Given a number (finish end point), then an array of scooters, where scooters represents the position of the
Each scooter can travel up to 10 points before the battery is fully discharged, and cannot go further. For example, if the scooter is at point 5, it can travel to points 5, 6, 7, …, ., up to point 15 (inclusive), but not to point 16 or beyond.
Calculate walking steps to reach the target.
finish=23, scooters[7, 4, 14] output -> solution(finish, scooters) = 4
finish=27, scooters[15, 7, 3, 10] output -> solution(finish, scooters) = 5
finish=23 scooters = [7, 4, 14] def solution(finish, scooters): sum = min(scooters) step = min(scooters) while sum < finish: step += 10 sum = sum + step return step solution(finish, scooters)
How to include
scooters[i] within the while loop to check for the next available scooter?
One way to solve this is by counting the number of steps which cannot be taken by scooter. You can do this by creating a list of all the steps that need to be made, populating it with 1s. Then set to 0 any step that can be taken by a scooter. The number of walking steps is then the sum of the list:
def solution(finish, scooters): steps = [1 for _ in range(finish+1)] zeros = [0 for _ in range(11)] for s in scooters: steps[s:s+11] = zeros return sum(steps) solution(27, [15, 7, 3, 10]) # 5 solution(23, [7,4,14]) # 4
Simplest way to achieve consistency without too much logic is first
sort()ing your list. After that you can calculate any distance between two scooters that has over 10 steps and add it to the sum:
def solution(finish, scooters): scooters.append(finish) scooters.sort() steps = scooters for i in range(1, len(scooters)): steps += max(0, scooters[i] - scooters[i - 1] - 10) if finish in (scooters[i], scooters[i - 1]): break return steps print(solution(27, [15, 7, 3, 10])) # 5 print(solution(23, [7, 4, 14])) # 4 print(solution(10, [15, 7, 3, 10])) # 3 print(solution(2, [15, 3])) # 2
As a simple approach, you can just iterate over the
scooters array, and increment steps based on the distance between the
i-th scooter and the
def solution(finish, scooters): i = 0 n = len(scooters) scooters.sort() steps = min(scooters, finish) for i in range(0, n-1): if scooters[i] + 10 >= finish: break elif scooters[i+1] > finish: steps += max(0, finish-scooters[i]-10) else: steps += max(0, scooters[i+1]-scooters[i]-10) if scooters[-1] < finish: steps += max(0, finish - scooters[-1] - 10) return steps print(solution(27, [15, 7, 3, 10])) print(solution(23, [7,4,14]))
n is the length of