这是针对代码战的挑战，我通过了所有测试，但它说我的代码运行时间太长？

Question

I am solving the CodeWars challenge Bouncing Balls :

A child is playing with a ball on the nth floor of a tall building. The height of this floor, h , is known.

He drops the ball out of the window. The ball bounces (for example), to two-thirds of its height (a bounce of 0.66).

His mother looks out of a window 1.5 meters from the ground.

How many times will the mother see the ball pass in front of her window (including when it's falling and bouncing?

Three conditions must be met for a valid experiment:

• Float parameter "h" in meters must be greater than 0
• Float parameter "bounce" must be greater than 0 and less than 1
• Float parameter "window" must be less than h.

If all three conditions above are fulfilled, return a positive integer, otherwise return -1.

Note:

The ball can only be seen if the height of the rebounding ball is strictly greater than the window parameter.

My code:

def bouncing_ball(h, bounce, window):
    ball_h = h
    num = 0 
    #initial drop
    if ball_h>window:
        num+=1
        ball_h = ball_h*bounce
    else:
        num = -1

    while ball_h > window:
        num +=2
        ball_h = ball_h*bounce
        
    return num

My code passes all the tests yet it says my code takes too long to run, what could be the problem?

Thanks

Answer 1

There is a closed formula for this (which avoids doing the loop in Python code -- and therefor can be expected to run faster).

Let n be the number of times that the ball reaches a relative peak height that is still high enough, then the following condition is true:

h * bounce ⁿ > window

From this we can derive n :

n = ^⌈ log _bounce (window / h)) ^⌉

We need to multiply this n with 2 (for up and down) and subtract one as the original position of the ball is at a peak (there is no corresponding "up").

This leads to the following code:

from math import log, ceil

def bouncing_ball(h, bounce, window):
    if not (0 <= bounce < 1) or h < 0: 
        return -1 
    return ceil(log(window / h, bounce)) * 2 - 1

Answer 2

You forgot to check the conditions and thus got into an infinite loop. If you add this at the start of your code, it easily gets accepted:

    if not (h > 0 and 0 < bounce < 1 and window < h):
        return -1

Often, such constraints are guarantees , ie, you can assume that every input fulfills them. But in this case, you're told to return -1 if they're not fulfilled, so you must not assume them to be.

That said, trincot's closed formula should be better. Maybe there's a harder version of the problem on codewars, where that is needed. But in this one, both your solution (with the above fix) and trincot's get accepted in about 0.5 seconds (so that seems to be the baseline, most of the time actually taken by the judge, your solution actually takes something like 0.000034 seconds and trincot's 0.000023 seconds (measured by running codewars' test cases elsewhere (you get to see them once you solved the kata)).

I also added assert num <= 87 before the return, that still got accepted. With 86 it failed. So your loop in fact only has very little to do. At first I suspected that maybe they're not testing harder cases because large results might require too much precision for float , but not even that seems to be the case. For example for arguments 100, 0.9999999, 1.5 , both you and trincot return the same result, 83994097. And yours takes several seconds while trincot's only takes microseconds:

83994097  3571749 us  Julian
83994097       61 us  trincot

Try it online!