具有相同奇偶校验的所有可能的数组元素对计数

给定整数数组A [] ，任务是找到对总数，以使每对包含偶数或奇数元素。仅当i！= j时，才能形成有冠词对( A [i] ， A [j] )。

例子：

Input: A[ ] = {1, 2, 3, 1, 3} 
Output: 6 
Explanation: 
Possible odd pairs = (1, 3), (1, 1), (1, 3), (3, 1), (3, 3), (1, 3) = 6 
Possible even pairs = 0 
Hence, total pairs = 6 + 0 = 6

Input: A[ ] = {8, 2, 3, 1, 4, 2} 
Output: 7 
Explanation: 
Possible odd pair = (3, 1) = 1 
Possible even pairs = (8, 2), (8, 4), (8, 2), (2, 4), (2, 2), (4, 2) = 6 
Hence, total pairs = 6 + 1 = 7 
 

天真的方法：
最简单的方法是生成所有可能的对。对于每对，检查两个元素是否都是奇数或均是偶数。如果是这样，请增加一个计数器。最终的数量将是必需的答案。

时间复杂度： O(N ² )

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to return the answer
int countPairs(int A[], int n)
{
    int count = 0, i, j;
     
    // Generate all possible pairs
    for(i = 0; i < n; i++)
    {
        for(j = i + 1; j < n; j++)
        {
             
            // Increment the count if
            // both even or both odd
            if ((A[i] % 2 == 0 &&
                 A[j] % 2 == 0) ||
                (A[i] % 2 != 0 &&
                 A[j] % 2 != 0))
                count++;
        }
    }
    return count;
}
 
// Driver Code
int main()
{
    int A[] = { 1, 2, 3, 1, 3 };
    int n = sizeof(A) / sizeof(int);
     
    cout << countPairs(A, n);
}
 
// This code is contributed by jrishabh99

Java

// Java program for above approach
 
import java.util.*;
 
class GFG {
 
    static int countPairs(
        int[] A, int n)
    {
        int count = 0, i, j;
 
        // Generate all possible pairs
        for (i = 0; i < n; i++) {
            for (j = i + 1; j < n; j++) {
 
                // Increment the count if
                // both even or both odd
                if ((A[i] % 2 == 0
                     && A[j] % 2 == 0)
                    || (A[i] % 2 != 0
                        && A[j] % 2 != 0))
                    count++;
            }
        }
        return count;
    }
 
    // Driver Code
    public static void main(
        String[] args)
    {
        int[] A = { 1, 2, 3, 1, 3 };
        int n = A.length;
        System.out.println(
            countPairs(A, n));
    }
}

Python3

# Python3 program for
# the above approach
 
# Function to return the answer
def countPairs(A, n):
 
    count = 0
     
    # Generate all possible pairs
    for i in range (n):
        for j in range (i + 1, n):
       
            # Increment the count if
            # both even or both odd
            if ((A[i] % 2 == 0 and
                 A[j] % 2 == 0) or
                (A[i] % 2 != 0 and
                 A[j] % 2 != 0)):
                count += 1
     
    return count
 
# Driver Code
if __name__ == "__main__":
   
    A = [1, 2, 3, 1, 3]
    n = len(A)
    print(countPairs(A, n))
     
# This code is contributed by Chitranayal

C#

// C# program for above approach
using System;
class GFG{
 
static int countPairs(int[] A, int n)
{
    int count = 0, i, j;
 
    // Generate all possible pairs
    for (i = 0; i < n; i++)
    {
        for (j = i + 1; j < n; j++)
        {
 
            // Increment the count if
            // both even or both odd
            if ((A[i] % 2 == 0 && A[j] % 2 == 0) ||
                (A[i] % 2 != 0 && A[j] % 2 != 0))
                count++;
        }
    }
    return count;
}
 
// Driver Code
public static void Main(String[] args)
{
    int[] A = { 1, 2, 3, 1, 3 };
    int n = A.Length;
    Console.Write(countPairs(A, n));
}
}
 
// This code is contributed by shivanisinghss2110

Javascript

 
// Javascript program for above approach
function countPairs(A, n)
{
    let count = 0, i, j;
 
