found an analytic solution to 1518!

this took way longer than i had hoped, but was kind of fun

Author: jeremymanning

problems/1518/jeremymanning.md

@@ -43,3 +43,58 @@ class Solution:
 ![Screenshot 2024-07-06 at 11 47 50 PM](https://github.com/ContextLab/leetcode-solutions/assets/9030494/66a9540c-5afc-4cd4-9652-6e8aa99ab59d)
 
 Slow, but I'll take it-- solved!
+
+---
+
+# Thinking more about this...
+
+- Let's see if we can come up with an analytic solution
+- We can initially drink `numBottles` (let's write this as $B$ to simplify notation)
+- After that, every `numExchange` bottles (we'll write this as $E$ to simplify notation) turns into a full bottle (which can be added to the total) and then an empty bottle (which can be exchanged in the next round)
+
+So the total drinkable number of bottles can be given by the series: $\text{total} = B + \left\lfloor \frac{B}{E} \right\rfloor + \left\lfloor \frac{\left\lfloor \frac{B}{E} \right\rfloor + (B \mod E)}{E} \right\rfloor + \left\lfloor \frac{\left\lfloor \frac{\left\lfloor \frac{B}{E} \right\rfloor + (B \mod E)}{E} \right\rfloor + ((B \mod E) \mod E)}{E} \right\rfloor + \ldots$
+
+In other words:
+  - Start by drinking $B$ bottles, which yields $B$ empty bottles
+  - Those empties can be exchanged for $\left\lfloor \frac{B}{E} \right\rfloor$ new full bottles
+  - There are $B \mod E$ additional empty bottles left after that exchange
+  - In each subsequent round, we can repeat this same process (divide by $E$, take the floor, add this to the total number of drinks, add the remainder to the total number of empties)
+
+Interestingly, we can see that $\left\lfloor \frac{\left\lfloor \frac{B}{E} \right\rfloor + (B \mod E)}{E} \right\rfloor$ simplifies to $\left\lfloor \frac{\left\lfloor \frac{B}{E} \right\rfloor}{E} \right\rfloor$:
+  - $B \mod E$ is always less than $E$
+  - Therefore $(B \mod E) / E$ is always less than 1, which means we can get rid of it in the "floor" operation
+
+So now we have $\left\lfloor \frac{\left\lfloor \frac{B}{E} \right\rfloor}{E} \right\rfloor$, which simplifies to $\left\lfloor \frac{B}{E^2} \right\rfloor$.  In general (as the number of exchanges approaches infinity) we can see that
+the exponent in the demoninator will keep growing (equal to the number of exchanges): $\sum_{k = 1}^\infty \left\lfloor \frac{B}{E^k} \right\rfloor$.  The demonitor is increasing exponentially, so the series will converge to...something.  I wish
+I remembered my calculus better 🙃.
+
+Doing some Googling, it looks like the series $\sum_{k = 0}^\infty x r^k$ converges to $\frac{x}{1 - r}$ if $|r| < 1$.  So...let's see...
+
+We can take $x = B$ and $r = \frac{1}{E}$ (which is less than 1, since we know that $E \geq 2$).  So the series will converge to 
+
+$\frac{B}{1 - \frac{1}{E}} = $B \frac{1}{1 - \frac{1}{E}} = B \frac{E}{E - 1}$.
+
+Now we can split the fraction:
+
+$B\frac{E}{E - 1} = B \left( 1 + \frac{1}{E - 1} \right)$
+
+Which simplifies to:
+
+$B \left( 1 + \frac{1}{E - 1} \right) = B + \frac{B}{E - 1}$.  Since we can only exchange "whole" bottles, we need to round down to the nearest integer:
+
+$\frac{BE}{E - 1} \approx \left\lfloor \frac{B -1}{E - 1} \right\rfloor$.  So the *total* number of drinks is $B + \left\lfloor \frac{B -1}{E - 1} \right\rfloor$.
+
+Finally(!!), we can turn this back into Python code:
+```python
+class Solution:
+    def numWaterBottles(self, numBottles: int, numExchange: int) -> int:        
+        return numBottles + (numBottles - 1) // (numExchange - 1)
+```
+- Test cases pass!!
+- Checking a random additional test case: `numBottles = 48, numExchange = 7` (passes!)
+- I'm just gonna submit this instead of checking in more detail...
+
+![Screenshot 2024-07-07 at 2 47 33 PM](https://github.com/ContextLab/leetcode-solutions/assets/9030494/b17fec58-14a2-4faa-9f22-0bf1cdc13455)
+
+Woo!  Go math 🎉!
+