All unique triplets that sum up to a given value in C++

C++Server Side ProgrammingProgramming

Here we will see one interesting problem. We have an array with some elements. One sum value is given. Our task is to find triplets from the array, and whose sum is same as the given sum. Suppose the array is {4, 8, 63, 21, 24, 3, 6, 1, 0}, and the sum value is S = 18. So the triplets will be {4, 6, 8}. If more than one triplet is present, it will show all of them.

Algorithm

getTriplets(arr, n, sum) −

Begin
   define one array to store triplets, say trip_arr
   define one set unique_trip to store unique triplets. No same triplets will be their.
   sort arr
   for i in range 0 to n-2, do
      j :- i + 1, and k := n – 1
      while(j < k), do
         if arr[i] + arr[j] + arr[k] = sum, then
            temp := arr[i] : arr[j] : arr[k]
            if temp is not present into the unique_trip, then
               insert temp into unique_trip set
               create newTriplet using arr[i] arr[j] and arr[k]
               add newTriplet into trip_arr
            endif
            increase j, and decrease k
         else if arr[i] + arr[j] + arr[k] > sum
            decrease k
         else
            increase j
         end if
      done
   done
   display all triplets
End

Example

#include <iostream>
#include <vector>
#include <set>
#include <algorithm>
using namespace std;
class triplet {
   public:
      int first, second, third;
      void display() {
         cout << "("<<first<<", "<<second<<", "<<third<<")" << endl;
      }
};
int getTriplets(int arr[], int n, int sum) {
   int i, j, k;
   vector <triplet> triplets;
   set <string> uniqTriplets; //use set to avoid duplicate triplets
   string temp_triplet;
   triplet newTriplet;
   sort(arr, arr + n); //sort the array
   for(i = 0; i < n - 2; i++) {
      j = i + 1;
      k = n - 1;
      while(j < k) {
         if(arr[i] + arr[j] + arr[k] == sum) {
            temp_triplet = to_string(arr[i]) + " : " + to_string(arr[j]) + " : " + to_string(arr[k]);
            if(uniqTriplets.find(temp_triplet) == uniqTriplets.end()) {
               uniqTriplets.insert(temp_triplet);
               newTriplet.first = arr[i];
               newTriplet.second = arr[j];
               newTriplet.third = arr[k];
               triplets.push_back(newTriplet);
            }
            j++;
            k--;
         } else if(arr[i] + arr[j] + arr[k] > sum)
               k--;
         else
            j++;
      }
   }
   if(triplets.size() == 0)
      return 0;
   for(i = 0; i < triplets.size(); i++) {
      triplets[i].display();
   }
}
int main() {
   int nums[] = {4, 8, 63, 21, 24, 3, 6, 1, 0, 5};
   int n = sizeof(nums) / sizeof(nums[0]);
   int sum = 27;
   if(!getTriplets(nums, n, sum))
      cout << "No triplets can be formed.";
}

Output

(0, 3, 24)
(0, 6, 21)
(1, 5, 21)
raja
Published on 20-Aug-2019 07:26:11
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