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  1. Practice
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  4. Birthday Cake Candles
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Birthday Cake Candles

  • georgechristied1 + 1 comment

    cheers for pointing this out, but is the solution provided by breakbadsp actually O(2^n)?

    • k_mouratidis + 0 comments

      No, it's not. Also, if this were an interview I would get extra points. It is actually 2*n (compared to n), so both are O(n). But you must keep in mind that this is Python, not C++. It IS actually faster to use its functions rather than looping through everything. In fact, it is almost 2x faster. Code for benchmarks:

      from timeit import timeit
from random import randint
from funtools import partial

f = [randint(0,5) for _ in range(10000)]

def breakbadsp(ar): 
    return ar.count(max(ar))
		
def aida71(ar):
    max_h = ar[0]
    max_n = 1
    for c in ar[1:]:
        if c > max_h:
            max_h = c
            max_n = 1
        elif c == max_h:
            max_n += 1
    return max_n
		
		
# 1.72946... seconds
timeit(partial(breakbadsp, f), number=10000)

# 3.10752... seconds
timeit(partial(aida71, f), number=10000)

      Also, this is Python. Even if the second was faster, you should pick the first for 90% of the cases. You need less time to figure what it's doing, and is less error-prone.