Time complexity for a recursive function

For problems like this, you should figure out the recurrence relation first (by looking at the code), then solve the recurrence relation (using mathematics).

To do step 1, we need to look at each line and see what it contributes to the overall running time T(n) of our algorithm:

int bss(int n){
    if(n <= 1)                        // contributes a constant time a
        return n;                     // contributes a constant time b in the base case only

    return bss(n/2) + bss(n/2);       // contributes a constant time c
                                      // for the two divisions and one addition,
                                      // plus 2T(n/2)
}

Adding up, we get two cases:

n <= 1: T(n) = a + b
n  > 1: T(n) = a + c + 2T(n/2)

To solve this system, we can start writing out terms for values of n. Because we divide n by 2, we might as well choose even n only. Also, it would be nice to have already calculated T(n/2) to calculate T(n); so, we can double our test values of n each time.

n    T(n)
---------
1    a + b
2    a + c + 2T(1) = a + c + 2a + 2b = 3a + 2b + c
4    a + c + 2T(2) = a + c + 6a + 4b + 2c = 7a + 4b + 3c
8    a + c + 2T(4) = a + c + 14a + 8b + 6c = 15a + 8b + 7c
16   a + c + 2T(8) = a + c + 30a + 16b + 14c = 31a + 16b + 15c
...
k    (2k - 1)a + kb + (k - 1)c

Based on the pattern we saw, it seems as though the solution for n = k is (2k - 1)a + kb + (k - 1)c. We can try to verify this by plugging it into our equations:

k = 1: (2k - 1)a + kb + (k - 1)c = a + b = T(1) ... correct
k > 1:

(2k - 1)a + kb + (k - 1)c ?= a + c + 2[(2k/2 - 1)a + (k/2)b + (k/2 - 1)c]
                          ?= a + c + (2k - 2)a + kb + (k - 2)c
                          ?= a + c + 2ka - 2a + kb + kc - 2c
                          ?= -a -c + 2ka + kb + kc
                          ?= (2k - 1)a + kb + (k - 1)c ... correct

So, we have found a valid solution to our recurrence relation. The solution is:

T(n) = (2n - 1)a + nb + (n - 1)c

Rearranging:

T(n) = (2a + c + 1)n - (a + c)

T(n) is the equation of a line.