15-liner: lightning - up to 10**10 speedy solution for Number Length by przemyslaw.daniel - python coding challenges

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15-liner: lightning - up to 10**10 solution in Speedy category for Number Length by przemyslaw.daniel

from bisect import bisect
K = [0, 
     10, 
     100, 
	 1000, 
	 10000, 
	 100000, 
	 1000000,
     10000000, 
	 100000000, 
	 1000000000,
     10000000000]

def number_length(n):
    return bisect(K, n)

Feb. 27, 2020

Comments:

_Chico__5d82cee4f5234e78a5c65e on May 26, 2021, 9:45 a.m.
Why not len(str(n))?

przemyslaw.daniel on May 26, 2021, 2:36 p.m.
the secret word is 'time' <pre class='brush: python'>import bisect, random, time EXP, SIZE = 10000, 50000 K = [0] + [10**x for x in range(1, EXP + 1)] S = [random.randint(0, 10**EXP) for _ in range(SIZE)] def number_length1(n): return bisect.bisect(K, n) def number_length2(n): return len(str(n)) def benchamrk(fun): exec_time = time.time() list(map(fun, S)) exec_time = time.time() - exec_time print(f"exec time for {fun.__name__}: {exec_time:.2f}") benchamrk(number_length1) benchamrk(number_length2) </pre> and the result on my machine is <pre class='brush: python'>exec time for number_length1: 0.04 exec time for number_length2: 77.58 </pre>

_Chico__5d82cee4f5234e78a5c65e on May 26, 2021, 2:45 p.m.
Aha...

oduvan on June 24, 2022, 9:13 a.m.
I'm so happy you exist.

przemyslaw.daniel on June 24, 2022, 1:35 p.m.
we are more worry about your and Ukraine existence

Strings and Integers

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