具有奇偶校验相邻元素的所有子序列的计数

给定大小为N的数组arr [] ，任务是从给定数组中找到非空子序列的数量，以使子序列中没有两个相邻元素具有相同的奇偶性。
例子：

Input: arr[] = [5, 6, 9, 7] 
Output: 9 
Explanation: 
All such subsequences of given array will be {5}, {6}, {9}, {7}, {5, 6}, {6, 7}, {6, 9}, {5, 6, 9}, {5, 6, 7}.
Input: arr[] = [2, 3, 4, 8] 
Output: 9 
 

幼稚的方法：生成所有非空子序列，并选择具有交替的奇偶或奇数的子序列，并对所有这些子序列进行计数以获得答案。
时间复杂度： O(2 ^N )
高效方法：
可以使用动态编程来优化上述方法。请按照以下步骤解决问题：

• 考虑一个尺寸为(N + 1)*(2)的dp []矩阵。
• dp [i] [0]存储子序列的计数，直到^第i^个索引以偶数元素结尾。
• dp [i] [1]存储子序列的计数，直到^第i^个索引以奇数元素结尾。
• 因此，对于每个^第i^个元素，请检查该元素是偶数还是奇数，然后通过包含和排除第i^个元素来进行操作。
• 因此，如果第i^个元素为奇数，则为递归关系：

 
 

dp[i][1] = dp[i – 1][0] (Including the i^th element by considering all subsequences ending with even element till (i – 1)^th index) + 1 + dp[i – 1][1] (Excluding the i^th element) 
 

 

• 同样，如果第i^个元素是偶数：

 
 

dp[i][0] = dp[i – 1][1] (Including the i^th element by considering all subsequences ending with odd element till (i – 1)^th index) + 1 + dp[i – 1][0] (Excluding the i^th element) 
 

 

• 最后，要求答案是dp [n] [0]的总和，它包含以偶数元素结尾的所有此类子序列，而dp [n] [1]包含以奇数元素结尾的所有此类子序列。

下面是上述方法的实现：

C++

// C++ Program to implement the 
// above approach 
#include  
using namespace std; 
  
// Function to find required subsequences 
int validsubsequences(int arr[], int n) 
{ 
    // dp[i][0]: Stores the number of 
    // subsequences till i-th index 
    // ending with even element 
    // dp[i][1]: Stores the number of 
    // subsequences till i-th index 
    // ending with odd element 
    long long int dp[n + 1][2]; 
  
    // Initialise the dp[][] with 0. 
    for (int i = 0; i < n + 1; i++) { 
        dp[i][0] = 0; 
        dp[i][1] = 0; 
    } 
  
    for (int i = 1; i <= n; i++) { 
  
        // If odd element is 
        // encountered 
        if (arr[i - 1] % 2) { 
  
            // Considering i-th element 
            // will be present in 
            // the subsequence 
            dp[i][1] += 1; 
  
            // Appending i-th element to all 
            // non-empty subsequences 
            // ending with even element 
            // till (i-1)th indexes 
            dp[i][1] += dp[i - 1][0]; 
  
            // Considering ith element will 
            // not be present in 
            // the subsequence 
            dp[i][1] += dp[i - 1][1]; 
  
            dp[i][0] += dp[i - 1][0]; 
        } 
        else { 
  
            // Considering i-th element 
            // will be present in 
            // the subsequence 
            dp[i][0] += 1; 
  
            // Appending i-th element to all 
            // non-empty subsequences 
            // ending with odd element 
            // till (i-1)th indexes 
            dp[i][0] += dp[i - 1][1]; 
  
            // Considering ith element will 
            // not be present in 
            // the subsequence 
            dp[i][0] += dp[i - 1][0]; 
            dp[i][1] += dp[i - 1][1]; 
        } 
    } 
  
    // Count of all valid subsequences 
    return dp[n][0] + dp[n][1]; 
} 
  
// Driver Code 
int main() 
{ 
    int arr[] = { 5, 6, 9, 7 }; 
    int n = sizeof(arr) / sizeof(arr[0]); 
  
    cout << validsubsequences(arr, n); 
  
    return 0; 
}

Java

// Java Program implementation
// of the approach
import java.util.*;
import java.io.*;
  
class GFG{
  
// Function to find required subsequences     
static int validsubsequences(int arr[], int n)
{
      
    // dp[i][0]: Stores the number of 
    // subsequences till i-th index 
    // ending with even element 
    // dp[i][1]: Stores the number of 
    // subsequences till i-th index 
    // ending with odd element 
    long dp[][] = new long [n + 1][2]; 
  
    // Initialise the dp[][] with 0. 
    for(int i = 0; i < n + 1; i++)
    { 
        dp[i][0] = 0; 
        dp[i][1] = 0; 
    }
      
    for(int i = 1; i <= n; i++)
    { 
          
        // If odd element is 
        // encountered 
        if (arr[i - 1] % 2 != 0) 
        { 
  
            // Considering i-th element 
            // will be present in 
            // the subsequence 
            dp[i][1] += 1; 
  
            // Appending i-th element to all 
            // non-empty subsequences 
            // ending with even element 
            // till (i-1)th indexes 
            dp[i][1] += dp[i - 1][0]; 
  
