Count of K length subsequence whose product is even

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.

  1. For making a product of the subsequence odd we must choose K numbers as odd.
  2. So the number of subsequences of length K whose product is odd is possibly finding k odd numbers, i.e., “o choose k” or   _{k}^{o}\textrm{C}
    where o is the count of odd numbers in the subsequence.
  3. \text{So count of a subsequence with even product  = }  _{k}^{n}\textrm{C} -  _{k}^{o}\textrm{C}
    where n and o is the count of total numbers and odd numbers respectively.

Below is the implementation of above program:

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// 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 nCr
int nCr(int n, int r)
{
    if (r > n)
        return 0;
    return fact(n)
           / (fact(r)
              * fact(n - r));
}
  
// Returns factorial of n
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 number
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] & 1)
            countOdd++;
    }
    int ans = nCr(n, k)
              - nCr(countOdd, k);
  
    return ans;
}
  
// Driver code
int main()
{
  
    int arr[] = { 2, 4 };
    int K = 1;
  
    int N = sizeof(arr) / sizeof(arr[0]);
  
    cout << countSubsequences(arr, n, k);
  
    return 0;
}

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Output:

2

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