Given an array arr[] of N integers, the task is to find the sum of all the pairs possible from the given array. Note that,
- (arr[i], arr[i]) is also considered as a valid pair.
- (arr[i], arr[j]) and (arr[j], arr[i]) are considered as two different pairs.
Examples:
Input: arr[] = {1, 2}
Output: 12
All valid pairs are (1, 1), (1, 2), (2, 1) and (2, 2).
1 + 1 + 1 + 2 + 2 + 1 + 2 + 2 = 12Input: arr[] = {1, 2, 3, 1, 4}
Output: 110
Naive approach: Find all the possible pairs and calculate the sum of the elements of each pair.
Below is the implementation of the above approach:
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;
// Function to return the sum of the elements // of all possible pairs from the array int sumPairs( int arr[], int n)
{ // To store the required sum
int sum = 0;
// Nested loop for all possible pairs
for ( int i = 0; i < n; i++) {
for ( int j = 0; j < n; j++) {
// Add the sum of the elements
// of the current pair
sum += (arr[i] + arr[j]);
}
}
return sum;
} // Driver code int main()
{ int arr[] = { 1, 2, 3 };
int n = sizeof (arr) / sizeof (arr[0]);
cout << sumPairs(arr, n);
return 0;
} |
// Java implementation of the approach import java.util.*;
class GFG
{ // Function to return the sum of the elements
// of all possible pairs from the array
static int sumPairs( int arr[], int n)
{
// To store the required sum
int sum = 0 ;
// Nested loop for all possible pairs
for ( int i = 0 ; i < n; i++)
{
for ( int j = 0 ; j < n; j++)
{
// Add the sum of the elements
// of the current pair
sum += (arr[i] + arr[j]);
}
}
return sum;
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 1 , 2 , 3 };
int n = arr.length;
System.out.println(sumPairs(arr, n));
}
} // This code is contributed by PrinciRaj1992 |
# Python3 implementation of the approach # Function to return the summ of the elements # of all possible pairs from the array def summPairs(arr, n):
# To store the required summ
summ = 0
# Nested loop for all possible pairs
for i in range (n):
for j in range (n):
# Add the summ of the elements
# of the current pair
summ + = (arr[i] + arr[j])
return summ
# Driver code arr = [ 1 , 2 , 3 ]
n = len (arr)
print (summPairs(arr, n))
# This code is contributed by Mohit Kumar |
// C# implementation of the approach using System;
class GFG
{ // Function to return the sum of the elements
// of all possible pairs from the array
static int sumPairs( int []arr, int n)
{
// To store the required sum
int sum = 0;
// Nested loop for all possible pairs
for ( int i = 0; i < n; i++)
{
for ( int j = 0; j < n; j++)
{
// Add the sum of the elements
// of the current pair
sum += (arr[i] + arr[j]);
}
}
return sum;
}
// Driver code
public static void Main(String[] args)
{
int []arr = {1, 2, 3};
int n = arr.Length;
Console.WriteLine(sumPairs(arr, n));
}
} // This code is contributed by PrinciRaj1992 |
<script> // Javascript implementation of the approach // Function to return the sum of the elements // of all possible pairs from the array function sumPairs(arr, n)
{ // To store the required sum
var sum = 0;
var i, j;
// Nested loop for all possible pairs
for (i = 0; i < n; i++)
{
for (j = 0; j < n; j++)
{
// Add the sum of the elements
// of the current pair
sum += (arr[i] + arr[j]);
}
}
return sum;
} // Driver code var arr = [ 1, 2, 3 ];
var n = arr.length;
document.write(sumPairs(arr, n)); // This code is contributed by ipg2016107 </script> |
36
Time Complexity: O(N2)
Auxiliary Space: O(1)
Efficient approach: It can be observed that each element appears exactly (2 * N) times as one of the elements of the pair (x, y). Exactly N times as x and exactly N times as y.
Below is the implementation of the above approach:
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;
// Function to return the sum of the elements // of all possible pairs from the array int sumPairs( int arr[], int n)
{ // To store the required sum
int sum = 0;
// For every element of the array
for ( int i = 0; i < n; i++) {
// It appears (2 * n) times
sum = sum + (arr[i] * (2 * n));
}
return sum;
} // Driver code int main()
{ int arr[] = { 1, 2, 3 };
int n = sizeof (arr) / sizeof (arr[0]);
cout << sumPairs(arr, n);
return 0;
} |
// Java implementation of the approach import java.util.*;
class GFG
{ // Function to return the sum of the elements // of all possible pairs from the array static int sumPairs( int arr[], int n)
{ // To store the required sum
int sum = 0 ;
// For every element of the array
for ( int i = 0 ; i < n; i++)
{
// It appears (2 * n) times
sum = sum + (arr[i] * ( 2 * n));
}
return sum;
} // Driver code static public void main(String []arg)
{ int arr[] = { 1 , 2 , 3 };
int n = arr.length;
System.out.println(sumPairs(arr, n));
} } // This code is contributed by 29AjayKumar |
# Python3 implementation of the approach # Function to return the sum of the elements # of all possible pairs from the array def sumPairs(arr, n) :
# To store the required sum
sum = 0 ;
# For every element of the array
for i in range (n) :
# It appears (2 * n) times
sum = sum + (arr[i] * ( 2 * n));
return sum ;
# Driver code if __name__ = = "__main__" :
arr = [ 1 , 2 , 3 ];
n = len (arr);
print (sumPairs(arr, n));
# This code is contributed by AnkitRai01 |
// C# implementation of the approach using System;
class GFG
{ // Function to return the sum of the elements // of all possible pairs from the array static int sumPairs( int []arr, int n)
{ // To store the required sum
int sum = 0;
// For every element of the array
for ( int i = 0; i < n; i++)
{
// It appears (2 * n) times
sum = sum + (arr[i] * (2 * n));
}
return sum;
} // Driver code static public void Main(String []arg)
{ int []arr = { 1, 2, 3 };
int n = arr.Length;
Console.WriteLine(sumPairs(arr, n));
} } // This code contributed by Rajput-Ji |
<script> // Javascript implementation of the approach // Function to return the sum of the elements // of all possible pairs from the array function sumPairs(arr, n)
{ // To store the required sum
let sum = 0;
// For every element of the array
for (let i = 0; i < n; i++) {
// It appears (2 * n) times
sum = sum + (arr[i] * (2 * n));
}
return sum;
} // Driver code let arr = [ 1, 2, 3 ];
let n = arr.length;
document.write(sumPairs(arr, n));
</script> |
36
Time Complexity: O(N)
Auxiliary Space: O(1)