fancyRide.md

Fancy Ride

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Being a new Uber user, you have $20 off your first ride. You want to make the most out of it and drive in the fanciest car you can afford, without spending any out-of-pocket money. There are 5 options, from the least to the most expensive: "UberX", "UberXL", "UberPlus", "UberBlack" and "UberSUV".

You know the length l of your ride in miles and how much one mile costs for each car. Find the best car you can afford.

Example

For l = 30 and fares = [0.3, 0.5, 0.7, 1, 1.3], the output should be
solution(l, fares) = "UberXL".

The cost for the ride in this car would be $15, which you can afford, but "UberPlus" would cost $21, which is too much for you.

Input/Output

• [execution time limit] 4 seconds (py3)

• [input] integer l

  A positive number representing the length of the ride.

  Guaranteed constraints:
  4 ≤ l ≤ 30.

• [input] array.float fares

  A strictly increasing array of 5 elements. fares[0] stands for dollars per mile in "UberX", fares[1] is the same for "UberXL", etc.

  Guaranteed constraints:
  0.3 ≤ fares[i] ≤ 5.0.

• [output] string

  The car that you should choose: "UberX", "UberXL", "UberPlus", "UberBlack" or "UberSUV". It is guaranteed that you can afford at least one of them.

[Python 3] Syntax Tips

# Prints help message to the console
# Returns a string
def helloWorld(name):
    print("This prints to the console when you Run Tests")
    return "Hello, " + name

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Solution

def solution(l, fares):
    res_price = 0
    res_uber_i = 0
    
    for i in range(0,len(fares)):
        p = l * fares[i]
        if p <= 20:
            if p > res_price:
                res_price = p
                res_uber_i = i
    
    ubers = ["UberX", "UberXL", "UberPlus", "UberBlack","UberSUV"]
    
    return ubers[res_uber_i]

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See on app.codesignal.com