Write a C++ program for a given array of integers, you have to find three numbers such that the sum of two elements equals the third element.
Examples:
Input: {5, 32, 1, 7, 10, 50, 19, 21, 2}
Output: 21, 2, 19Input: {5, 32, 1, 7, 10, 50, 19, 21, 0}
Output: no such triplet exist
Question source: Arcesium Interview Experience | Set 7 (On campus for Internship)
Simple approach:
Run three loops and check if there exists a triplet such that sum of two elements equals the third element.
Below is the implementation of the above approach:
// CPP program to find three numbers // such that sum of two makes the // third element in array #include <bits/stdc++.h> using namespace std;
// Utility function for finding // triplet in array void findTriplet( int arr[], int n)
{ for ( int i = 0; i < n; i++) {
for ( int j = i + 1; j < n; j++) {
for ( int k = j + 1; k < n; k++) {
if ((arr[i] + arr[j] == arr[k])
|| (arr[i] + arr[k] == arr[j])
|| (arr[j] + arr[k] == arr[i])) {
// printing out the first triplet
cout << "Numbers are: " << arr[i] << " "
<< arr[j] << " " << arr[k];
return ;
}
}
}
}
// No such triplet is found in array
cout << "No such triplet exists" ;
} // driver program int main()
{ int arr[] = { 5, 32, 1, 7, 10, 50, 19, 21, 2 };
int n = sizeof (arr) / sizeof (arr[0]);
findTriplet(arr, n);
return 0;
} |
Numbers are: 5 7 2
Time complexity: O(n^3)
Auxiliary Space: O(1)
Efficient approach:
The idea is similar to Find a triplet that sum to a given value.
Step-by-step approach:
- Sort the given array first.
- Start fixing the greatest element of three from the back and traverse the array to find the other two numbers which sum up to the third element.
- Take two pointers j(from front) and k(initially i-1) to find the smallest of the two number and from i-1 to find the largest of the two remaining numbers
- If the addition of both the numbers is still less than A[i], then we need to increase the value of the summation of two numbers, thereby increasing the j pointer, so as to increase the value of A[j] + A[k].
- If the addition of both the numbers is more than A[i], then we need to decrease the value of the summation of two numbers, thereby decrease the k pointer so as to decrease the overall value of A[j] + A[k].
Below image is a dry run of the above approach:
Below is the implementation of the above approach:
// C++ program to find three numbers // such that sum of two makes the // third element in array #include <bits/stdc++.h> using namespace std;
// Utility function for finding // triplet in array void findTriplet( int arr[], int n)
{ // Sort the array
sort(arr, arr + n);
// For every element in arr check
// if a pair exist(in array) whose
// sum is equal to arr element
for ( int i = n - 1; i >= 0; i--)
{
int j = 0;
int k = i - 1;
// Iterate forward and backward to
// find the other two elements
while (j < k)
{
// If the two elements sum is
// equal to the third element
if (arr[i] == arr[j] + arr[k])
{
// Pair found
cout << "numbers are " << arr[i] <<
" " << arr[j] << " " <<
arr[k] << endl;
return ;
}
// If the element is greater than
// sum of both the elements, then try
// adding a smaller number to reach the
// equality
else if (arr[i] > arr[j] + arr[k])
j += 1;
// If the element is smaller, then
// try with a smaller number
// to reach equality, so decrease K
else
k -= 1;
}
}
// No such triplet is found in array
cout << "No such triplet exists" ;
} // Driver code int main()
{ int arr[] = {5, 32, 1, 7, 10,
50, 19, 21, 2};
int n = sizeof (arr) / sizeof (arr[0]);
findTriplet(arr, n);
return 0;
} |
Output:
numbers are 21 2 19
Time complexity: O(N^2)
Auxiliary Space: O(1)
C++ Program to Find a triplet such that sum of two equals to third element using Binary Search.
- Sort the given array.
- Start a nested loop, fixing the first element i(from 0 to n-1) and moving the other one j (from i+1 to n-1).
- Take the sum of both the elements and search it in the remaining array using Binary Search.
Below is the implementation of the above approach:
// C++ program to find three numbers // such that sum of two makes the // third element in array #include <bits/stdc++.h> #include <iostream> using namespace std;
// Function to perform binary search bool search( int sum, int start,
int end, int arr[])
{ while (start <= end)
{
int mid = (start + end) / 2;
if (arr[mid] == sum)
{
return true ;
}
else if (arr[mid] > sum)
{
end = mid - 1;
}
else
{
start = mid + 1;
}
}
return false ;
} // Function to find the triplets void findTriplet( int arr[], int n)
{ // Sorting the array
sort(arr, arr + n);
// Initialising nested loops
for ( int i = 0; i < n; i++)
{
for ( int j = i + 1; j < n; j++)
{
// Finding the sum of the numbers
if (search((arr[i] + arr[j]),
j, n - 1, arr))
{
// Printing out the first triplet
cout << "Numbers are: " << arr[i] <<
" " << arr[j] << " " <<
(arr[i] + arr[j]);
return ;
}
}
}
// If no such triplets are found
cout << "No such numbers exist" << endl;
} // Driver code int main()
{ int arr[] = {5, 32, 1, 7, 10,
50, 19, 21, 2};
int n = sizeof (arr) / sizeof (arr[0]);
findTriplet(arr, n);
return 0;
} // This code is contributed by Sarthak Delori |
Time Complexity: O(N^2*log N)
Auxiliary Space: O(1)
Please refer complete article on Find a triplet such that sum of two equals to third element for more details!