使数组的所有元素即使具有给定的操作

给定一个正整数的数组arr [] ，找到使所有数组元素都需要的最小操作数，即使：

1. 如果存在奇数，则将该元素和下一个相邻元素加1。
2. 每次增量花费一次操作。

注意：如果arr []中的任何数字在所有操作之后都是奇数，则打印-1。

例子

Input: arr[] = {2, 3, 4, 5, 6}
Output: 4
Explanation:
Add 1 to 3(at 1st index) and add 1 to its adjacent element 4(2nd index).
Now the array becomes {2, 4, 5, 5, 6}.
Add 1 to 5(at 2nd index) and add 1 to its adjacent element 5(3rd index).
Now the array becomes {2, 4, 6, 6, 6}.
The resultant array has all even numbers.
The total number of operations for 4 increments is 4.

Input: arr[] = {5, 6}
Output: -1
Explanation:
Adding 1 to 5(0th index) then we have to increment 1 to its adjacent element 6(1st index).
Now the array becomes {6, 7}.
And we have 1 odd number left after all possible increments. Therefore we can’t make all array elements even.

方法：
可以使用贪婪方法解决此问题。步骤如下：

1. 遍历给定数组arr [] 。
2. 如果出现奇数元素，则将该元素加1使其变为偶数，将下一个相邻元素加1。
3. 对给定数组arr []的所有奇数元素重复上述步骤。
4. 如果arr []中的所有元素均是偶数，则打印操作数。
5. 其他打印-1。

下面是上述方法的实现：

C++

// C++ program to make all array
// element even
#include "bits/stdc++.h"
using namespace std;
  
// Function to count the total
// number of operations needed to make
// all array element even
int countOperations(int arr[], int n)
{
    int count = 0;
  
    // Traverse the given array
    for (int i = 0; i < n - 1; i++) {
  
        // If an odd element occurs
        // then increment that element
        // and next adjacent element
        // by 1
        if (arr[i] & 1) {
            arr[i]++;
            arr[i + 1]++;
            count += 2;
        }
    }
  
    // Traverse the array if any odd
    // element occurs then return -1
    for (int i = 0; i < n; i++) {
        if (arr[i] & 1)
            return -1;
    }
  
    // Returns the count of operations
    return count;
}
  
int main()
{
    int arr[] = { 2, 3, 4, 5, 6 };
    int n = sizeof(arr) / sizeof(int);
    cout << countOperations(arr, n);
    return 0;
}

Java

// Java program to make all array
// element even
class GFG
{
  
// Function to count the total
// number of operations needed to make
// all array element even
static int countOperations(int arr[], int n)
{
    int count = 0;
  
    // Traverse the given array
    for (int i = 0; i < n - 1; i++)
    {
  
        // If an odd element occurs
        // then increment that element
        // and next adjacent element
        // by 1
        if (arr[i] % 2 == 1) 
        {
            arr[i]++;
            arr[i + 1]++;
            count += 2;
        }
    }
  
    // Traverse the array if any odd
    // element occurs then return -1
    for (int i = 0; i < n; i++)
    {
        if (arr[i] % 2 == 1)
            return -1;
    }
  
    // Returns the count of operations
    return count;
}
  
// Driver code
public static void main(String[] args)
{
    int arr[] = { 2, 3, 4, 5, 6 };
    int n = arr.length;
    System.out.print(countOperations(arr, n));
}
}
  
// This code is contributed by 29AjayKumar

Python3

# Python3 program to make all array 
# element even 
  
# Function to count the total 
# number of operations needed to make 
# all array element even 
def countOperations(arr, n) :
  
    count = 0; 
  
    # Traverse the given array 
    for i in range(n - 1) :
  
        # If an odd element occurs 
        # then increment that element 
        # and next adjacent element 
        # by 1 
        if (arr[i] & 1) :
            arr[i] += 1; 
            arr[i + 1] += 1; 
            count += 2; 
  
    # Traverse the array if any odd 
    # element occurs then return -1 
    for i in range(n) :
        if (arr[i] & 1) :
            return -1; 
  
    # Returns the count of operations 
    return count; 
  
if __name__ == "__main__" : 
  
    arr = [ 2, 3, 4, 5, 6 ]; 
    n = len(arr); 
    print(countOperations(arr, n)); 
      
    # This code is contributed by AnkitRai01

C#

// C# program to make all array
// element even
using System;
  
class GFG
{
  
// Function to count the total
// number of operations needed to make
// all array element even
static int countOperations(int []arr, int n)
{
    int count = 0;
  
    // Traverse the given array
    for (int i = 0; i < n - 1; i++)
    {
  
        // If an odd element occurs
        // then increment that element
        // and next adjacent element
        // by 1
        if (arr[i] % 2 == 1) 
        {
            arr[i]++;
            arr[i + 1]++;
            count += 2;
        }
    }
  
    // Traverse the array if any odd
    // element occurs then return -1
    for (int i = 0; i < n; i++)
    {
        if (arr[i] % 2 == 1)
            return -1;
    }
  
    // Returns the count of operations
    return count;
}
  
// Driver code
public static void Main()
{
    int []arr = { 2, 3, 4, 5, 6 };
    int n = arr.Length;
    Console.Write(countOperations(arr, n));
}
}
  
// This code is contributed by AnkitRai01

输出：

4

时间复杂度： O(N)，其中N是数组中元素的数量。