    // Generate all possible pairs
    for(i = 0; i < n; i++)
    {
        for(j = i + 1; j < n; j++)
        {
             
            // Increment the count if
            // both even or both odd
            if ((A[i] % 2 == 0 && A[j] % 2 == 0) ||
                (A[i] % 2 != 0 && A[j] % 2 != 0))
                count++;
        }
    }
    return count;
}
 
// Driver code
let A = [ 1, 2, 3, 1, 3 ];
let n = A.length;
 
document.write(countPairs(A, n));
 
// This code is contributed by rameshtravel07
 

输出：

6

高效方法：
遍历数组并计数并将偶数和奇数存储在数组中，并根据各自的计数计算可能的对，并显示其总和。

Let the count of even and odd elements in the array be EC and OC respectively. 
Count of even pairs = ( EC * ( EC – 1 ) ) / 2 
Count of odd pairs = ( OC * ( OC – 1 ) ) / 2 
Hence, total number of possible pairs = (( EC * ( EC – 1 ) ) + ( OC * ( OC – 1 ) ))/ 2 
 

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
int countPairs(int A[], int n)
{
     
    // Store count of
    // even and odd elements
    int even = 0, odd = 0;
     
    for(int i = 0; i < n; i++)
    {
        if (A[i] % 2 == 0)
            even++;
        else
            odd++;
    }
 
    return (even * (even - 1)) / 2 +
             (odd * (odd - 1)) / 2;
}
 
// Driver code
int main()
{
    int A[] = { 1, 2, 3, 1, 3 };
    int n = sizeof(A) / sizeof(int);
     
    cout << countPairs(A, n);
}
 
// This code is contributed by jrishabh99

Java

// Java program for the above approach
 
import java.util.*;
 
class GFG {
 
    static int countPairs(
        int[] A, int n)
    {
        // Store count of
        // even and odd elements
        int even = 0, odd = 0;
 
        for (int i = 0; i < n; i++) {
 
            if (A[i] % 2 == 0)
                even++;
            else
                odd++;
        }
 
        return (even * (even - 1)) / 2
            + (odd * (odd - 1)) / 2;
    }
 
    // Driver Program
    public static void main(
        String[] args)
    {
 
        int[] A = { 1, 2, 3, 1, 3 };
        int n = A.length;
        System.out.println(
            countPairs(A, n));
    }
}

Python3

# Python3 program for the above approach
 
# Function to return count of pairs
def countPairs(A, n):
     
    # Store count of
    # even and odd elements
    even, odd = 0, 0
     
    for i in range(0, n):
        if A[i] % 2 == 0:
            even = even + 1
        else:
            odd = odd + 1
 
    return ((even * (even - 1)) // 2 +
              (odd * (odd - 1)) // 2)
 
# Driver code
A = [ 1, 2, 3, 1, 3 ]
n = len(A)
 
print(countPairs(A, n))
 
# This code is contributed by jrishabh99

C＃

// C# program for the above approach
using System;
 
class GFG{
 
static int countPairs(int[] A, int n)
{
     
    // Store count of
    // even and odd elements
    int even = 0, odd = 0;
 
    for(int i = 0; i < n; i++)
    {
    if (A[i] % 2 == 0)
        even++;
    else
        odd++;
    }
 
    return (even * (even - 1)) / 2 +
            (odd * (odd - 1)) / 2;
}
 
// Driver code
public static void Main()
{
    int[] A = { 1, 2, 3, 1, 3 };
    int n = A.Length;
     
    Console.Write(countPairs(A, n));
}
}
 
// This code is contributed by nidhi_biet

Java脚本

 
    // Javascript program for the above approach
     
    function countPairs(A, n)
    {
 
        // Store count of
        // even and odd elements
        let even = 0, odd = 0;
 
        for(let i = 0; i < n; i++)
        {
            if (A[i] % 2 == 0)
                even++;
            else
                odd++;
        }
 
        return (even * (even - 1)) / 2 + (odd * (odd - 1)) / 2;
    }
     
    let A = [ 1, 2, 3, 1, 3 ];
    let n = A.length;
      
    document.write(countPairs(A, n));
     

输出：

6

时间复杂度： O(N)