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 elements of the subsequence have the same parity.
Examples:
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
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 <bits/stdc++.h> 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 approachimport 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 subsequencesdef 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 codeif __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 approachusing 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 |
Javascript
<script>// Javascript program for the above approach // Function to find required subsequences function 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 let dp = new Array(n + 1); for (var i = 0; i < dp.length; i++) { dp[i] = new Array(2); } // Initialise the dp[][] with 0. for(let i = 0; i < n + 1; i++) { dp[i][0] = 0; dp[i][1] = 0; } for(let 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 (dp[n][0] + dp[n][1]); }// Driver Code let arr = [ 5, 6, 9, 7 ]; let n = arr.length; document.write(validsubsequences(arr, n));</script> |
Output:
9
Time complexity: O(N)
Auxiliary Space complexity: O(N)
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