feat: add LeetCode 1518 Water Bottles solution

- Implement simulation approach with O(log n) time complexity
- Include alternative mathematical O(1) solution in comments
- Add comprehensive problem explanation and examples
- Addresses GitHub issue #22

Author: URAYUSHJAIN

1518. Water Bottles.cpp

@@ -0,0 +1,45 @@
+class Solution {
+public:
+    int numWaterBottles(int numBottles, int numExchange) {
+        int totalDrunk = numBottles;  
+        int emptyBottles = numBottles;          
+      
+        while (emptyBottles >= numExchange) {
+            int newBottles = emptyBottles / numExchange;  
+            totalDrunk += newBottles; 
+            emptyBottles = emptyBottles % numExchange + newBottles;  
+        }
+        
+        return totalDrunk;
+    }
+};
+
+/* 
+Alternative Mathematical Solution (More Efficient):
+class Solution {
+public:
+    int numWaterBottles(int numBottles, int numExchange) {
+        // Mathematical formula: numBottles + (numBottles - 1) / (numExchange - 1)
+        return numBottles + (numBottles - 1) / (numExchange - 1);
+    }
+};
+
+Problem Explanation:
+- You have numBottles full water bottles
+- After drinking, you get numBottles empty bottles
+- You can exchange numExchange empty bottles for 1 new full bottle
+- Find maximum bottles you can drink
+
+Example: numBottles = 9, numExchange = 3
+- Drink 9 bottles → 9 empty bottles
+- Exchange 9 empty → 3 new bottles, 0 empty remaining
+- Drink 3 bottles → 3 empty bottles  
+- Exchange 3 empty → 1 new bottle, 0 empty remaining
+- Drink 1 bottle → 1 empty bottle
+- Cannot exchange anymore (need 3 empty for 1 new)
+- Total drunk: 9 + 3 + 1 = 13
+
+Time Complexity: O(log(numBottles)) - simulation approach
+                 O(1) - mathematical approach
+Space Complexity: O(1)
+*/