通过增加奇数长度子数组的奇数索引元素使数组的所有元素成为奇数

给定大小为N的数组arr[] ，任务是通过选择arr[]的奇数长度子数组使所有数组元素为奇数，并将此子数组中所有奇数定位元素递增1 。打印所需的此类操作的计数。

例子：

Input: arr[] = {2, 3, 4, 3, 5, 3, 2}
Output: 2
Explanation:
In first operation, choose the subarray {2, 3, 4} and increment all its elements at odd positions, i.e., 1 and 3 of this subarray. The updated array is {3, 3, 5, 3, 5, 3, 2}.
In second operation, choose the last index which is subarray of length 1 and increment its value. The updated array is {3, 3, 5, 3, 5, 3, 3}

Input: arr[] = {1, 5, 7}
Output: 0
Explanation: Since all array elements are odd, no changes required.

Approach: The idea is based on the observation that whenever a subarray is chosen, either the odd positioned values are changed or the even positioned values in the original array.该问题可以通过在每次操作中贪婪地选择子数组来解决。首先，遍历所有奇数索引，一旦找到偶数就标记子数组的开始，当找到奇数时结束子数组，同时更新操作次数。对偶数索引重复相同的过程。请按照以下步骤解决问题：

1. 初始化一个变量，比如flips ，以存储所需的最少操作数。
2. 遍历数组arr[] 的偶数索引执行以下步骤：
   • 如果当前元素是奇数，则继续迭代。
   • 否则，从该索引开始迭代每个第二个元素，直到遇到偶数元素。完成遍历数组后或遇到偶数元素后，将flips增加1 。
3. 对奇数索引也重复步骤 2 。
4. 完成上述步骤后，根据需要将翻转的值打印为最小操作次数。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to count minimum subarrays
// whose odd-indexed elements need to
// be incremented to make array odd
void minOperations(int arr[], int n)
{
    // Stores the minimum number of
    // operations required
    int flips = 0;
 
    // Iterate over even-indices
    for (int i = 0; i < n; i += 2) {
 
        // Check if the current
        // element is odd
        if (arr[i] % 2 == 1) {
 
            // If true, continue
            continue;
        }
 
        // Otherwise, mark the starting
        // of the subarray and iterate
        // until i < n and arr[i] is even
        while (i < n && arr[i] % 2 == 0) {
            i += 2;
        }
 
        // Increment number of operations
        flips++;
    }
 
    // Iterate over odd indexed
    // positions of arr[]
    for (int i = 1; i < n; i += 2) {
 
        // Check if the current
        // element is odd
        if (arr[i] % 2 == 1) {
 
            // If true, continue
            continue;
        }
 
        // Otherwise, mark the starting
        // of the subarray and iterate
        // until i < n and arr[i] is even
        while (i < n && arr[i] % 2 == 0) {
            i += 2;
        }
 
        // Increment the number
        // of operations
        flips++;
    }
 
    // Print the number of operations
    cout << flips;
}
 
// Driver Code
int main()
{
    int arr[] = { 2, 3, 4, 3, 5, 3, 2 };
    int N = sizeof(arr) / sizeof(int);
 
    // Function Call
    minOperations(arr, N);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
       
// Function to count minimum subarrays
// whose odd-indexed elements need to
// be incremented to make array odd
static void minOperations(int arr[], int n)
{
     
    // Stores the minimum number of
    // operations required
    int flips = 0;
     
    // Iterate over even-indices
    for(int i = 0; i < n; i += 2)
    {
         
        // Check if the current
        // element is odd
        if (arr[i] % 2 == 1)
        {
             
            // If true, continue
            continue;
        }
         
        // Otherwise, mark the starting
        // of the subarray and iterate
        // until i < n and arr[i] is even
        while (i < n && arr[i] % 2 == 0)
        {
            i += 2;
        }
         
        // Increment number of operations
        flips++;
    }
     
    // Iterate over odd indexed
    // positions of arr[]
    for(int i = 1; i < n; i += 2)
    {
         
        // Check if the current
        // element is odd
        if (arr[i] % 2 == 1)
        {
             
            // If true, continue
            continue;
        }
  
        // Otherwise, mark the starting
        // of the subarray and iterate
        // until i < n and arr[i] is even
        while (i < n && arr[i] % 2 == 0)
        {
            i += 2;
        }
         
        // Increment the number
        // of operations
        flips++;
    }
     
    // Print the number of operations
    System.out.println(flips);
}
 
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 2, 3, 4, 3, 5, 3, 2 };
    int N = arr.length;
     