            // Considering ith element will 
            // not be present in 
            // the subsequence 
            dp[i][1] += dp[i - 1][1]; 
            dp[i][0] += dp[i - 1][0]; 
        } 
        else
        { 
  
            // Considering i-th element 
            // will be present in 
            // the subsequence 
            dp[i][0] += 1; 
  
            // Appending i-th element to all 
            // non-empty subsequences 
            // ending with odd element 
            // till (i-1)th indexes 
            dp[i][0] += dp[i - 1][1]; 
  
            // Considering ith element will 
            // not be present in 
            // the subsequence 
            dp[i][0] += dp[i - 1][0]; 
            dp[i][1] += dp[i - 1][1]; 
        } 
    } 
  
    // Count of all valid subsequences 
    return (int)(dp[n][0] + dp[n][1]); 
}
  
// Driver code 
public static void main(String[] args) 
{
    int arr[] = { 5, 6, 9, 7 };
    int n = arr.length;
          
    System.out.print(validsubsequences(arr, n));
}
}
  
// This code is contributed by code_hunt

Python3

# Python3 program to implement the
# above approach
  
# Function to find required subsequences
def validsubsequences(arr, n):
  
    # dp[i][0]: Stores the number of
    # subsequences till i-th index
    # ending with even element
    # dp[i][1]: Stores the number of
    # subsequences till i-th index
    # ending with odd element
  
    # Initialise the dp[][] with 0.
    dp = [[0 for i in range(2)]
             for j in range(n + 1)]
               
    for i in range(1, n + 1):
  
        # If odd element is
        # encountered
        if(arr[i - 1] % 2):
  
            # Considering i-th element
            # will be present in
            # the subsequence
            dp[i][1] += 1
  
            # Appending i-th element to all
            # non-empty subsequences
            # ending with even element
            # till (i-1)th indexes
            dp[i][1] += dp[i - 1][0]
  
            # Considering ith element will
            # not be present in
            # the subsequence
            dp[i][1] += dp[i - 1][1]
            dp[i][0] += dp[i - 1][0]
  
        else:
  
            # Considering i-th element
            # will be present in
            # the subsequence
            dp[i][0] += 1
  
            # Appending i-th element to all
            # non-empty subsequences
            # ending with odd element
            # till (i-1)th indexes
            dp[i][0] += dp[i - 1][1]
  
            # Considering ith element will
            # not be present in
            # the subsequence
            dp[i][0] += dp[i - 1][0]
            dp[i][1] += dp[i - 1][1]
  
    # Count of all valid subsequences 
    return dp[n][0] + dp[n][1]
  
# Driver code
if __name__ == '__main__':
  
    arr = [ 5, 6, 9, 7 ]
    n = len(arr)
  
    print(validsubsequences(arr, n))
  
# This code is contributed by Shivam Singh

C#

// C# program implementation
// of the approach
using System;
  
class GFG{
  
// Function to find required subsequences     
static int validsubsequences(int[] arr, int n)
{
      
    // dp[i][0]: Stores the number of 
    // subsequences till i-th index 
    // ending with even element 
    // dp[i][1]: Stores the number of 
    // subsequences till i-th index 
    // ending with odd element 
    long[,] dp = new long [n + 1, 2]; 
  
    // Initialise the dp[][] with 0. 
    for(int i = 0; i < n + 1; i++)
    { 
        dp[i, 0] = 0; 
        dp[i, 1] = 0; 
    }
      
    for(int i = 1; i <= n; i++)
    { 
          
        // If odd element is 
        // encountered 
        if (arr[i - 1] % 2 != 0) 
        { 
  
            // Considering i-th element 
            // will be present in 
            // the subsequence 
            dp[i, 1] += 1; 
  
            // Appending i-th element to all 
            // non-empty subsequences 
            // ending with even element 
            // till (i-1)th indexes 
            dp[i, 1] += dp[i - 1, 0]; 
  
            // Considering ith element will 
            // not be present in 
            // the subsequence 
            dp[i, 1] += dp[i - 1, 1]; 
            dp[i, 0] += dp[i - 1, 0]; 
        } 
        else
        { 
  
            // Considering i-th element 
            // will be present in 
            // the subsequence 
            dp[i, 0] += 1; 
  
            // Appending i-th element to all 
            // non-empty subsequences 
            // ending with odd element 
            // till (i-1)th indexes 
            dp[i, 0] += dp[i - 1, 1]; 
  
            // Considering ith element will 
            // not be present in 
            // the subsequence 
            dp[i, 0] += dp[i - 1, 0]; 
            dp[i, 1] += dp[i - 1, 1]; 
        } 
    } 
  
    // Count of all valid subsequences 
    return (int)(dp[n, 0] + dp[n, 1]); 
}
  
// Driver code 
public static void Main() 
{
    int[] arr = { 5, 6, 9, 7 };
    int n = arr.Length;
          
    Console.Write(validsubsequences(arr, n));
}
}
  
// This code is contributed by chitranayal

输出：

9

时间复杂度： O(N)
辅助空间复杂度： O(N)