最多一次转换，最大化给定数组中奇数对的数量

给定一个大小为 n的整数数组，通过最多更改一个数字，从数组中找到最大对数，使得它们的和为奇数。

例子：

Input: N = 6, arr = [1, 5, 3, 6, 8, 0]
Output: 3
Explanation: There are 3 even and 3 odd numbers

Input: N = 8, arr = [1, 5, 3, 6, 8, 0, 2, 4]
Output: 4
Explanation: There are 5 even and 3 odd numbers, and an even number can be converted to an odd number. Hence final output will be 4.

方法：要使元素之和为奇数，一个数应为奇数，另一个数应为偶数。可以按照以下步骤进行：

• 计算x等于奇数元素的数量
• 计算 Y 等于偶数元素的数量。
• 用 x 和 y 的最小值初始化 answer 变量
• 如果 |x – y| >= 2，然后将答案加 1（即，如果 y – x >= 2 则我们可以将一个奇数转换为偶数，如果 x – y >= 2 那么我们可以将一个偶数转换为奇数）
• 结果答案变量是我们需要的答案。

下面是上述方法的实现：

C++

// C++ code implementation for above approach
#include 
using namespace std;
 
// To find maximum number of pairs in
// array with conversion of at most one element
int maximumNumberofpairs(int n, int arr[])
{
    // Initialize count of even elements
    int x = 0;
 
    // Initialize count of odd elements
    int y = 0;
    for (int i = 0; i < n; i++) {
 
        // If current number is even
        // then increment x by 1
        if (arr[i] % 2 == 0) {
            x++;
        }
 
        // If current number is odd
        // then increment y by 1
        else {
            y++;
        }
    }
 
    // Initialize the answer by min(x, y)
    int answer = min(x, y);
 
    // If difference in count of odd and even
    // is more than 2 than increment answer
    if (abs(x - y) >= 2) {
        answer++;
    }
 
    // Return final answer
    return answer;
}
// Driver code
int main()
{
    // Given array
    int arr[] = { 1, 2, 4, 6, 5, 10, 12 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << maximumNumberofpairs(n, arr) << endl;
 
    return 0;
}

Java

// Java implementation for the above approach
import java.io.*;
 
class GFG
 {
 
  // To find maximum number of pairs in
// array with conversion of at most one element
static int maximumNumberofpairs(int n, int arr[])
{
 
    // Initialize count of even elements
    int x = 0;
   
    // Initialize count of odd elements
    int y = 0;
    for (int i = 0; i < n; i++) {
   
        // If current number is even
        // then increment x by 1
        if (arr[i] % 2 == 0) {
            x++;
        }
   
        // If current number is odd
        // then increment y by 1
        else {
            y++;
        }
    }
   
    // Initialize the answer by min(x, y)
    int answer = Math.min(x, y);
   
    // If difference in count of odd and even
    // is more than 2 than increment answer
    if (Math.abs(x - y) >= 2) {
        answer++;
    }
   
    // Return final answer
    return answer;
}
 
// Driver code
    public static void main (String[] args) {
      // Given array
    int arr[] = { 1, 2, 4, 6, 5, 10, 12 };
    int n = arr.length;
 
        System.out.println( maximumNumberofpairs(n, arr));
    }
}
 
// This code is contributed by Potta Lokesh

Python

# python 3 code implementation for above approach
 
# To find maximum number of pairs in
# array with conversion of at most one element
def maximumNumberofpairs(n, arr):
    # Initialize count of even elements
    x = 0
 
    # Initialize count of odd elements
    y = 0
    for i in range(n):
        # If current number is even
        # then increment x by 1
        if (arr[i] % 2 == 0):
            x += 1
 
        # If current number is odd
        # then increment y by 1
        else:
            y += 1
 
    # Initialize the answer by min(x, y)
    answer = min(x, y)
 
    # If difference in count of odd and even
    # is more than 2 than increment answer
    if (abs(x - y) >= 2):
        answer += 1
 
    # Return final answer
    return answer
 
# Driver code
if __name__ == '__main__':
    # Given array
    arr = [1, 2, 4, 6, 5, 10, 12]
    n = len(arr)
    print(maximumNumberofpairs(n, arr))
 
# This code is contributed by ipg2016107.

C#

// C# implementation for the above approach
 
using System;
 
public class GFG
 {
 
  // To find maximum number of pairs in
// array with conversion of at most one element
static int maximumNumberofpairs(int n, int []arr)
{
 
    // Initialize count of even elements
    int x = 0;
   
    // Initialize count of odd elements
    int y = 0;
    for (int i = 0; i < n; i++) {
   
        // If current number is even
        // then increment x by 1
        if (arr[i] % 2 == 0) {
            x++;
        }
   
        // If current number is odd
        // then increment y by 1
        else {
            y++;
        }
    }
   
    // Initialize the answer by min(x, y)
    int answer = Math.Min(x, y);
   
    // If difference in count of odd and even
    // is more than 2 than increment answer
    if (Math.Abs(x - y) >= 2) {
        answer++;
    }
   
    // Return final answer
    return answer;
}
 
// Driver code
    public static void Main (string[] args) {
     
    // Given array
    int []arr = { 1, 2, 4, 6, 5, 10, 12 };
    int n = arr.Length;
 
        Console.WriteLine(maximumNumberofpairs(n, arr));
    }
}
 
// This code is contributed by AnkThon

JavaScript

// JavaScript implementation for the above approach
 
// To find maximum number of pairs in
// array with conversion of at most one element
function maximumNumberofpairs(n, arr)
{
 
    // Initialize count of even elements
    var x = 0;
 
    // Initialize count of odd elements
    var y = 0;
    for (var i = 0; i < n; i++) {
 
        // If current number is even
        // then increment x by 1
        if (arr[i] % 2 == 0) {
            x++;
        }
 
        // If current number is odd
        // then increment y by 1
        else {
            y++;
        }
    }
 
    // Initialize the answer by min(x, y)
    var answer = Math.min(x, y);
 
    // If difference in count of odd and even
    // is more than 2 than increment answer
    if (Math.abs(x - y) >= 2) {
        answer++;
    }
 
    // Return final answer
    return answer;
}
 
// Driver code
// Given array
var arr = [ 1, 2, 4, 6, 5, 10, 12 ];
var n = arr.length;
 
document.write( maximumNumberofpairs(n, arr));
     
 
// This code is contributed by shivanisinghss2110
 

输出：

3

时间复杂度： O(N)
辅助空间： O(1)