1
$\begingroup$ 
fun(int n)
{
    if (n == 1)
       S1 // S1 takes O(N) time
    else 
       S2 + fun(n-1) // S2 takes O(1) time
}

What can be said about the time complexity of given Code snippet ?

I think the recurrence equation can be written as $T(N) = T(N-1) + N$ and this says Time complexity = $O(N^2)$

But I am not sure about this ?

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    Which language? In C, you get undefined behaviour on calling `fun(3)` for example. – 2017-01-05
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    @DanielFischer, I am just interested in recurrence equation for such problems . – 2017-01-05
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    @DanielFischer,Is my recurrence correct in case "The statement inside IF block takes O(N) time " ? – 2017-01-05
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    The problem is that not all inputs lead to a predictable recurrence. If we use hypothetical truly mathematical integers, you get an infinite loop for inputs $< 4$. – 2017-01-05
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    @DanielFischer I have tried to ease the code !! Can you please check Now ? – 2017-01-05
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    Oops, you only go inside the `if` for one particular value of `n`, so it doesn't make sense to speak of $O(n)$ for that, and that work is only done once. But you still have the problem that inputs $\leqslant 0$ aren't well-behaved. – 2017-01-05

1 Answers 1

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It is linear in $N$. The recurrence is $T(n) = T(n-1)+C$ with $T(4)$ being a constant, since you do a constant amount of work (the if statement, and then the return) at each step before calling the recursion.

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    What will be recurrence in case "The statement inside IF block takes O(N) time " ? – 2017-01-05
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    Surely it would still be O(n). The function fun(n) is O(n) (for n > 1) since it does n-1 recursive calls to itself, and S1 is O(n) and is done exactly 1 time. O(n) + O(n) (in informal terms) is still O(n). – 2017-01-05