We want to make a row of bricks that have long goal inches. We have a number of small bricks (1 inch each) and large bricks (5 inches each). Go back True if it is possible to make the goal by choosing between the given bricks.
Function:
def make_bricks(small, big, goal):
.
.
.
Examples:
make_bricks(3, 1, 8) → True
make_bricks(3, 1, 9) → False
make_bricks(3, 2, 10) → True
Even though I think you have not counted the full statement of the problem, a solution without using loops would be using combinations :
from itertools import combinations
def make_bricks(small, big, goal):
bricks = [1]*small+[5]*big
return any(sum(lst)==goal
for n in range(len(bricks)+1)
for lst in combinations(bricks,n) )
It is a brute force algorithm, very inefficient. But it is a solution for simple cases.
Edited
Let's solve it without loops or recursion:
With enough small bricks the solution would be trivial:
goal <= small
Let's calculate how many large bricks we are going to use:
nbig = min(big, goal // 5)
What we have left to reach the goal has to be made with small bricks:
(goal - nbig*5) <= small
Finally:
def make_bricks(small, big, goal):
nbig = min(big, goal // 5)
return (goal - nbig*5) <= small
An example using only small bricks to check this solution:
make_bricks(30, 0, 10) --> True