The problem
The depth of an integer n is outlined to be what number of multiples of n it’s essential to compute earlier than all 10 digits have appeared at the least as soon as in some a number of.
Instance:
let see n=42
A number of worth digits remark
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
Wanting on the above desk underneath digits column yow will discover all of the digits from “ to 9, Therefore it required 9 multiples of 42 to get all of the digits. So the depth of 42 is 9. Write a operate named computeDepth which computes the depth of its integer argument. Solely optimistic numbers higher than zero will probably be handed as an enter.
The answer in Python code
Possibility 1:
def compute_depth(n):
i = 0
digits = set()
whereas len(digits) < 10:
i += 1
digits.replace(str(n * i))
return i
Possibility 2:
def compute_depth(n):
s, i = set(str(n)), 1
whereas len(s) < 10:
i += 1
s |= set(str(n*i))
return i
Possibility 3:
from itertools import depend
def compute_depth(n):
discovered = set()
replace = discovered.replace
return subsequent(i for i,x in enumerate(depend(n, n), 1) if replace(str(x)) or len(discovered) == 10)
Check instances to validate our answer
check.it("Fundamental checks")
check.assert_equals(compute_depth(1),10)
check.assert_equals(compute_depth(42),9)
