Space complexity of recursive algorithm

Question

I was asked at an interview, the efficient way to solve a problem checking for pallindrome.

Now i can do two things:

starting from i = 0 to i = n/2 and comparing ith and n-ith character to be equal.
I can use recursion to check if first and last are same and the rest of the string is a pallindrome.

The second is recursive. My question is what is the difference in the space complexity of an algorithm's recursive and non-recursive versions?

Community · Accepted Answer · 2017-05-23 12:16:29Z

10

Have a read at

Basically, a recursive algorithm will add overhead since you store recursive calls in the execution stack.

But if the recursive function is the last line of the call (tail recursion) then there is no additional penalty.

That is of course both algorithms are the same.

edited May 23, 2017 at 12:16

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answered May 30, 2012 at 17:22

Yannis

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Ja͢ck · Accepted Answer · 2012-05-30 17:20:37Z

1

In theory they have the same space complexity; it largely depends on whether tail recursion can be optimized.

If so, the stack gets replaced at every recursion so it doesn't incur a penalty.

answered May 30, 2012 at 17:20

Ja͢ck

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4 Comments

Sudhakar Over a year ago

For large data sets , even with tail recursion , will lead to StackOverflow error . No?

Ja͢ck Over a year ago

@Sudhakar That's exactly what tail recursion optimisation prevents :)

Sudhakar Over a year ago

Is this program a valid example of Tail Recursion. This produces stackoverflow . Pls correct me if iam wrong. <pre> public class TailTest { public static void main(String[] args) { System.out.println(TailTest.iterate(100000)); } public static long iterate(int n){ if(n==1) return 1; return iterate(n-1); } } </pre>

Ja͢ck Over a year ago

@Sudhakar In that case, there's no tail recursion optimisation being done.

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