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All unique triplets that sum up to a given value in C++
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)
Updated on: 20-Aug-2019
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