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:

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 `` 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

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

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)

Test cases to validate our solution

test.it("Basic tests")
test.assert_equals(compute_depth(1),10)
test.assert_equals(compute_depth(42),9)