Time complexity of mixedsort (a modification on bubblesort)

Question

So i tried solving this with master theorem but unsure if my solution is correct:

function MixedSort(A[1..n]){
    if (n==1) {
        return A[1];
    }
    m = n/2;
    B[1 ... m] = bubbleSort(A[1...m]) //normal Bubblesort function O(m^2)
    B[m+1 ... n] = MixedSort(A[m+1 ... n])
    return Merge(B[1 ... m], B[m+1 .. n]) //merges 2 sorted arrays in O(n)
}

• T(1) = 1
• T(n) = T(n/2)+n^2+n = T(n/2)+n^2

then used mastertheorem and got O(n^2) because 1<2^2=4

Answer 1

It is O(n^2), yes. You don't even need the master theorem to see this. The recurrence relation is the following: (The O((n/2)^2) and O(n) terms are from the bubble sort and the merge.)

T(n) = T(n/2) + O((n/2)^2) + O(n)

But O((n/2)^2) is equivalent to O(n^2) and O(n^2) dwarfs O(n) such that the O(n) term can just be omitted yielding

T(n) = T(n/2) + O(n^2).

The recurrence T(n) = T(n/2) + n^2 can be solved using the recursive tree method. At each level of the recursion, the problem size is halved (n/2), and it takes O(n^2) time to process each level.

The total time complexity is the sum of the work done at each level, which forms the series:

T(n) = n^2 + (n^2 / 4) + (n^2 / 16) + ... + (n^2 / 2^k)

Since all the terms of that series are constants times n^2 we can rearrange as

T(n) = n^2 * (1 + 1/4 + 1/16 + ...)

That series converges to a constant so the running time is O(n^2).