diff --git a/Problem1.py b/Problem1.py
new file mode 100644
index 00000000..7d41f341
--- /dev/null
+++ b/Problem1.py
@@ -0,0 +1,73 @@
+## Problem1 (https://leetcode.com/problems/coin-change/)
+
+# Time Complexity: O(M*N) where m is the amount and n is the number of coins
+# class Solution:
+#     def coinChange(self, coins: List[int], amount: int) -> int:
+        # Region Recursive solution
+
+        # def findCoins(coins: List[int], amount: int, index: int, coinsUsed: int):
+        #     # base
+        #     if index >= len(coins) or amount < 0:
+        #         return -1
+        #     if amount == 0:
+        #         return coinsUsed
+             
+        #     # case choose the coin
+        #     case0 = findCoins(coins, amount, index + 1, coinsUsed)
+        #     # case not choose the coin
+        #     case1 = findCoins(coins, amount - coins[index], index, coinsUsed + 1)
+
+        #     if case0 == -1:
+        #         return case1
+        #     if case1 == -1:
+        #         return case0
+            
+        #     return min(case0, case1)
+        
+        # return findCoins(coins, amount, 0, 0)
+        # End Region Recursive solution
+
+class Solution:
+    def coinChange(self, coins: List[int], amount: int) -> int:
+        """
+        Algorithm: Dynamic Programming (1D Tabulation)
+        We use a 1D array `minCoins` where the index `i` represents the target amount, 
+        and the value at `minCoins[i]` represents the minimum number of coins needed 
+        to make that amount.
+        
+        Time Complexity: O(C * A)
+        Where C is the length of `coins` and A is the `amount`.
+        We have a nested loop: the outer loop runs C times, and the inner loop runs A + 1 times.
+        
+        Space Complexity: O(A)
+        Where A is the `amount`. We use a single 1D array of size A + 1 to store the subproblem results.
+        """
+        
+        # Initialize the DP array with `amount + 1` (a placeholder for "infinity").
+        # The maximum possible coins we could ever use is `amount` (if using only 1-value coins), 
+        # so `amount + 1` acts as an unreachable max value.
+        minCoins = [(amount + 1) for _ in range(amount + 1)]
+        
+        # Base case: 0 coins are needed to make an amount of 0
+        minCoins[0] = 0
+        
+        # Iterate through each coin denomination available
+        for c in coins:
+            # For each coin, compute the minimum coins needed for every amount up to `amount`
+            for i in range(amount + 1):
+                if i < c:
+                    # If the current amount `i` is strictly less than the coin's value, 
+                    # we can't use this coin. We keep the previous best count.
+                    continue
+                else:
+                    # If we CAN use the coin, the optimal choice is the minimum between:
+                    # 1. Not using the coin at all (keeping the current `minCoins[i]`)
+                    # 2. Using the coin (1 + the optimal solution for the remaining amount `i - c`)
+                    minCoins[i] = min(minCoins[i], 1 + minCoins[i - c])
+                    
+        # The final answer for the requested amount is at the end of our DP array
+        result = minCoins[-1]
+        
+        # If the result is still >= amount + 1, it means the amount cannot be made 
+        # with any combination of the given coins, so we return -1.
+        return -1 if result >= amount + 1 else result
\ No newline at end of file
diff --git a/Problem2.py b/Problem2.py
new file mode 100644
index 00000000..204a1046
--- /dev/null
+++ b/Problem2.py
@@ -0,0 +1,34 @@
+## Problem2 (https://leetcode.com/problems/house-robber/)
+
+# Time Complexity: O(n)
+# Space Complexity: O(1)
+class Solution:
+    def rob(self, nums: List[int]) -> int:
+        n = len(nums)
+        
+        # Guard clause / Edge cases: 
+        # If there are 1 or 2 houses, just take the maximum available.
+        # This prevents an IndexError when trying to access nums[1] later.
+        if n <= 2:
+            return max(nums) if nums else 0
+            
+        # 'previous' tracks the max money we can rob up to two houses ago (i - 2)
+        previous = nums[0]
+        
+        # 'current' tracks the max money we can rob up to the previous house (i - 1)
+        current = max(previous, nums[1])
+        
+        # Iterate starting from the 3rd house (index 2)
+        for i in range(2, n):
+            # Temporarily store 'current' so we can update 'previous' afterwards
+            temp = current
+            
+            # Core DP Logic: Decide whether to rob or skip the current house.
+            # 1. Skip: Keep the max money from the previous house ('temp').
+            # 2. Rob: Add this house's value to the max from two houses ago ('previous').
+            current = max(temp, nums[i] + previous)
+            
+            # Shift our two tracking variables forward for the next iteration
+            previous = temp
+            
+        return current
\ No newline at end of file