hellothere1 - 6 months ago 46

Ruby Question

currently in the process of learning ruby/ programming in general, came across this question:

*Your task is to construct a building which will be a pile of n cubes. The cube at the bottom will have a volume of n^3, the cube above will have volume of (n-1)^3 and so on until the top which will have a volume of 1^3.
You are given the total volume m of the building. Being given m can you find the number n of cubes you will have to build?
The parameter of the function findNb (find_nb, find-nb) will be an integer m and you have to return the integer n such as n^3 + (n-1)^3 + ... + 1^3 = m if such a n exists or -1 if there is no such n*.

and heres my attempt to solve this:

`def find_nb(m)`

(1..Float::INFINITY).each do |n|

if (1..n).inject(0) {|sum, value| sum + value**3} == m

return p n

else

next

end

end

end

This seems to work ok with inputs that i know will work such as

`find_nb(4183059834009)`

find_nb(135440716410000)

find_nb(40539911473216)

Areas i need help in:

- I don't know how i would get it to understand when there is no value of n that would satisfy the equation and therefore output '-1' for an input such as

`find_nb(24723578342962)`

- Any tips on how to make the existing code better would be greatly appreciated

Answer

Hint 1: You don't need to go to infinity: after a certain `n`

, the sum will be greater than `m`

, and rapidly getting further away.

Hint 2: If the `n`

is found, the function will never reach its last line, because of `return`

.

Hint 3: `next`

is automatic if you reach the end of `each`

block.

Hint 4: The sum of cubes does not need to be recalculated from scratch each time. You are not making a whole new building, you're just putting a larger cube underneath.

So...

```
def find_nb(m)
n = 1
sum = 1
while sum < m
n += 1
sum += n**3
end
return sum == m ? n : -1
end
```

Edit: Here's a functional version, but I think the plain `while`

above is still much clearer (and probably faster, too):

```
def find_nb(m)
sum = 0
sizes = 1.upto(Float::INFINITY)
.lazy
.map { |n| sum += n ** 3 }
.take_while { |x| x <= m }
.to_a
sizes.last == m ? sizes.length : -1
end
```

Source (Stackoverflow)