Question

Here I come again with basic questions 🙁

If I have the next pseudo-code:

iterate over set (A)
    //some *O(1)* operations

iterate over set (B)
    //another *O(1)* operations

From what I have learned, the time would be O(numberOfElementsInA + numberOfElementsInB)

However, If I know that B is subset of A and numberOfElementsInA is always greater or equal than
numberOfElementsInB, can I simplify the time by writing just O(numberOfElementsInA) ?

Answer 1

Yes, you are correct.

This is because numberOfElementsInA + numberOfElementsInB <= 2 * numberOfElementsInA, and from definition of big O notation it makes it O(numberOfElementsInA) (with c=2, and for every N)

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EDIT: To be exact, each loop is O(numberOfElementsInSet_i) – thus there are constants c_i, N_i for each loop such that T(loop_i) <= numberOfElementsInSet_i * c_i for each numberOfElementsInSet_i > N_i.

Thus:

for each numberOfElementsInSet_1 > max{N1,N2}:
T(loop_1) + T(loop_2) <= numberOfElementsInSet_1 * c_1 + numberOfElementsInSet_2 * c_2
<= numberOfElementsInSet_1 * c_1 + numberOfElementsInSet_1 * c_2 //set1 is bigger
<= 2 * numberOfElementsInSet_1 * max{c_1,c_2}

And now we have a formal proof that the loops together are also O(numberOfElementsInSet_1) with N = max{N1,N2} and c = max{c_1,c_2} * 2