总和等于乘积的数组中的对数

给定数组arr [] ，任务是查找数组中的对数(arr [i]，arr [j]) ，以使arr [i] + arr [j] = arr [i] * arr [j ]
例子：

Input: arr[] = {2, 2, 3, 4, 6} 
Output: 1 
(2, 2) is the only possible pair as (2 + 2) = (2 * 2) = 4.
Input: arr[] = {1, 2, 3, 4, 5} 
Output: 0 

方法：满足给定条件的唯一可能的整数对是(0，0)和(2，2) 。因此，现在的任务是计算数组中0和2的数量，并将它们分别存储在cnt0和cnt2中，然后所需的计数将是(cnt0 *(cnt0 – 1))/ 2 +(cnt2 *(cnt2 – 1) ))/ 2 。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the count
// of the required pairs
int sumEqualProduct(int a[], int n)
{
    int zero = 0, two = 0;
 
    // Find the count of 0s
    // and 2s in the array
    for (int i = 0; i < n; i++) {
        if (a[i] == 0) {
            zero++;
        }
        if (a[i] == 2) {
            two++;
        }
    }
 
    // Find the count of required pairs
    int cnt = (zero * (zero - 1)) / 2
              + (two * (two - 1)) / 2;
 
    // Return the count
    return cnt;
}
 
// Driver code
int main()
{
    int a[] = { 2, 2, 3, 4, 2, 6 };
    int n = sizeof(a) / sizeof(a[0]);
 
    cout << sumEqualProduct(a, n);
 
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
 
class GFG {
    // Function to return the count
    // of the required pairs
    static int sumEqualProduct(int a[], int n)
    {
        int zero = 0, two = 0;
 
        // Find the count of 0s
        // and 2s in the array
        for (int i = 0; i < n; i++) {
            if (a[i] == 0) {
                zero++;
            }
            if (a[i] == 2) {
                two++;
            }
        }
 
        // Find the count of required pairs
        int cnt = (zero * (zero - 1)) / 2
                  + (two * (two - 1)) / 2;
 
        // Return the count
        return cnt;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int a[] = { 2, 2, 3, 4, 2, 6 };
        int n = a.length;
 
        System.out.print(sumEqualProduct(a, n));
    }
}
 
// This code is contributed by Rajput-Ji

Python3

# Python 3 implementation of the approach
 
# Function to return the count
# of the required pairs
def sumEqualProduct(a, n):
    zero = 0
    two = 0
     
    # Find the count of 0s
    # and 2s in the array
    for i in range(n):
        if a[i] == 0:
            zero += 1
        if a[i] == 2:
            two += 1
             
    # Find the count of required pairs
    cnt = (zero * (zero - 1)) // 2 + \
            (two * (two - 1)) // 2
     
    # Return the count
    return cnt
     
# Driver code
a = [ 2, 2, 3, 4, 2, 6 ]
n = len(a)
 
print(sumEqualProduct(a, n))
 
# This code is contributed by Ankit kumar

C#

// C# implementation of the approach
using System;
 
class GFG
{
 
// Function to return the count
// of the required pairs
static int sumEqualProduct(int []a, int n)
{
    int zero = 0, two = 0;
 
    // Find the count of 0s
    // and 2s in the array
    for (int i = 0; i < n; i++)
    {
        if (a[i] == 0)
        {
            zero++;
        }
        if (a[i] == 2)
        {
            two++;
        }
    }
 
    // Find the count of required pairs
    int cnt = (zero * (zero - 1)) / 2 +
               (two * (two - 1)) / 2;
 
    // Return the count
    return cnt;
}
 
// Driver code
public static void Main(String[] args)
{
    int []a = { 2, 2, 3, 4, 2, 6 };
    int n = a.Length;
 
    Console.Write(sumEqualProduct(a, n));
}
}
 
// This code is contributed by 29AjayKumar

输出：

3

时间复杂度： O(N)

辅助空间： O(1)