奇数乘积的子阵列数

给定大小为N的整数数组arr [] ，任务是计算具有奇积的子数组的数量。

例子：

Input : arr[] = {5, 1, 2, 3, 4}
Output : 4
Explanation: The sub-arrays with odd product are-
{5}, {1}, {3}, {5, 1}. Hence the count is 4.

Input : arr[] = {12, 15, 7, 3, 25, 6, 2, 1, 1, 7}
Output : 16

天真的方法：一个简单的解决方案是计算每个子数组的乘积，并检查其是否为奇数，然后相应地计算计数。
时间复杂度： O(N ² )

有效的方法：奇数乘积只能由奇数的乘积来实现。因此，对于数组中每K个连续的奇数，具有奇数乘积的子数组的数量增加K *(K + 1)/ 2 。计算连续奇数的一种方法是计算每两个连续偶数的索引之间的差，并将其相减1 。为了进行计算，将-1和N视为偶数索引。

下面是上述方法的实现：

C++

// C++ program to find the count of
// sub-arrays with odd product
#include 
using namespace std;
  
// Function that returns the count of
// sub-arrays with odd product
int countSubArrayWithOddProduct(int* A, int N)
{
    // Initialize the count variable
    int count = 0;
  
    // Initialize variable to store the
    // last index with even number
    int last = -1;
  
    // Initialize variable to store
    // count of continuous odd numbers
    int K = 0;
  
    // Loop through the array
    for (int i = 0; i < N; i++) {
        // Check if the number
        // is even or not
        if (A[i] % 2 == 0) {
            // Calculate count of continuous
            // odd numbers
            K = (i - last - 1);
  
            // Increase the count of sub-arrays
            // with odd product
            count += (K * (K + 1) / 2);
  
            // Store the index of last
            // even number
            last = i;
        }
    }
  
    // N considered as index of
    // even number
    K = (N - last - 1);
  
    count += (K * (K + 1) / 2);
  
    return count;
}
  
// Driver Code
int main()
{
    int arr[] = { 12, 15, 7, 3, 25,
                  6, 2, 1, 1, 7 };
  
    int n = sizeof(arr) / sizeof(arr[0]);
  
    // Function call
    cout << countSubArrayWithOddProduct(arr, n);
  
    return 0;
}

Java

// Java program to find the count of 
// sub-arrays with odd product 
class GFG {
  
// Function that returns the count of 
// sub-arrays with odd product 
static int countSubArrayWithOddProduct(int A[], 
                                       int N)
{
      
    // Initialize the count variable 
    int count = 0;
  
    // Initialize variable to store the 
    // last index with even number 
    int last = -1;
  
    // Initialize variable to store 
    // count of continuous odd numbers 
    int K = 0;
  
    // Loop through the array 
    for(int i = 0; i < N; i++) 
    {
  
       // Check if the number 
       // is even or not 
       if (A[i] % 2 == 0)
       {
  
           // Calculate count of continuous 
           // odd numbers 
           K = (i - last - 1);
             
           // Increase the count of sub-arrays 
           // with odd product 
           count += (K * (K + 1) / 2);
             
           // Store the index of last 
           // even number 
           last = i;
       }
    }
  
    // N considered as index of 
    // even number 
    K = (N - last - 1);
    count += (K * (K + 1) / 2);
      
    return count;
}
  
// Driver Code
public static void main(String args[]) 
{
    int arr[] = { 12, 15, 7, 3, 25, 6, 2, 1, 1, 7 };
    int n = arr.length;
  
    // Function call
    System.out.println(countSubArrayWithOddProduct(arr, n));
}
}
  
// This code is contributed by rutvik_56

Python3

# Python3 program to find the count of
# sub-arrays with odd product
  
# Function that returns the count of
# sub-arrays with odd product
def countSubArrayWithOddProduct(A, N):
      
    # Initialize the count variable
    count = 0
  
    # Initialize variable to store the
    # last index with even number
    last = -1
  
    # Initialize variable to store
    # count of continuous odd numbers
    K = 0
  
    # Loop through the array
    for i in range(N):
          
        # Check if the number
        # is even or not
        if (A[i] % 2 == 0):
              
            # Calculate count of continuous
            # odd numbers
            K = (i - last - 1)
  
            # Increase the count of sub-arrays
            # with odd product
            count += (K * (K + 1) / 2)
  
            # Store the index of last
            # even number
            last = i
  
    # N considered as index of
    # even number
    K = (N - last - 1)
  
    count += (K * (K + 1) / 2)
    return count
  
# Driver Code
if __name__ == '__main__':
      
    arr = [ 12, 15, 7, 3, 25, 6, 2, 1, 1, 7 ]
    n = len(arr)
  
    # Function call
    print(int(countSubArrayWithOddProduct(arr, n)))
  
# This code is contributed by Bhupendra_Singh

C#

// C# program to find the count of 
// sub-arrays with odd product 
using System;
class GFG{
  
// Function that returns the count of 
// sub-arrays with odd product     
static int countSubArrayWithOddProduct(int[] A, 
                                       int N) 
{
          
    // Initialize the count variable 
    int count = 0; 
      
    // Initialize variable to store the 
    // last index with even number 
    int last = -1; 
      
    // Initialize variable to store 
    // count of continuous odd numbers 
    int K = 0; 
      
    // Loop through the array 
    for(int i = 0; i < N; i++) 
    { 
         
       // Check if the number 
       // is even or not 
       if (A[i] % 2 == 0) 
       { 
             
           // Calculate count of continuous 
           // odd numbers 
           K = (i - last - 1); 
             
           // Increase the count of sub-arrays 
           // with odd product 
           count += (K * (K + 1) / 2); 
             
           // Store the index of last 
           // even number 
           last = i; 
       } 
    } 
      
    // N considered as index of 
    // even number 
    K = (N - last - 1); 
    count += (K * (K + 1) / 2); 
      
    return count; 
} 
  
// Driver code
static void Main()
{
    int[] arr = { 12, 15, 7, 3, 25,
                  6, 2, 1, 1, 7 }; 
    int n = arr.Length; 
      
    // Function call 
    Console.WriteLine(countSubArrayWithOddProduct(arr, n));
}
}
  
// This code is contributed by divyeshrabadiya07

输出：

16

时间复杂度： O(N)
辅助空间： O(1)