从一个数组中计数对，它们的和与差之积等于1

给定大小为N的数组arr [] ，任务是对可能的成对数组元素(arr [i]，arr [j])进行计数，使得(arr [i] + arr [j])*(arr [i] – arr [j])为1。

例子：

Input: arr[] = {3, 1, 1, 0} 
Output: 2 
Explanation: 
The two possible pairs are: 

1. (arr[1] + arr[3]) * (arr[1] – arr[3]) = 1
2. (arr[2] + arr[3]) * (arr[2] – arr[3]) = 1

Input: arr[] = {12, 0, 1, 1, 14, 0, 9, 0} 
Output: 4 
Explanation: 
The four possible pairs are as follows: 

1. (3, 6): (arr[3] + arr[6]) * (arr[3] – arr[6]) = 1
2. (4, 6): (arr[4] + arr[6]) * (arr[4] – arr[6]) = 1
3. (3, 8): (arr[3] + arr[8]) * (arr[3] – arr[8]) = 1
4. (3, 6): (arr[4] + arr[8]) * (arr[4] – arr[8]) = 1

天真的方法：最简单的方法是遍历数组，并从给定的数组中生成所有可能的对，并对那些总和与差之和为1的对进行计数。最后，打印完成上述步骤后获得的最终计数。

时间复杂度： O(N ² )
辅助空间： O(N)

高效的方法：可以将任意一对数组元素(arr [i]，arr [j])的给定条件表示为：

(arr[i] + arr[j]) * (arr[i] – arr[j]) = 1 
=> (arr[i]² – arr[j]²) = 1 

因此，可以得出结论，仅对条件arr [i] = 1和arr [j] = 0且i 可以满足要求的条件。请按照以下步骤解决问题：

• 初始化两个变量oneCount和desirePairs分别存储1的计数和所需的对。
• 遍历给定的数组并检查以下内容：
  • 如果arr [i] = 1：将oneCount加1 。
  • 如果arr [i] = 0：将到目前为止获得的1 s的计数加到requiredPairs中。
• 完成上述步骤后，将desirablePairs的值打印为所需答案。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to count the desired
// number of pairs
int countPairs(int arr[], int n)
{
    // Initialize oneCount
    int oneCount = 0;
 
    // Initialize the desiredPair
    int desiredPair = 0;
 
    // Traverse the given array
    for (int i = 0; i < n; i++) {
 
        // If 1 is encountered
        if (arr[i] == 1) {
            oneCount++;
        }
 
        // If 0 is encountered
        if (arr[i] == 0) {
 
            // Update count of pairs
            desiredPair += oneCount;
        }
    }
 
    // Return the final count
    return desiredPair;
}
 
// Driver Code
int main()
{
    // Given array arr[]
    int arr[] = { 3, 1, 1, 0 };
    int N = sizeof(arr) / sizeof(arr[0]);
    // Function Call
    cout << countPairs(arr, N);
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
 
class GFG{
 
// Function to count the desired
// number of pairs
static int countPairs(int arr[], int n)
{
     
    // Initialize oneCount
    int oneCount = 0;
 
    // Initialize the desiredPair
    int desiredPair = 0;
 
    // Traverse the given array
    for(int i = 0; i < n; i++)
    {
         
        // If 1 is encountered
        if (arr[i] == 1)
        {
            oneCount++;
        }
 
        // If 0 is encountered
        if (arr[i] == 0)
        {
             
            // Update count of pairs
            desiredPair += oneCount;
        }
    }
 
    // Return the final count
    return desiredPair;
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given array arr[]
    int arr[] = { 3, 1, 1, 0 };
    int N = arr.length;
 
    // Function call
    System.out.println(countPairs(arr, N));
}
}
 
// This code is contributed by sanjoy_62

Python3

# Python3 program for the above approach
 
# Function to count the desired
# number of pairs
def countPairs(arr, n):
     
    # Initialize oneCount
    oneCount = 0
 
    # Initialize the desiredPair
    desiredPair = 0
 
    # Traverse the given array
    for i in range(n):
 
        # If 1 is encountered
        if (arr[i] == 1):
            oneCount += 1
 
        # If 0 is encountered
        if (arr[i] == 0):
 
            # Update count of pairs
            desiredPair += oneCount
 
    # Return the final count
    return desiredPair
 
# Driver Code
if __name__ == '__main__':
     
    # Given array arr[]
    arr = [ 3, 1, 1, 0 ]
    N = len(arr)
     
    # Function call
    print(countPairs(arr, N))
 
# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Function to count the desired
// number of pairs
static int countPairs(int []arr, int n)
{
     
    // Initialize oneCount
    int oneCount = 0;
 
    // Initialize the desiredPair
    int desiredPair = 0;
 
    // Traverse the given array
    for(int i = 0; i < n; i++)
    {
         
        // If 1 is encountered
        if (arr[i] == 1)
        {
            oneCount++;
        }
 
        // If 0 is encountered
        if (arr[i] == 0)
        {
             
            // Update count of pairs
            desiredPair += oneCount;
        }
    }
 
    // Return the readonly count
    return desiredPair;
}
 
// Driver Code
public static void Main(String[] args)
{
     
    // Given array []arr
    int []arr = { 3, 1, 1, 0 };
    int N = arr.Length;
 
    // Function call
    Console.WriteLine(countPairs(arr, N));
}
}
 
// This code is contributed by gauravrajput1

输出：

2

时间复杂度： O(N)
辅助空间： O(N)