From 73b3a85e740e065e843538e888d1215d76bb8580 Mon Sep 17 00:00:00 2001
From: Priya <40591285+lakshmiPriya99@users.noreply.github.com>
Date: Sat, 8 Aug 2026 14:54:06 -0500
Subject: [PATCH] Complete DP-1 exercises

---
 Exercise_1.py | 27 +++++++++++++++++++++++++++
 Exercise_2.py | 21 +++++++++++++++++++++
 2 files changed, 48 insertions(+)
 create mode 100644 Exercise_1.py
 create mode 100644 Exercise_2.py

diff --git a/Exercise_1.py b/Exercise_1.py
new file mode 100644
index 00000000..4a02866a
--- /dev/null
+++ b/Exercise_1.py
@@ -0,0 +1,27 @@
+# Time Complexity : O(amount * n), where n is the number of coin types.
+# Space Complexity : O(amount)
+# Did this code successfully run on Leetcode : Yes
+# Any problem you faced while coding this : I had to initialize unreachable amounts clearly so they do not look like valid answers.
+# Approach: I used a one-dimensional DP array where dp[i] stores the fewest coins needed for amount i.
+# For each amount, I tried every coin that can fit and used the best previously solved amount.
+# If the final amount is still unreachable, I returned -1.
+
+
+class Solution:
+
+    def coinChange(self, coins, amount):
+        dp = [amount + 1] * (amount + 1)
+        dp[0] = 0
+
+        for current_amount in range(1, amount + 1):
+            for coin in coins:
+                if coin <= current_amount:
+                    dp[current_amount] = min(
+                        dp[current_amount],
+                        dp[current_amount - coin] + 1
+                    )
+
+        if dp[amount] == amount + 1:
+            return -1
+
+        return dp[amount]
diff --git a/Exercise_2.py b/Exercise_2.py
new file mode 100644
index 00000000..05cbda82
--- /dev/null
+++ b/Exercise_2.py
@@ -0,0 +1,21 @@
+# Time Complexity : O(n)
+# Space Complexity : O(1)
+# Did this code successfully run on Leetcode : Yes
+# Any problem you faced while coding this : The main decision was keeping only the previous two states instead of a full DP array.
+# Approach: At each house, I choose between robbing it with the best value from two houses back or skipping it.
+# Two variables are enough because the current answer only depends on the previous two answers.
+# After processing all houses, the latest value is the maximum money that can be robbed.
+
+
+class Solution:
+
+    def rob(self, nums):
+        two_back = 0
+        one_back = 0
+
+        for money in nums:
+            current = max(one_back, two_back + money)
+            two_back = one_back
+            one_back = current
+
+        return one_back