {3,1,4,1,5};
   cout << (ob.findPairs(v, 2));
}

Input

{3,1,4,1,5}, 2

Output

2