Determining complexity for recursive functions (Big O notation)

The time complexity, in Big O notation, for each function:

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int recursiveFun1(int n)
{
    if (n <= 0)
        return 1;
    else
        return 1 + recursiveFun1(n-1);
}

This function is being called recursively n times before reaching the base case so its O(n), often called linear.

---

int recursiveFun2(int n)
{
    if (n <= 0)
        return 1;
    else
        return 1 + recursiveFun2(n-5);
}

This function is called n-5 for each time, so we deduct five from n before calling the function, but n-5 is also O(n). (Actually called order of n/5 times. And, O(n/5) = O(n) ).

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int recursiveFun3(int n)
{
    if (n <= 0)
        return 1;
    else
        return 1 + recursiveFun3(n/5);
}

This function is log(n) base 5, for every time we divide by 5 before calling the function so its O(log(n))(base 5), often called logarithmic and most often Big O notation and complexity analysis uses base 2.

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void recursiveFun4(int n, int m, int o)
{
    if (n <= 0)
    {
        printf("%d, %d\n",m, o);
    }
    else
    {
        recursiveFun4(n-1, m+1, o);
        recursiveFun4(n-1, m, o+1);
    }
}

Here, it’s O(2^n), or exponential, since each function call calls itself twice unless it has been recursed n times.

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int recursiveFun5(int n)
{
    for (i = 0; i < n; i += 2) {
        // do something
    }

    if (n <= 0)
        return 1;
    else
        return 1 + recursiveFun5(n-5);
}

And here the for loop takes n/2 since we’re increasing by 2, and the recursion takes n/5 and since the for loop is called recursively, therefore, the time complexity is in

(n/5) * (n/2) = n^2/10,

due to Asymptotic behavior and worst-case scenario considerations or the upper bound that big O is striving for, we are only interested in the largest term so O(n^2).

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Good luck on your midterms 😉