计算包含至少一个偶数值的数组中的对

给定一个数组arr[] ，任务是计算对，使得每一对 (arr[i], arr[j]) 中至少包含一个偶数元素，其中i != j 。

例子：

Input: arr[] = {1, 2, 3, 1, 3} 
Output: 4 
Explanation: 
Possible pairs are: (1, 2), (2, 3), (2, 1), (2, 3).

Input: arr[] = {8, 2, 3, 1, 4, 2} 
Output: 14 
Explanation: 
Possible pairs are: (8, 2), (8, 3), (8, 1), (8, 4), (8, 2), (2, 3), (2, 1), (2, 4), (2, 2), (3, 4), (3, 2), (1, 4), (1, 2), (4, 2). 

一个简单的方法是运行两个循环。一个一个选择每个元素，并为每个元素在数组右侧找到包含条件的元素，然后增加计数。
时间复杂度：

$O(N^{2})$

下面是上述方法的实现：

C++

// C++ implementation to count
// pairs in an array such that
// each pair contains at
// least one even element
#include
using namespace std;
  
// Function to count the pairs in
// the array such as there is at
// least one even element in each pair
int CountPairs(int arr[], int n)
{
    int count = 0;
  
    // Generate all possible pairs
    // and increment then count
    // if the condition is satisfied
    for(int i = 0; i < n; i++)
    {
       for(int j = i + 1; j < n; j++)
       {
          if (arr[i] % 2 == 0 ||
              arr[j] % 2 == 0)
              count++;
       }
    }
    return count;
}
  
// Driver code
int main()
{
    int arr[] = { 8, 2, 3, 1, 4, 2 };
    int n = sizeof(arr) / sizeof(int);
  
    // Function call
    cout << (CountPairs(arr, n));
}
  
// This code is contributed by rock_cool

Java

// Java implementation to Count
// pairs in an array such that
// each pair contains at
// least one even element
import java.util.*;
 
class GFG {
 
    // Function to count the pairs in
    // the array such as there is at
    // least one even element in each pair
    static int CountPairs(int[] arr, int n)
    {
 
        int count = 0;
 
        // Generate all possible pairs
        // and increment then count
        // if the condition is satisfied
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
 
                if (arr[i] % 2 == 0
                    || arr[j] % 2 == 0)
                    count++;
            }
        }
 
        return count;
    }
 
    // Driver code
    public static void main(String[] args)
    {
 
        int[] arr = { 8, 2, 3, 1, 4, 2 };
        int n = arr.length;
 
        // Function Call
        System.out.println(CountPairs(arr, n));
    }
}

Python3

# Python3 implementation to count
# pairs in an array such that
# each pair contains at
# least one even element
def CountPairs(arr, n):
     
    count = 0
     
    # Generate all possible pairs
    # and increment then count
    # if the condition is satisfied
    for i in range(n):
        for j in range(i + 1, n):
            if (arr[i] % 2 == 0 or
                arr[j] % 2 == 0):
                count += 1
                 
    return count
     
# Driver code
arr = [ 8, 2, 3, 1, 4, 2 ]
n = len(arr)
 
# Function call
print(CountPairs(arr, n))
 
# This code is contributed by rutvik_56

C#

// C# implementation to count
// pairs in an array such that
// each pair contains at
// least one even element
using System;
  
class GFG{
  
// Function to count the pairs in
// the array such as there is at
// least one even element in each pair
static int CountPairs(int[] arr, int n)
{
  
    int count = 0;
  
    // Generate all possible pairs
    // and increment then count
    // if the condition is satisfied
    for(int i = 0; i < n; i++)
    {
       for(int j = i + 1; j < n; j++)
       {
          if (arr[i] % 2 == 0 ||
              arr[j] % 2 == 0)
              count++;
       }
    }
    return count;
}
  
// Driver code
public static void Main(String[] args)
{
    int[] arr = { 8, 2, 3, 1, 4, 2 };
    int n = arr.Length;
  
    // Function Call
    Console.WriteLine(CountPairs(arr, n));
}
}
  
// This code is contributed by PrinciRaj1992

Javascript

 
// Javascript implementation to count
// pairs in an array such that each
// pair contains at least one even element
 
// Function to count the pairs in
// the array such as there is at
// least one even element in each pair
function CountPairs(arr, n)
{
    let count = 0;
 
    // Generate all possible pairs
    // and increment then count
    // if the condition is satisfied
    for(let i = 0; i < n; i++)
    {
       for(let j = i + 1; j < n; j++)
       {
          if (arr[i] % 2 == 0 ||
              arr[j] % 2 == 0)
              count++;
       }
    }
    return count;
}
 
// Driver code
let arr = [ 8, 2, 3, 1, 4, 2 ];
let n = arr.length;
 
