Skip to main content

Chocolate Feast - Problem Solving - Hacker Rank Solution.

The expectation is to find the total number of choclate one can consume by taking full advantage of the offer, Here there are 3 inputs n which holds the value of initial amount of money for buying choclate, c is the cost price of each candy if paid by cash and m is the exchange rate for the candy.

Inputs

  • n Initial cash to buy candy.
  • c Coast of each candy if paid by cas.h
  • m Exchange rate for a new candy in offer.

The initial count of choclate will be the cash / coast and the wrappers in hand will be the same value of choclate, and from there we loop through until the wrap count is less than the exchange rate, inside the loop the choclate count will still hold the same fourmula as before but divided with exchange rate. The wrap count is the tricky part... the wrap will be wrap/ exchange rate(the no. choclate) + the remainder of this division(THIS IS VERY IMPORTANT) because for example if the count of wrapper is 3 and the exchange rate is 2 you can only buy 1 choclate but the remainder 1 will not be included and will pop errors in the test cases.

Algorithm

  • declare 2 new variables choclate and wrappers.
  • Initilize the choclate to n/m.
  • Initilize the wrapper to choclate.
  • Run a while loop until the wraper &ls;= exchange rate(or m)
  • Add the value of choclate to wrapper/exchange rate(or m)
  • Set the value of wrapper to the number of choclates exchanged and the balance un-used wrapper.
  • return the count of choclates consumed.

I Highly recommend you dont go through the below code now but try to solve it in your own to get the ability to solve problems but im just pasting it here for those who hasn't manged to find the solution despite many trys 

  	public static int chocolateFeast(int n, int c, int m) {
		int cho = n / c, wrap = cho;
		
		while (wrap >= m) {
			cho += wrap / m;
			wrap = wrap / m + wrap % m;

		}

		return cho;
	}
Labels:

Comments

Post a Comment

Popular posts from this blog

Designer PDF Viewer - HackerRank Problems

Difficulty: EASY Problem : The objective here is to find the size of the highlighted area, and we are given the size's of all the alphabets, we have to find the largest alphabet in the highlighted word and then calculate the size of the rectangle so if the tallest character is 3 then the size of the box will be 3 * number of characters given. Visual representation of the selection : abc def ghij Inputs An array with the sizes of all alphabets a-z in order. A String of highlighted words. Important points to note The array which holds the height of each character in ascending order which means the arrays 0th index will have the height of a 1st index will have the height of b and so on and so forth in the end the hight of z will be there so it's easy to locate each character. A String with the highlighted word. This means we have got the characters inside the rectangle, all we have to find is ...
Post a Comment
Read more

Literals of Base numbers in Java ( Octal , Hexadecimal, Decimal)

1. Overview: A literal key indicates the compiler how to interpret the value of the given data type, for numbers we can calculate the value by using Octal representation or hexadecimal representation but just typing out a hexadecimal value to an int will throw us an error because the compiler has no idea how to handle it but if we assign the java specified prefix for the required bases with some literals then the compiler will not throw us any error as it understands how to interpret the value.  Base Litrals Values Example Eg. Value Decimal none 0-9 int x = 10; x is 10 Octal 0 (zero) as the prefix 0-7 int x = 12; x is 10 Hexadecimal 0x or (zero) along with an x 0-9 and a-f or A-F int x = 0XA; x is 10 Binary or Base(2) Allowed Digits 0 and 1 int i = 10; and now the variable i has value 10. i...
Post a Comment
Read more

Array List - Collections Framework in Java - DSA

Gist: An array list can store individual objects by following insertion order, here the initial capacity is 10 by default but can be modified as per the requirement, once the array list reaches its load factor then internal all the elements of the current array is copied to a new array with the new capacity and the reference variable will now be referring to this new array list and the old array will be dealt by the garbage collector. Hierarchical order Type of constructors Empty argument constructor or the default constructor is the same as invoking any other object here a new ArrayList is created with a default size of 10. Below is the most commonly used constructor by beginners and others alike. ArrayList array = new ArrayList(); //array has a capacity of 10 The default constructor above will allot only 10 slots but if you want the initial size to be 20 or 1000 you can do so with the following constructor this is ideal w...
Post a Comment
Read more