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haykhvs.haykh

Find if at least 2 out of 3 values === True.

a bit shorter version (but slighly slower)

Code
Diff
  • def solution(p):
    return (r := 0b11) and [(r := (r >> (_ is True))) for _ in p] and not r
    • def solution(p):
    • r = 0b11
    • for _ in p:
    • if _ is True:
    • r >>= 0b1
    • return not r
    • return (r := 0b11) and [(r := (r >> (_ is True))) for _ in p] and not r
haykhvs.user4931254

Find if at least 2 out of 3 values === True.

i just like bitwise operations.

ps. this is the fastest solution i found, even though it's not the shortest one.

Code
Diff
  • def solution(p):
    r = 0b11
    for _ in p:
        if _ is True:
            r >>= 0b1
    return not r
    • def solution(p):
    • return sum(1 for val in p if isinstance(val, bool) and val) >= 2
    • def solution(p):
    • r = 0b11
    • for _ in p:
    • if _ is True:
    • r >>= 0b1
    • return not r