查找给定数组中偶数对的计数

给定一个数组arr[] ，任务是在数组中找到计数的奇偶对。
例子：

Input: arr[] = { 1, 2, 1, 3 } 
Output: 2 
Explanation: 
The 2 pairs of the form (even, odd) are {2, 1} and {2, 3}.
 

Input: arr[] = { 5, 4, 1, 2, 3} 
Output: 3   

天真的方法：

1. 运行两个嵌套循环以获取数组中所有可能的数字对。
2. 对于每一对，检查 a[i] 是否为偶数且 a[j] 是否为奇数。
3. 如果是，则将所需计数加一。
4. 检查完所有对后，最后打印计数。

下面是上述方法的实现：

C++

// C++ program to count the pairs in array
// of the form (even, odd)
 
#include 
using namespace std;
 
// Function to count the pairs in array
// of the form (even, odd)
int findCount(int arr[], int n)
{
    // variable to store count of such pairs
    int res = 0;
 
    // Iterate through all pairs
    for (int i = 0; i < n - 1; i++)
        for (int j = i + 1; j < n; j++)
 
            // Increment count
            // if condition is satisfied
            if ((arr[i] % 2 == 0)
                && (arr[j] % 2 == 1)) {
                res++;
            }
 
    return res;
}
 
// Driver code
int main()
{
    int a[] = { 5, 4, 1, 2, 3 };
    int n = sizeof(a) / sizeof(a[0]);
    cout << findCount(a, n);
    return 0;
}

Java

// Java program to count the pairs in array
// of the form (even, odd)
import java.util.*;
 
class GFG{
 
// Function to count the pairs in array
// of the form (even, odd)
static int findCount(int arr[], int n)
{
    // Variable to store count of such pairs
    int res = 0;
 
    // Iterate through all pairs
    for (int i = 0; i < n - 1; i++)
        for (int j = i + 1; j < n; j++)
 
            // Increment count
            // if condition is satisfied
            if ((arr[i] % 2 == 0) &&
                (arr[j] % 2 == 1))
            {
                res++;
            }
    return res;
}
 
// Driver code
public static void main(String[] args)
{
    int a[] = { 5, 4, 1, 2, 3 };
    int n = a.length;
    System.out.print(findCount(a, n));
}
}
 
// This code is contributed by Rohit_ranjan

Python3

# Python3 program to count the pairs
# in array of the form (even, odd)
 
# Function to count the pairs in 
# array of the form (even, odd)
def findCount(arr, n):
 
    # Variable to store count
    # of such pairs
    res = 0
 
    # Iterate through all pairs
    for i in range(0, n - 1):
        for j in range(i + 1, n):
 
            # Increment count
            # if condition is satisfied
            if ((arr[i] % 2 == 0) and
                (arr[j] % 2 == 1)):
                res = res + 1
 
    return res
 
# Driver code
a = [ 5, 4, 1, 2, 3 ]
n = len(a)
 
print(findCount(a, n))
 
# This code is contributed by PratikBasu

C#

// C# program to count the pairs in array
// of the form (even, odd)
using System;
 
class GFG{
 
// Function to count the pairs in array
// of the form (even, odd)
static int findCount(int []arr, int n)
{
 
    // Variable to store count of such pairs
    int res = 0;
 
    // Iterate through all pairs
    for(int i = 0; i < n - 1; i++)
       for(int j = i + 1; j < n; j++)
        
          // Increment count
          // if condition is satisfied
          if ((arr[i] % 2 == 0) &&
              (arr[j] % 2 == 1))
          {
              res++;
          }
           
    return res;
}
 
// Driver code
public static void Main(String[] args)
{
    int []a = { 5, 4, 1, 2, 3 };
    int n = a.Length;
     
    Console.Write(findCount(a, n));
}
}
 
// This code is contributed by Rohit_ranjan

Javascript

 
// Javascript program to count the pairs in array
// of the form (even, odd)
 
// Function to count the pairs in array
// of the form (even, odd)
function findCount(arr, n)
{
    // variable to store count of such pairs
    let res = 0;
 
