double foo(int n){int i;double sum;if(n == 0){return 1.0;}else{sum = 0.0;for(i = 0; i andlt; n; i++){sum += foo(i);}return sum;}}Suppose we modify the above function foo() and stores the value of foo(i) 0 andlt;= i andlt; n, as and when they are computed. With this modification the time complexity for function foo() is significantly reduced. The space complexity of the modified function would be:a)b)c)d)Correct answer is option 'B'. Can you explain this answer? - EduRev Computer Science Engineering (CSE) Question

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double foo(int n)

{

int i;

double sum;

if(n == 0)

{return 1.0;

}

else

{

sum = 0.0;

for(i = 0; i < n; i++)

{

sum += foo(i);

}

return sum;

}
}

Suppose we modify the above function foo() and stores the value of foo(i) 0 <= i < n, as and when they are computed. With this modification the time complexity for function foo() is significantly reduced. The space complexity of the modified function would be:

Correct answer is option 'B'. Can you explain this answer?

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double foo(int n){int i;double sum;if(n == 0){return 1.0;}else{sum = 0...

The modification being proposed is tabulation method of Dynamic Programming.
So let we have an auxillary array which stores the value of foo(i) once it is computed and can be used at later stage .So foo(i) is not called again , instead lookup from the table takes place. It would be something like :

#include <stdio.h>
#define max 1000
int foo[max];

foo(int n) {
if(foo[n] == -1) //compute it and store at foo[n] and return
else // return foo[n] directly
}

int main(void) {
//initialize all elements of foo[max] to -1
return 0;
}Using this approach , we need an auxillary array of size n to store foo(i) where i ranges from 0 to n-1.

So space complexity of this method = O(n) And function foo(n) computes 2^{n - 1}

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double foo(int n){int i;double sum;if(n == 0){return 1.0;}else{sum = 0.0;for(i = 0; i < n; i++){sum += foo(i);}return sum;}}Suppose we modify the above function foo() and stores the value of foo(i) 0 <= i < n, as and when they are computed. With this modification the time complexity for function foo() is significantly reduced. The space complexity of the modified function would be:a)b)c)d)Correct answer is option 'B'. Can you explain this answer?

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