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I guess you are doing way to much unnecessary work. Heree is my while loop
while(m<=wrappers){ wrappers = wrappers - m; totalchocs++; wrappers++; }
This approach is clean and easy to follow, which is great.
However, it results in O(N) time (input of 10000 1 2 results in the loop running 9999 times.)
If you switch the loop to the following you end up with O( lg(N)) (or 14 in the worst case, since you are using integer math).
def chocolateFeast(cash, cost, wrapper_cost): chocolate = cash//cost wrappers = chocolate while wrappers // wrapper_cost > 0: new_chocolate = wrappers // wrapper_cost wrappers -= new_chocolate * wrapper_cost wrappers += new_chocolate chocolate += new_chocolate return chocolate
Hi Jason I kind of did the same thing what you did but mine didn't pass the test cases how whats the difference
import time
start_time = time.time()
def chocolateFeast1(he_has,cost,wrapper):
current_chocates = int(he_has / cost) current_wrappers = current_chocates while current_wrappers >= wrapper: current_chocates += 1 current_wrappers -= cost current_wrappers += 1 print(f"Took { time.time() - start_time}") print (current_chocates)
start_time = time.time() def chocolateFeast(cash, cost, wrapper_cost):
chocolate = cash//cost wrappers = chocolate while wrappers // wrapper_cost > 0: new_chocolate = wrappers // wrapper_cost wrappers -= new_chocolate * wrapper_cost wrappers += new_chocolate chocolate += new_chocolate print(f"Took { time.time() - start_time}") print (chocolate)
wrappers = wrappers % wrapper_cost instead of wrappers -= new_chocolate * wrapper_cost
was better understandable for me.
wrappers %= wrapper_cost would be even shorter
your code will result in performance issues when the count is too high. Thats the reason suggested to do with / and % operators
Chocolate Feast
You are viewing a single comment's thread. Return to all comments →
I guess you are doing way to much unnecessary work. Heree is my while loop
This approach is clean and easy to follow, which is great.
However, it results in O(N) time (input of 10000 1 2 results in the loop running 9999 times.)
If you switch the loop to the following you end up with O( lg(N)) (or 14 in the worst case, since you are using integer math).
Hi Jason I kind of did the same thing what you did but mine didn't pass the test cases how whats the difference
import time
start_time = time.time()
def chocolateFeast1(he_has,cost,wrapper):
start_time = time.time() def chocolateFeast(cash, cost, wrapper_cost):
wrappers = wrappers % wrapper_cost instead of wrappers -= new_chocolate * wrapper_cost
was better understandable for me.
wrappers %= wrapper_cost would be even shorter
your code will result in performance issues when the count is too high. Thats the reason suggested to do with / and % operators