Tortoise racing Python Codewars

Task:

Two tortoises named A and B must run a race. A starts with an average speed of 720 feet per hour. Young B knows she runs faster than A, and furthermore has not finished her cabbage.

When she starts, at last, she can see that A has a 70 feet lead but B's speed is 850 feet per hour. How long will it take B to catch A?

More generally: given two speeds v1 (A's speed, integer > 0) and v2 (B's speed, integer > 0) and a lead g (integer > 0) how long will it take B to catch A?

The result will be an array [hour, min, sec] which is the time needed in hours, minutes and seconds (round down to the nearest second) or a string in some languages.

If v1 >= v2 then return nil, nothing, null, None or {-1, -1, -1} for C++, C, Go, Nim, Pascal, COBOL, [-1, -1, -1] for Perl,[] for Kotlin or "-1 -1 -1".

Examples:

(form of the result depends on the language)

race(720, 850, 70) => [0, 32, 18] or "0 32 18"
race(80, 91, 37)   => [3, 21, 49] or "3 21 49"

Solution:

def race(v1, v2, g):
    if v1 >= v2:
        return None
    else:
        time = g / (v2 - v1)
        h = int(time)
        mn = int((time * 60) % 60)
        s = int((time*3600) % 60)
        time_required = [h, mn, s]
    return time_required

def race(v1, v2, g):
    t = 3600 * g/(v2-v1)
    return [t/3600, t/60%60, t%60] if v2 > v1 else None

print(race(720, 850, 70))

Training on Tortoise racing | Codewars

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