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You need to burn the rope6/30/2023 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.
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