##### Short Problem Definition:

Little Bob loves chocolates, and goes to a store with $N in his pocket. The price of each chocolate is $C. The store offers a discount: for every M wrappers he gives to the store, he gets one chocolate for free. How many chocolates does Bob get to eat?

##### Link

##### Complexity:

time complexity is O(N);

space complexity is O(1)

##### Execution:

Evaluate the number of wraps after each step. Do this until you have enough wraps to buy new chocolates.

If someone knows how to solve this in O(1) with a mathematical formula, let me know!

##### Solution:

int eatenChocolades(int availableCash, int price, int wrapperDiscount){ int eaten = 0; int wraps = 0; wraps = eaten = availableCash/price; while (wraps >= wrapperDiscount){ int newlyEaten = wraps/wrapperDiscount; eaten += newlyEaten; wraps %= wrapperDiscount; wraps += newlyEaten; } return eaten; }

def chocolateFeast(availableCash, price, wrapperDiscount): eaten = availableCash // price wraps = eaten while wraps >= wrapperDiscount: newlyEaten = wraps // wrapperDiscount eaten += newlyEaten wraps = wraps % wrapperDiscount + newlyEaten return eaten

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