Given an array S of n integers, are there elements a, b, c in S such that a + b

Q: Given an array S of n integers, are there elements a, b, c in S such that a + b

C++ code: #include using namespace std; // function to print triplets with 0 sum vector > Triplets(int arr[], int n) { bool found = false; vector output; for (int i=0; i

ID: 3866574 • Letter: G

Question

Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? write a function to Find all unique triplets in the array which gives the sum of zero.

Note: The solution set must not contain duplicate triplets.

For example, given array S = [-1, 0, 1, 2, -1, -4],

A solution set is:

[ [-1, 0, 1], [-1, -1, 2] ]

return the set for this as an array or arrays

return null for fail cases

keep in mind time comlpexity and explain if using a hashset is in the better interest for time complexity

write written algorithm to how you solved it with words DO THIS PART TOO PLEASE!

type please

comment each line of code

c++

give test cases

Explanation / Answer

C++ code:

#include<bits/stdc++.h>
using namespace std;

// function to print triplets with 0 sum
vector <set < int > > Triplets(int arr[], int n)
{
bool found = false;
vector< set<int> > output;
for (int i=0; i<n-1; i++)
{
// Find all pairs with sum equals to
// "-arr[i]"

vector <int> s;
for (int j=i+1; j<n; j++)
{
int x = -(arr[i] + arr[j]);
if (find(s.begin(),s.end(), x) != s.end())
{
set<int> myset2; myset2.insert(x);myset2.insert(arr[i]);myset2.insert(arr[j]);
bool flag = true;

for (int w = 0; w < output.size(); ++w)
{
set<int> myset1 = output[w];
if(myset2 == myset1) //check duplicates
{
flag = false;
break;
}
}
if(flag == true)
{
output.push_back(myset2);
printf("Possible Triplet %d %d %d ", x, arr[i], arr[j]);
}
found = true;
}
else
s.push_back(arr[j]);
}
}

if (found == false)
{
cout << " No Triplet with sum equal to 0 exist in this array!" << endl;
}
return output;
}

int main()
{
int arr[] = {-1, 0, 1, 2, -1, -4}; //Input array
int n = 6; //size of the array
Triplets(arr, n);
return 0;
}

Sample Output:

Possible Triplet 0 -1 1
Possible Triplet 2 -1 -1

Time complexity: O(n^2)

For each element in the input array x, we are finding a pair of other two elements which have sum equal to -x.So, this is O(n^2).

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