Encuentre la suma de todos los pares posibles en una array de N elementos

Dada una array arr[] de N enteros, la tarea es encontrar la suma de todos los pares posibles de la array dada. Tenga en cuenta que, 

1. (arr[i], arr[i]) también se considera un par válido.
2. (arr[i], arr[j]) y (arr[j], arr[i]) se consideran dos pares diferentes.

Ejemplos: 

Entrada: arr[] = {1, 2} 
Salida: 12 
Todos los pares válidos son (1, 1), (1, 2), (2, 1) y (2, 2). 
1 + 1 + 1 + 2 + 2 + 1 + 2 + 2 = 12

Entrada: arr[] = {1, 2, 3, 1, 4} 
Salida: 110 
 

Enfoque ingenuo: encuentre todos los pares posibles y calcule la suma de los elementos de cada par.

A continuación se muestra la implementación del enfoque anterior:  

// 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

Complejidad del tiempo: O(N ² )

Espacio Auxiliar: O(1)

Enfoque eficiente: Se puede observar que cada elemento aparece exactamente (2*N) veces como uno de los elementos del par (x, y) . Exactamente N veces como x y exactamente N veces como y .

A continuación se muestra la implementación del enfoque anterior:  

// 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

Complejidad temporal: O(N)
 Espacio auxiliar: O(1)