Add implementation for leetcode problems 198, 322

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
+
+

leetcode322.py

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+#!/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
+