Algorithms | Analysis of Algorithms (Recurrences) | Question 3

What is the worst case time complexity of following implementation of subset sum problem.

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// Returns true if there is a subset of set[] with sun equal to given sum
bool isSubsetSum(int set[], int n, int sum)
{
   // Base Cases
   if (sum == 0)
     return true;
   if (n == 0 && sum != 0)
     return false;
   
   // If last element is greater than sum, then ignore it
   if (set[n-1] > sum)
     return isSubsetSum(set, n-1, sum);
   
   /* else, check if sum can be obtained by any of the following
      (a) including the last element
      (b) excluding the last element   */
   return isSubsetSum(set, n-1, sum) || 
          isSubsetSum(set, n-1, sum-set[n-1]);
}

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(A) O(n * 2^n)
(B) O(n^2)
(C) O(n^2 * 2^n)
(D) O(2^n)


Answer: (D)

Explanation: Following is the recurrence for given implementation of subset sum problem

T(n) = 2T(n-1) + C1
T(0) = C1

Where C1 and C2 are some machine specific constants.

The solution of recurrence is O(2^n)

We can see it with the help of recurrence tree method

           C1
       /       \
    T(n-1)     T(n-1) 


                    C1
                /       \
              C1           C1
           /     \        /    \
      T(n-2)  T(n-2)   T(n-2)  T(n-2)

                    C1
                /       \
              C1           C1
           /     \        /    \
          C1     C1      C1     C1
        /   \   /  \    /  \   /  \

       
If we sum the above tree level by level, we get the following series
T(n) = C1 + 2C1 + 4C1 + 8C1 + ...
The above series is Geometrical progression and there will be n terms in it.
So T(n) = O(2^n)    


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