在N个元素的数组中找到所有可能的对的总和

给定N个整数的数组arr [] ，任务是从给定数组中找到所有可能的对的和。注意，

1. (arr [i]，arr [i])也被视为有效对。
2. (arr [i]，arr [j])和(arr [j]，arr [i])被视为两个不同的对。

例子：

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 = 12

Input: arr[] = {1, 2, 3, 1, 4}
Output: 110

天真的方法：找到所有可能的对，并计算每对元素的总和。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
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

// 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

# 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#

// 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

输出：

36

时间复杂度： O(N ² )

高效的方法：可以观察到，每个元素作为对中的一个元素(x，y)的出现正好是(2 * N)次。正好是x的N倍，正好是y的N倍。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
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

// 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

# 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＃

// 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

输出：

36

时间复杂度： O(N)