二进制数组的所有对的乘积之和

给定大小为N的二进制数组arr [] ，任务是打印给定数组元素的所有对的乘积之和。

注意：二进制数组仅包含0和1。

例子：

Input: arr[] = {0, 1, 1, 0, 1}
Output: 3
Explanation: Sum of product of all possible pairs are: {0 × 1 + 0 × 1 + 0 × 0 + 0 × 1 + 1 × 1 + 1 × 0 + 1 × 1 + 1 × 0 + 1 × 1 + 0 × 1}.
Therefore, the required output is 3.

Input: arr[] = {1, 1, 1, 1}
Output: 6

天真的方法：解决问题的最简单方法是使用从数组生成所有可能的对并计算其乘积之和。

时间复杂度： O(N ² )
辅助空间： O(1)

高效方法：要优化上述方法，请仅考虑两个元素均为1的那些对。以下是观察结果：

If there is a pair (arr[i], arr[j]) where arr[i] × arr[j] = 1, then arr[i] and arr[j] must be 1.

Total number of pairs that satisfy (arr[i] × arr[j] = 1) are: 
=> 

$\binom{cntOne}{2}$
 

=> cntOne × (cntOne – 1) / 2 
where, cntOne is the count of 1s in the given array

 

请按照以下步骤解决问题：

• 初始化变量cntOne，以存储给定数组中1 s的计数。
• 最后，返回cntOne *(cntOne – 1)/ 2的值。

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
#include 
using namespace std;
 
// Function to print the sum of product
// of all pairs of the given array
int productSum(int arr[], int N)
{
     
    // Stores count of one in
    // the given array
    int cntOne = 0;
  
    for(int i = 0; i < N; i++)
    {
         
        // If current element is 1
        if (arr[i] == 1)
  
            // Increase count
            cntOne++;
    }
  
    // Return the sum of product
    // of all pairs
    return cntOne * (cntOne - 1) / 2;
}
  
// Driver Code
int main()
{
    int arr[] = { 0, 1, 1, 0, 1 };
     
    // Stores the size of
    // the given array
    int n = sizeof(arr) / sizeof(arr[0]);
     
    cout << productSum(arr, n) << endl;
}
 
// This code is contributed by code_hunt

Java

// Java program to implement
// the above approach
import java.util.*;
 
class GFG {
 
    // Function to print the sum of product
    // of all pairs of the given array
    static int productSum(int[] arr)
    {
 
        // Stores count of one in
        // the given array
        int cntOne = 0;
 
        // Stores the size of
        // the given array
        int N = arr.length;
 
        for (int i = 0; i < N; i++) {
 
            // If current element is 1
            if (arr[i] == 1)
 
                // Increase count
                cntOne++;
        }
 
        // Return the sum of product
        // of all pairs
        return cntOne * (cntOne - 1) / 2;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
 
        int[] arr = { 0, 1, 1, 0, 1 };
 
        System.out.println(productSum(arr));
    }
}

Python3

# Python3 program to implement
# the above approach
 
# Function to prthe sum of product
# of all pairs of the given array
def productSum(arr):
  
    # Stores count of one in
    # the given array
    cntOne = 0
     
    # Stores the size of
    # the given array
    N = len(arr)
  
    for i in range(N):
  
        # If current element is 1
        if (arr[i] == 1):
  
            # Increase count
            cntOne += 1
     
    # Return the sum of product
    # of all pairs
    return cntOne * (cntOne - 1) // 2
 
# Driver Code
arr = [ 0, 1, 1, 0, 1 ]
 
print(productSum(arr))
 
# This code is contributed by code_hunt

C#

// C# program to implement
// the above approach
using System;
 
class GFG{
  
// Function to print the sum of product
// of all pairs of the given array
static int productSum(int[] arr)
{
     
    // Stores count of one in
    // the given array
    int cntOne = 0;
 
    // Stores the size of
    // the given array
    int N = arr.Length;
 
    for(int i = 0; i < N; i++)
    {
         
        // If current element is 1
        if (arr[i] == 1)
 
            // Increase count
            cntOne++;
    }
 
    // Return the sum of product
    // of all pairs
    return cntOne * (cntOne - 1) / 2;
}
 
// Driver Code
public static void Main()
{
    int[] arr = { 0, 1, 1, 0, 1 };
 
    Console.Write(productSum(arr));
}
}
 
// This code is contributed by code_hunt

Javascript

 
// Javascript program to implement
// the above approach
 
// Function to print the sum of product
// of all pairs of the given array
function productSum(arr, N)
{
     
    // Stores count of one in
    // the given array
    let cntOne = 0;
 
    for(let i = 0; i < N; i++)
    {
         
        // If current element is 1
        if (arr[i] == 1)
 
            // Increase count
            cntOne++;
    }
 
    // Return the sum of product
    // of all pairs
    return cntOne * (cntOne - 1) / 2;
}
 
// Driver Code
 
    let arr = [ 0, 1, 1, 0, 1 ];
     
    // Stores the size of
    // the given array
    let n = arr.length;
     
    document.write(productSum(arr, n) + "<br>");
 
 
// This code is contributed by Mayank Tyagi
 

输出：

3

时间复杂度： O(N)
辅助空间： O(1)