    // Function Call
    minOperations(arr, N);
}
}
 
// This code is contributed by jana_sayantan

C#

// C# program for the above approach
using System;
class GFG
{
       
// Function to count minimum subarrays
// whose odd-indexed elements need to
// be incremented to make array odd
static void minOperations(int []arr, int n)
{
     
    // Stores the minimum number of
    // operations required
    int flips = 0;
     
    // Iterate over even-indices
    for(int i = 0; i < n; i += 2)
    {
         
        // Check if the current
        // element is odd
        if (arr[i] % 2 == 1)
        {
             
            // If true, continue
            continue;
        }
         
        // Otherwise, mark the starting
        // of the subarray and iterate
        // until i < n and arr[i] is even
        while (i < n && arr[i] % 2 == 0)
        {
            i += 2;
        }
         
        // Increment number of operations
        flips++;
    }
     
    // Iterate over odd indexed
    // positions of []arr
    for(int i = 1; i < n; i += 2)
    {
         
        // Check if the current
        // element is odd
        if (arr[i] % 2 == 1)
        {
             
            // If true, continue
            continue;
        }
  
        // Otherwise, mark the starting
        // of the subarray and iterate
        // until i < n and arr[i] is even
        while (i < n && arr[i] % 2 == 0)
        {
            i += 2;
        }
         
        // Increment the number
        // of operations
        flips++;
    }
     
    // Print the number of operations
    Console.WriteLine(flips);
}
 
// Driver Code
public static void Main(String[] args)
{
    int []arr = { 2, 3, 4, 3, 5, 3, 2 };
    int N = arr.Length;
     
    // Function Call
    minOperations(arr, N);
}
}
 
// This code is contributed by Princi Singh

Python3

# Python3 program for the above approach
 
# Function to count minimum subarrays
# whose odd-indexed elements need to
# be incremented to make array odd
def minOperations(arr, n) :
 
    # Stores the minimum number of
    # operations required
    flips = 0;
    i = 0;
     
    # Iterate over even-indices
    while i < n :
 
        # Check if the current
        # element is odd
        if (arr[i] % 2 == 1) :           
            i += 2;
             
            # If true, continue
            continue;
 
        # Otherwise, mark the starting
        # of the subarray and iterate
        # until i < n and arr[i] is even
        while (i < n and arr[i] % 2 == 0) :
            i += 2;
 
        # Increment number of operations
        flips += 1;       
        i += 2;
 
    # Iterate over odd indexed
    # positions of arr[]
    i = 1
    while i < n :
 
        # Check if the current
        # element is odd
        if (arr[i] % 2 == 1) :
            i += 2;
             
            # If true, continue
            continue;
 
        # Otherwise, mark the starting
        # of the subarray and iterate
        # until i < n and arr[i] is even
        while (i < n and arr[i] % 2 == 0) :
            i += 2;
 
        # Increment the number
        # of operations
        flips += 1;       
        i += 2;
 
    # Print the number of operations
    print(flips);
 
# Driver Code
if __name__ == "__main__" :
 
    arr = [ 2, 3, 4, 3, 5, 3, 2 ];
    N = len(arr);
 
    # Function Call
    minOperations(arr, N);
 
    # This code is contributed by AnkThon

Javascript

// Javascript program for the above approach
 
// Function to count minimum subarrays
// whose odd-indexed elements need to
// be incremented to make array odd
function minOperations(arr, n)
{
      
    // Stores the minimum number of
    // operations required
    let flips = 0;
      
    // Iterate over even-indices
    for(let i = 0; i < n; i += 2)
    {
          
        // Check if the current
        // element is odd
        if (arr[i] % 2 == 1)
        {
              
            // If true, continue
            continue;
        }
          
        // Otherwise, mark the starting
        // of the subarray and iterate
        // until i < n and arr[i] is even
        while (i < n && arr[i] % 2 == 0)
        {
            i += 2;
        }
          
        // Increment number of operations
        flips++;
    }
      
    // Iterate over odd indexed
    // positions of arr[]
    for(let i = 1; i < n; i += 2)
    {
          
        // Check if the current
        // element is odd
        if (arr[i] % 2 == 1)
        {
              
            // If true, continue
            continue;
        }
   
        // Otherwise, mark the starting
        // of the subarray and iterate
        // until i < n and arr[i] is even
        while (i < n && arr[i] % 2 == 0)
        {
            i += 2;
        }
          
        // Increment the number
        // of operations
        flips++;
    }
      
    // Prlet the number of operations
    document.write(flips);
}
 
    // Driver Code
     
    let arr = [ 2, 3, 4, 3, 5, 3, 2 ];
    let N = arr.length;
      
    // Function Call
    minOperations(arr, N);
     
// This code is contributed by souravghosh0416.

输出：

2

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