Question

Hi, I need help in solving Python programming problem below:

When John gambles at the casino, he always uses a special system of tactics that he devised himself. It's based on always betting in one of two ways in each game: • betting exactly one chip (to test his luck); • betting all-in (he bets everything he has). Wins in the casino are paid equal to the wager, so if he bets C chips and wins, he gets 2C chips back. If he loses, he doesn't get any chips back. It was a very lucky day yesterday and John won every game he played, starting with one chip and leaving the casino with N chips. We also know that John played all-in no more than K times. Your task is to calculate the smallest possible number of rounds he could have played. Note: betting just one chip is never considered playing all-in. Write a function: def solution (N, K) that, given an integer N and an integer K, returns the minimum number of rounds that are necessary for John to leave the casino with N chips, having played all-in no more than K times.

Thank you.

Best Regards

Answer 1

Answer: I am posting code and its outputs. Please comment in case of any modifications.

def fun(n,k):
    allin = 0
    d1 = dict()
    while(1):
        if n==2:
            d1[n] = "he bet 1"
            break
        if n%2==0 and allin<k:
            d1[n] = "all in"
            allin+=1
            n=int(n/2)
        else:
            d1[n] = "he bet 1"
            n-=1
    print("at the beginning: 1\n")
    count=1
    for key in sorted(d1):
        if count%10==2:
            temp = str(count)+"nd"
        elif count%10==3:
            temp = str(count)+"rd"
        elif count==1:
            temp = str(count)+"st"
        else:
            temp = str(count)+"th"
        print("After {} round: {} ({})".format(temp,key,d1[key]))
        count+=1

    print("\nHe played all-in {} times".format(allin))


print("Enter N and K: ")
a = [int(x) for x in input().split()]
fun(a[0],a[1])

Code Pic:

Output:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 def fun(n, k): allin = 0 d1 = dict() while(1): if n==2: d1[n] = "he bet 1" break if n%2==0 and allin<k: d1[n] = "all in" allin+=1 n=int(n/2) else: d1[n] = "he bet 1" n-=1 print("at the beginning: 1\n") count=1 for key in sorted(d1): if count%10==2: temp = str(count)+"nd" elif count%10==3: temp str(count)+"rd" elif count==1: temp = str(count)+"st" else: temp str(count)+"th" print("After {} round: {} ({})".Format(temp, key, d1[key])) count+=1 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 print("\nHe played all-in {} times".format(allin)) za

Enter N and K: 234 23 at the beginning: 1 After 1st round: 2 (he bet 1) After 2nd round: 3 (he bet 1) After 3rd round: 6 (all in) After 4th round: 7 (he bet 1) After 5th round: 14 (all in) After 6th round: 28 (all in) After 7th round: 29 (he bet 1) After 8th round: 58 (all in) After 9th round: 116 (all in) After 10th round: 117 (he bet 1) After 11th round: 234 (all in) He played all-in 6 times

Answer 2

public class Solution3 
{
   public int solution(int N, int K)
   {
      int n=N;
      int k=K;
      int rounds=0;
      System.out.println(n);
      while ( n>1 )
      {
         if (n % 2 == 0 && k>0)
         {
            n >>= 1;
            k--;
         }
         else
         {
            n--;
         }
         System.out.println(n);
         rounds++;
      }
      return rounds;
   }