From 51252f28fc2fc20de96a8785d19b56dd4509435f Mon Sep 17 00:00:00 2001
From: rishigoswamy <goswamyrishi@gmail.com>
Date: Thu, 12 Feb 2026 15:34:51 -0800
Subject: [PATCH] Add implementation for leetcode problems 198, 322

---
 leetcode198.py | 44 ++++++++++++++++++++++++++++++++++++++++++++
 leetcode322.py | 39 +++++++++++++++++++++++++++++++++++++++
 2 files changed, 83 insertions(+)
 create mode 100644 leetcode198.py
 create mode 100644 leetcode322.py

diff --git a/leetcode198.py b/leetcode198.py
new file mode 100644
index 00000000..8eab499e
--- /dev/null
+++ b/leetcode198.py
@@ -0,0 +1,44 @@
+#!/usr/bin/env python3
+# -*- coding: utf-8 -*-
+"""
+Created on Thu Feb 12 15:33:04 2026
+
+@author: rishigoswamy
+
+    Problem: House Robber
+    Link: https://leetcode.com/problems/house-robber/
+
+    Description:
+    You are given an integer array nums representing the amount of money
+    at each house. Adjacent houses cannot be robbed.
+    Return the maximum amount of money you can rob.
+
+    Approach:
+    Optimized Dynamic Programming (Space Optimized).
+
+    Let:
+    r1 = max money till house i-2
+    r2 = max money till house i-1
+
+    For each house:
+        current = max(r1 + num, r2)
+
+    Shift:
+        r1 = r2
+        r2 = current
+
+    Time Complexity: O(n)
+    Space Complexity: O(1)
+
+"""
+
+class Solution:
+    def rob(self, nums: List[int]) -> int:
+        r1, r2 = 0, 0 
+        for num in nums:
+            temp = max(r1 + num, r2)
+            r1 = r2
+            r2 = temp
+        return r2
+        
+
diff --git a/leetcode322.py b/leetcode322.py
new file mode 100644
index 00000000..79c50a51
--- /dev/null
+++ b/leetcode322.py
@@ -0,0 +1,39 @@
+#!/usr/bin/env python3
+# -*- coding: utf-8 -*-
+"""
+Created on Thu Feb 12 15:30:44 2026
+
+@author: rishigoswamy
+
+    Problem: Coin Change
+    Link: https://leetcode.com/problems/coin-change/
+
+    Description:
+    Given an integer array coins representing coin denominations
+    and an integer amount representing a total amount of money,
+    return the fewest number of coins needed to make up that amount.
+    If that amount cannot be made up, return -1.
+
+    Approach:
+    Bottom-Up Dynamic Programming.
+
+    - dp[i] represents the minimum coins needed to make amount i.
+    - Initialize dp with a large value (amount + 1).
+    - Base case: dp[0] = 0.
+    - For each amount, try every coin and update dp.
+
+    Time Complexity: O(amount * number_of_coins)
+    Space Complexity: O(amount)
+
+"""
+
+class Solution:
+    def coinChange(self, coins, amounts: int) -> int:
+        dp = [amounts + 1] * (amounts + 1)
+        dp[0] = 0
+        for amount in range(1, amounts + 1):
+            for coin in coins:
+                if amount - coin >= 0:
+                    dp[amount] = min(dp[amount], 1 + dp[amount - coin])
+        return dp[-1] if dp[-1] != amounts+1 else -1
+        
\ No newline at end of file