要删除的最小元素，使得相邻元素的总和始终是偶数

给定一个包含 N 个整数的数组。任务是消除最小数量的元素，使得在结果数组中任何两个相邻值的总和是偶数。

例子：

Input : arr[] = {1, 2, 3}
Output : 1
Remove 2 from the array.

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

方法：两个数字的总和是偶数，即使它们都是奇数或它们都是偶数。这意味着对于具有不同奇偶性的每一对连续数字，消除其中一个。
因此，要使相邻元素之和为偶数，所有元素都应该是奇数或偶数。所以下面的贪心算法有效：

• 按顺序浏览所有元素。
• 计算数组中的奇数和偶数元素。
• 返回最小计数。

下面是上述方法的实现：

C++

// C++ implementation of the above approach
#include 
using namespace std;
 
// Function to find minimum number of eliminations
// such that sum of all adjacent elements is even
int min_elimination(int n, int arr[])
{
    int countOdd = 0;
 
    // Stores the new value
    for (int i = 0; i < n; i++)
 
        // Count odd numbers
        if (arr[i] % 2)
            countOdd++;
 
    // Return the minimum of even and
    // odd count
    return min(countOdd, n - countOdd);
}
 
// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 7, 9 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << min_elimination(n, arr);
 
    return 0;
}

Java

// Java implementation of the above approach
class GFG
{
     
// Function to find minimum number of
// eliminations such that sum of all
// adjacent elements is even
static int min_elimination(int n, int arr[])
{
    int countOdd = 0;
 
    // Stores the new value
    for (int i = 0; i < n; i++)
 
        // Count odd numbers
        if (arr[i] % 2 == 1)
            countOdd++;
 
    // Return the minimum of even
    // and odd count
    return Math.min(countOdd, n - countOdd);
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 1, 2, 3, 7, 9 };
    int n = arr.length;
 
    System.out.println(min_elimination(n, arr));
}
}
 
// This code is contributed by Code_Mech

Python3

# Python 3 implementation of the
# above approach
 
# Function to find minimum number of
# eliminations such that sum of all
# adjacent elements is even
def min_elimination(n, arr):
    countOdd = 0
 
    # Stores the new value
    for i in range(n):
         
        # Count odd numbers
        if (arr[i] % 2):
            countOdd += 1
 
    # Return the minimum of even and
    # odd count
    return min(countOdd, n - countOdd)
 
# Driver code
if __name__ == '__main__':
    arr = [1, 2, 3, 7, 9]
    n = len(arr)
 
    print(min_elimination(n, arr))
 
# This code is contributed by
# Surendra_Gangwar

C#

// C# implementation of the above approach
using System;
 
class GFG
{
     
// Function to find minimum number of
// eliminations such that sum of all
// adjacent elements is even
static int min_elimination(int n, int[] arr)
{
    int countOdd = 0;
 
    // Stores the new value
    for (int i = 0; i < n; i++)
 
        // Count odd numbers
        if (arr[i] % 2 == 1)
            countOdd++;
 
    // Return the minimum of even
    // and odd count
    return Math.Min(countOdd, n - countOdd);
}
 
// Driver code
public static void Main()
{
    int[] arr = { 1, 2, 3, 7, 9 };
    int n = arr.Length;
 
    Console.WriteLine(min_elimination(n, arr));
}
}
 
// This code is contributed by Code_Mech

Javascript

 
 
// Function to find minimum number of eliminations
// such that sum of all adjacent elements is even
function min_elimination(n, arr)
{
    let countOdd = 0;
 
    // Stores the new value
    for (let i = 0; i < n; i++)
 
        // Count odd numbers
        if (arr[i] % 2)
            countOdd++;
 
    // Return the minimum of even and
    // odd count
    return Math.min(countOdd, n - countOdd);
}
 
// Driver code
 
let arr= [1, 2, 3, 7, 9];
let n = arr.length;
 
document.write(min_elimination(n, arr));
 

输出：

1

时间复杂度： O(N)

辅助空间： O(1)