You half to burn the rope8/3/2023 The overall algorithm seems to be O(n^2). You can probably get O(1) run time for solution contains by using a boolean array for lookup. #then add the achievable times with i ropes #update the start_idx to be the starting index of times achievable with If solutions does not contain abs( x/2-time )Īppend abs( x/2-time ) to solutions_to_add If solutions does not contain abs( x-time ) #then continuously add additional ropes until we have n ropes #of the first recorded time measurable with 1 rope. Let start_idx be an index in the solutions array #we will store the index of the first time that was recorded with n-1 ropes #we can be efficient by only checking the times achievable with n-1 ropes So, in pseudocode, perhaps something like this let solutions be a list of measurable times We do not care about a time t that was achieved with m with m<=n-1 ropes because we would have considered these four cases when we were adding the m+1-th rope. This means we can achieve abs( x/2 - x_prev) minutes. Start burning the x_prev minutes as we burn the next rope from 2 ends.This means we can achieve abs( x - x_prev ) minutes. Start burning the x_prev minutes as we burn the next rope from 1 end.This means we can achieve x_prev + x/2 minutes. Wait for the whole x_prev minutes to expire, then burn the next rope from 2 ends.This means we can achieve x_prev + x minutes. Wait for the whole x_prev minutes to expire, then burn the next rope from 1 end.Then, consider what happens if we add the n+1th rope. Now, suppose it is possible to measure x_prev minutes with n ropes. We can start with a base state of 1 rope yields x minutes or x/2 minutes. I might have overlooked something, so be wary even if it seems to make sense. Well, here is my attempt to solve the problem with greater efficiency. until n = 0 (All ropes finished burning)įor n = 2 and x = 60, I've found that the following time period can be measured: 30, 60, 90, 120, 45 and 15.Īs suggested, I posted the question on cs.: Repeat the step 1 argument with x + y + z = n - 1 (with constraints imposed on x, y, and z since some ropes are still burning and we cannot set the fire off) and add all the newly generated cases to the stack/queue. Now we have another scenarios with certain amount of ropes that are being burnt. Output the time that has passed (calculated based on how long the finished rope has burnt, and which ends were burnt at what time). For each item in the stack/queue, determine how many minutes have passed when there is a rope finishes burning.Consider all possible cases for x, y and z and add those cases to a stack/queue. We have x + y + z = n and that x,y,z are positive integers and z != 0. Let number of ropes that will not be burnt at this stage be x, number of ropes that will be burnt one end be y, and number of ropes that will not be burnt at all be z. For a given rope, we have choices either to burn both ends, one end, or not burning the rope at all. Start at minute 0, we have n ropes, each takes x minutes to burn.I imagine the solution to this would involve dynamic programming, but I am not quite sure. Of course my aim would be finding an algorithm with minimal complexity. (burning one end of the rope), or 30 minute period (burning both ends Using these n ropes, what time quantity can you measure?įor example, with n = 1 and x = 60, I can measure 60 minute period But the ropes have different densities at different points, so There are n ropes, each rope takes x minutes toīurn (for simplicity assume x is positive integer). How do you use these two ropes to measure 45 minutes? There’s no guarantee of consistency in the time it takes different But either rope has different densities at different points, so Original question: There are two ropes, each rope takes 1 hour toīurn. Generalized from a technical interview question:
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