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  1. Practice
  2. Algorithms
  3. Implementation
  4. Chocolate Feast
  5. Discussions

Chocolate Feast

  • jason_e_reynolds + 5 comments

    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

    • akonibrahim1 + 0 comments

      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)

    • rampappula + 2 comments

      wrappers = wrappers % wrapper_cost instead of wrappers -= new_chocolate * wrapper_cost

      was better understandable for me.

      • gwarks + 0 comments

        wrappers %= wrapper_cost would be even shorter

      • chay_29 + 0 comments

        thanks.anyways, i got it.

    • pcortezzi + 0 comments

      Your code is flawless. Quick, simple and completely readable! Congratulations!

    • frankdeadlycaste + 1 comment

      see this

      def chocolateFeast(n, c, m):

      ch = n/c
all_ch = ch
while (ch>=1):
        all_ch += ch/m
        ch = ch/m + ch%m
return all_ch

      • prathameshbagek1 + 0 comments

        Did almost the same thing in Java8

        static int chocolateFeast(int n, int c, int m) {
        int sum = n / c;
        if (sum < m) return sum;
        if (sum == m)   return sum + 1;
        int choc = sum, wrap = choc;
        do {
            choc += wrap / m;
            wrap = (wrap / m) + (wrap % m);
        } while (wrap >= m);
        return choc;
    }

    • ___GHOST___ + 0 comments
      [deleted]