Given an array arr[] and an integer K, the task is to find number of non empty subsequence of length K from the given array arr of size N such that the product of subsequence is a even number.
Example:
Input: arr[] = [2, 3, 1, 7], K = 3
Output: 3
Explanation:
There are 3 subsequences of length 3 whose product is even number {2, 3, 1}, {2, 3, 7}, {2, 1, 7}.
Input: arr[] = [2, 4], K = 1
Output: 2
Explanation:
There are 2 subsequence of length 1 whose product is even number {2} {4}.
Approach:
To solve the problem mentioned above we have to find the total number of subsequence of length K and subtract the count of K length subsequence whose product is odd.
- For making a product of the subsequence odd we must choose K numbers as odd.
- So the number of subsequences of length K whose product is odd is possibly finding k odd numbers, i.e., “o choose k” or
where o is the count of odd numbers in the subsequence. 
where n and o is the count of total numbers and odd numbers respectively.
Below is the implementation of above program:
C++
// C++ implementation to Count of K// length subsequence whose// Product is even#include <bits/stdc++.h>using namespace std;int fact(int n);// Function to calculate nCrint nCr(int n, int r){ if (r > n) return 0; return fact(n) / (fact(r) * fact(n - r));}// Returns factorial of nint fact(int n){ int res = 1; for (int i = 2; i <= n; i++) res = res * i; return res;}// Function for finding number// of K length subsequences// whose product is even numberint countSubsequences( int arr[], int n, int k){ int countOdd = 0; // counting odd numbers in the array for (int i = 0; i < n; i++) { if (arr[i] & 1) countOdd++; } int ans = nCr(n, k) - nCr(countOdd, k); return ans;}// Driver codeint main(){ int arr[] = { 2, 4 }; int K = 1; int N = sizeof(arr) / sizeof(arr[0]); cout << countSubsequences(arr, N, K); return 0;} |
Java
// Java implementation to count of K// length subsequence whose product // is evenimport java.util.*;class GFG{ // Function to calculate nCrstatic int nCr(int n, int r){ if (r > n) return 0; return fact(n) / (fact(r) * fact(n - r));}// Returns factorial of nstatic int fact(int n){ int res = 1; for(int i = 2; i <= n; i++) res = res * i; return res;}// Function for finding number// of K length subsequences// whose product is even numberstatic int countSubsequences(int arr[], int n, int k){ int countOdd = 0; // Counting odd numbers in the array for(int i = 0; i < n; i++) { if (arr[i] % 2 == 1) countOdd++; } int ans = nCr(n, k) - nCr(countOdd, k); return ans;}// Driver codepublic static void main(String args[]){ int arr[] = { 2, 4 }; int K = 1; int N = arr.length; System.out.println(countSubsequences(arr, N, K));}}// This code is contributed by ANKITKUMAR34 |
Python3
# Python3 implementation to Count of K# length subsequence whose# Product is even# Function to calculate nCrdef nCr(n, r): if (r > n): return 0 return fact(n) // (fact(r) * fact(n - r))# Returns factorial of ndef fact(n): res = 1 for i in range(2, n + 1): res = res * i return res# Function for finding number# of K length subsequences# whose product is even numberdef countSubsequences(arr, n, k): countOdd = 0 # Counting odd numbers in the array for i in range(n): if (arr[i] & 1): countOdd += 1; ans = nCr(n, k) - nCr(countOdd, k); return ans # Driver codearr = [ 2, 4 ]K = 1N = len(arr)print(countSubsequences(arr, N, K))# This code is contributed by ANKITKUAR34 |
C#
// C# implementation to count of K// length subsequence whose product // is evenusing System;class GFG{ // Function to calculate nCrstatic int nCr(int n, int r){ if (r > n) return 0; return fact(n) / (fact(r) * fact(n - r));}// Returns factorial of nstatic int fact(int n){ int res = 1; for(int i = 2; i <= n; i++) res = res * i; return res;}// Function for finding number// of K length subsequences// whose product is even numberstatic int countSubsequences(int []arr, int n, int k){ int countOdd = 0; // Counting odd numbers in the array for(int i = 0; i < n; i++) { if (arr[i] % 2 == 1) countOdd++; } int ans = nCr(n, k) - nCr(countOdd, k); return ans;}// Driver codepublic static void Main(String []args){ int []arr = { 2, 4 }; int K = 1; int N = arr.Length; Console.WriteLine(countSubsequences(arr, N, K));}}// This code is contributed by Princi Singh |
Javascript
<script>// javascript implementation to Count of K// length subsequence whose// Product is even// Function to calculate nCrfunction nCr(n, r){ if (r > n) return 0; return fact(n) / (fact(r) * fact(n - r));}// Returns factorial of nfunction fact(n){ var res = 1; for (var i = 2; i <= n; i++) res = res * i; return res;}// Function for finding number// of K length subsequences// whose product is even numberfunction countSubsequences( arr, n, k){ var countOdd = 0; // counting odd numbers in the array for (var i = 0; i < n; i++) { if (arr[i] & 1) countOdd++; } var ans = nCr(n, k) - nCr(countOdd, k); return ans;}// Driver codevar arr = [ 2, 4 ];var K = 1;var N = arr.length;document.write( countSubsequences(arr, N, K));</script> |
Output:
2
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