##### 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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