    // Iterate through all pairs
    for (let i = 0; i < n - 1; i++)
        for (let j = i + 1; j < n; j++)
 
            // Increment count
            // if condition is satisfied
            if ((arr[i] % 2 == 0)
                && (arr[j] % 2 == 1)) {
                res++;
            }
 
    return res;
}
 
// Driver code
let a = [ 5, 4, 1, 2, 3 ];
let n = a.length;
document.write(findCount(a, n));
 

输出：

3

时间复杂度： O(n ² )

辅助空间： O(1)

有效的方法：

1. 对于从索引 0 开始的每个元素，我们将检查它是否为偶数。
2. 如果是偶数，我们将计数增加 1。
3. 否则，我们将通过计数增加我们的最终答案。

下面是上述方法的实现：

C++

// C++ program to count the pairs in array
// of the form (even, odd)
 
#include 
using namespace std;
 
// Function to count the pairs in array
// of the form (even, odd)
int findCount(int arr[], int n)
{
    int count = 0, ans = 0;
    for (int i = 0; i < n; i++) {
 
        // check if number is even or not
        if (arr[i] % 2 == 0)
            count++;
        else
            ans = ans + count;
    }
    return ans;
}
 
// Driver code
int main()
{
    int a[] = { 5, 4, 1, 2, 3 };
    int n = sizeof(a) / sizeof(a[0]);
    cout << findCount(a, n);
    return 0;
}

Java

// Java program to count the pairs
// in array of the form (even, odd)
class GFG{
 
// Function to count the pairs in 
// array of the form (even, odd)
static int findCount(int arr[], int n)
{
    int count = 0, ans = 0;
    for(int i = 0; i < n; i++)
    {
         
       // Check if number is even
       // or not
       if (arr[i] % 2 == 0)
       {
           count++;
       }
       else
       {
           ans = ans + count;
       }
    }
    return ans;
}
 
// Driver code
public static void main(String[] args)
{
    int a[] = { 5, 4, 1, 2, 3 };
    int n = a.length;
     
    System.out.print(findCount(a, n));
}
}
 
// This code is contributed by amal kumar choubey

蟒蛇3

# Python3 program to count the pairs
# in array of the form (even, odd)
 
# Function to count the pairs in
# array of the form (even, odd)
def findCount(arr, n):
 
    count = 0
    ans = 0
     
    for i in range(0, n):
 
        # Check if number is even or not
        if (arr[i] % 2 == 0):
            count = count + 1
        else:
            ans = ans + count
             
    return ans
 
# Driver code
a = [ 5, 4, 1, 2, 3 ]
n = len(a)
 
print(findCount(a, n))
 
# This code is contributed by PratikBasu

C＃

// C# program to count the pairs in
// array of the form (even, odd)
using System;
 
class GFG{
 
// Function to count the pairs in
// array of the form (even, odd)
static int findCount(int []arr, int n)
{
    int count = 0, ans = 0;
    for(int i = 0; i < n; i++)
    {
 
       // Check if number is even or not
       if (arr[i] % 2 == 0)
       {
           count++;  
       }
       else
       {
           ans = ans + count;
       }
    }
    return ans;
}
 
// Driver code
public static void Main()
{
    int []a = { 5, 4, 1, 2, 3 };
    int n = a.Length;
     
    Console.WriteLine(findCount(a, n));
}
}
 
// This code is contributed by Code_Mech

Javascript

// Javascript program to count the pairs in array
// of the form (even, odd)
 
// Function to count the pairs in array
// of the form (even, odd)
function findCount(arr, n)
{
    let count = 0, ans = 0;
    for (let i = 0; i < n; i++) {
 
        // check if number is even or not
        if (arr[i] % 2 == 0)
            count++;
        else
            ans = ans + count;
    }
    return ans;
}
 
// Driver code
let a = [ 5, 4, 1, 2, 3 ];
let n = a.length;
document.write(findCount(a, n));
 
// This code is contributed by subhammahato348.

输出：

3

时间复杂度： O(n)

辅助空间： 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