The challenge
The depth
of an integer n
is defined to be how many multiples of n
it is necessary to compute before all 10
digits have appeared at least once in some multiple.
Example:
Code language: plaintext (plaintext)let see n=42 Multiple value digits comment 42*1 42 2,4 42*2 84 8 4 existed 42*3 126 1,6 2 existed 42*4 168 - all existed 42*5 210 0 2,1 existed 42*6 252 5 2 existed 42*7 294 9 2,4 existed 42*8 336 3 6 existed 42*9 378 7 3,8 existed
Looking at the above table under digits
column you can find all the digits from 0
to 9
, Hence it required 9
multiples of 42
to get all the digits. So the depth of 42
is 9
. Write a function named computeDepth
which computes the depth of its integer argument. Only positive numbers greater than zero will be passed as an input.
The solution in Python code
Option 1:
def compute_depth(n):
i = 0
digits = set()
while len(digits) < 10:
i += 1
digits.update(str(n * i))
return i
Code language: Python (python)
Option 2:
def compute_depth(n):
s, i = set(str(n)), 1
while len(s) < 10:
i += 1
s |= set(str(n*i))
return i
Code language: Python (python)
Option 3:
from itertools import count
def compute_depth(n):
found = set()
update = found.update
return next(i for i,x in enumerate(count(n, n), 1) if update(str(x)) or len(found) == 10)
Code language: Python (python)
Test cases to validate our solution
test.it("Basic tests")
test.assert_equals(compute_depth(1),10)
test.assert_equals(compute_depth(42),9)
Code language: Python (python)