// Function call
document.write(CountPairs(arr, n));
 
// This code is contributed by divyeshrabadiya07
 

输出：

14

有效的方法：这个想法是计算数组中的偶数和奇数元素，并包括只有一个偶数元素的对或两个对都是偶数元素。

• 恰好有一个偶数元素的对：恰好有一个偶数元素的对的计数将是：

$(even*odd)$

• 恰好具有两个偶数元素的对：恰好具有两个偶数元素的对的计数将是：

$\frac{(even*(even-1))}{2}$

因此，具有至少一个偶数元素的对的计数将是
$\frac{(even*(even-1))}{2} + (even*odd)$

下面是上述方法的实现：

C++

// C++ implementation to Count
// pairs in an array such that
// each pair contains at
// least one even element
#include 
using namespace std;
 
// Function to count the pairs in
// the array such as there is at
// least one even element in each pair
int CountPairs(int arr[], int n)
{
     
    // Store count of even
    // and odd elements
    int even = 0, odd = 0;
    for(int i = 0; i < n; i++)
    {
         
       // Check element is
       // even or odd
       if (arr[i] % 2 == 0)
           even++;
       else
           odd++;
    }
    return (even * (even - 1)) / 2 +
           (even * odd);
}
 
// Driver Code
int main()
{
    int arr[] = { 8, 2, 3, 1, 4, 2 };
    int n = sizeof(arr) / sizeof(int);
     
    cout << CountPairs(arr, n);
}
 
// This code is contributed by jrishabh99

Java

// Java implementation to Count
// pairs in an array such that
// each pair contains at
// least one even element
import java.util.*;
 
class GFG {
 
    // Function to count the pairs in
    // the array such as there is at
    // least one even element in each pair
    static int CountPairs(int[] arr, int n)
    {
        // strore count of even
        // and odd elements
        int even = 0, odd = 0;
 
        for (int i = 0; i < n; i++) {
 
            // check element is
            // even or odd
            if (arr[i] % 2 == 0)
                even++;
            else
                odd++;
        }
 
        return (even * (even - 1)) / 2
            + (even * odd);
    }
 
    // Driver Code
    public static void main(String[] args)
    {
 
        int[] arr = { 8, 2, 3, 1, 4, 2 };
        int n = arr.length;
        System.out.println(CountPairs(arr, n));
    }
}

蟒蛇3

# Python3 implementation to count
# pairs in an array such that
# each pair contains at
# least one even element
   
# Function to count the pairs in
# the array such as there is at
# least one even element in each pair
def CountPairs(arr, n):
       
    # Store count of even
    # and odd elements
    even = 0
    odd = 0
     
    for i in range(n):
           
       # Check element is
       # even or odd
       if (arr[i] % 2 == 0):
           even += 1
       else:
           odd += 1
      
    return ((even * (even - 1)) // 2 +
            (even * odd))
  
# Driver Code
arr = [ 8, 2, 3, 1, 4, 2 ]
n = len(arr)
       
print(CountPairs(arr, n))
 
# This code is contributed by code_hunt

C＃

// C# implementation to Count
// pairs in an array such that
// each pair contains at
// least one even element
using System;
class GFG{
  
// Function to count the pairs in
// the array such as there is at
// least one even element in each pair
static int CountPairs(int[] arr, int n)
{
      
    // Store count of even
    // and odd elements
    int even = 0, odd = 0;
  
    for(int i = 0; i < n; i++)
    {
         
       // Check element is
       // even or odd
       if (arr[i] % 2 == 0)
           even++;
       else
           odd++;
    }
    return (even * (even - 1)) / 2 +
           (even * odd);
}
  
// Driver Code
public static void Main()
{
    int[] arr = { 8, 2, 3, 1, 4, 2 };
    int n = arr.Length;
      
    Console.Write(CountPairs(arr, n));
}
}
  
// This code is contributed by Nidhi_biet

Javascript

    // Javascript implementation to Count
    // pairs in an array such that
    // each pair contains at
    // least one even element
     
    // Function to count the pairs in
    // the array such as there is at
    // least one even element in each pair
    function CountPairs(arr, n)
    {
 
        // Store count of even
        // and odd elements
        let even = 0, odd = 0;
        for(let i = 0; i < n; i++)
        {
 
           // Check element is
           // even or odd
           if (arr[i] % 2 == 0)
               even++;
           else
               odd++;
        }
        return (even * (even - 1)) / 2 + (even * odd);
    }
     
    let arr = [ 8, 2, 3, 1, 4, 2 ];
    let n = arr.length;
      
    document.write(CountPairs(arr, n));
     
    // This code is contributed by divyesh072019.

输出：

14

时间复杂度： O(N)
空间复杂度： O(1)

如果您想与行业专家一起参加直播课程，请参阅Geeks Classes Live