Question

f the number of subsets of a set with 2006 elements having an even number of elements is 2k, then find the sum of digits of k.

Answer 1

Answer:

Input: arr[] = {2, 2, 3}

Output: 6

All possible sub-sets are {2}, {2}, {2, 2}, {2, 3}, {2, 3} and {2, 2, 3}

Input: arr[] = {3, 3, 3}

Output: 0

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: We already know that :

Even * Even = Even

Odd * Even = Even

Odd * Odd = Odd

Now, we need to count the total subsets in which at least a single even element is present in order for the product of the elements to be even.

Now, Total number of sub-sets having at least one even element = Total possible sub-sets of n – Total sub-sets having all odd elements

i.e. (2n – 1) – (2totalOdd – 1)

Below is the implementation of the above approach:

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// C++ implementation of above approach  

 

#include <iostream>  

#include<bits/stdc++.h>  

 

using namespace std;  

 

// Function to find total number of subsets  

// in which product of the elements is even  

void find(int a[], int n)  

{  

   int count_odd = 0;  

     

   for(int i = 0; i < n ; i++)  

   {  

       // counting number of odds elements  

       if (i % 2 != 0)  

       count_odd += 1;  

   }  

 

   int result = pow(2, n) - 1 ;  

   result -= (pow(2, count_odd) - 1) ;  

   cout << result << endl;  

     

}  

 

// Driver code  

int main()  

{  

  int a[] = {2, 2, 3} ;  

  int n = sizeof(a)/sizeof(a[0]) ;  

     

  // function calling  

  find(a,n);  

     

  return 0;  

  // This code is contributed by ANKITRAI1;  

}

Explanation: