Gf_4
If the running time of an algorithm A is always O(f(n)), then the worst-case running time of A is also O(f(n)).

Question

Gf_4
If the running time of an algorithm A is always O(f(n)), then the worst-case running time of A is also O(f(n)).

BlackTom AI · Accepted Answer

Consider the statement about running times in algorithm analysis. If an algorithm A has its running time always bounded by O(f(n)), this means there exists a constant c and a threshold n0 such that for all n ≥ n0, T_A(n) ≤ c·f(n).

Option 1: False. This would imply that the worst-case time could exceed a bound of O(f(n)) in some inputs, which contradicts the given condition that every running time is within O(f(n)). Hence this option is incorrect.

Option 2: True. Since every possible running time T_A(n) is bounded above by c·f(n) for all sufficiently large n, the maximum (i.e., the worst-case) over all inputs cannot exceed c·f(n) either. Therefore the worst-case running time is also O(f(n)).

In essence, a universal upper bound on all inputs naturally transfers to the maximum over inputs, because the worst case is simply the largest instance within that universal bound.

Gf_4 If the running time of an algorithm A is always O(f(n)), then the worst-case running time of A is also O(f(n)). Single choice

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