# Triplet Sum in Array | Amazon | Samsung

### Question

Write a program to find triplet sum in array. Given an array A[] of N numbers and another number x, determine whether or not there exist three elements in A[] whose sum is exactly x.

### Logic

• Get the array numbers and x from the user.
• Use nested for loop to generate each combination with three elements.
• If the sum of three elements equals the given x then Print the combination.

### Algorithm

1. We create three for loop because we find triplet sum.
2. From first for loop we get arr[i]. We keep this constant and change j and k.
3. From second for loop we get arr[j]. We keep this constant and change k.
4. We check sum by adding arr[i], arr[j], arr[k].

### Program

```include<iostream>
using namespace std;
int main()
{
int x = 10;
int i,j,sum,n,k;
int arr;
bool flag = true;
cin>>x>>n;
//get n numbers
for(i=0; i<n; i++)
{
cin>>arr[i];

}
/*Let the array= {1,2,4,3,6}*/
for(i=0; i<n-2; i++)                    //loop till n-2 items
{
/*EX: In first iteration we arr=1 in first iteration
once all the possibilities for combination
starting with 1 complete, we increment*/

for(j=i+1; j<n-1&&flag; j++)        //loop till n-1 items
{
/*EX: In first iteration of this loop we check
arr=2 with other combinations
so 1+2+arr[k]*/

for(k=j+1; k<n&&flag; k++)
{
sum=arr[i]+arr[j]+arr[k];
/*If sum equals given number*/
if(sum==x)
{
cout<<arr[i]<<"+"<<arr[j]"+"<<arr[k]<<"="<<sum;
flag = false;
break;

}

}

}

}
if(flag)
{
}

return 0;
}
```

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