# Count of all subsequences having adjacent elements with different parity

Given an array arr[] of size N, the task is to find the number of non-empty subsequences from the given array such that no two adjacent element of the subsequence have same parity.
Examples:

Input: arr[] = [5, 6, 9, 7]
Output:
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:

Naive Approach: Generate all non-empty subsequences and select the ones with alternate odd-even or even-odd numbers and count all such subsequences to obtain the answer.
Time Complexity: O(2N)
Efficient Approach:
The above approach can be optimized using Dynamic Programming. Follow the steps below to solve the problem:

• Consider a dp[] matrix of dimensions (N+1)*(2).
• dp[i][0] stores the count of subsequences till ith index ending with an even element.
• dp[i][1] stores the count of subsequences till ith index ending with an odd element.
• Hence, for every ith element, check if the element is even or odd and proceed by including as well as excluding the ith element.
• Hence, the recurrence relation if the ith element is odd:

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

• Similarly, if the ith element is even:

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

• Finally, the sum of dp[n][0], which contains all such subsequences ending with an even element, and dp[n][1], which contains all such subsequences ending with an odd element, is the required answer.

Below is the implementation of the above approach:

## 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 `

Output:

```9
```

Time complexity: O(N)
Auxiliary Space complexity: O(N)

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