From 286dbb6a9538e0d9544fb1cd57459d6055c18682 Mon Sep 17 00:00:00 2001
From: MeghaN28 <n.megha82@gmail.com>
Date: Wed, 22 Apr 2026 12:16:34 -0700
Subject: [PATCH] DP1

---
 HouseRobber.java | 41 ++++++++++++++++++++++++++++++++++++
 Sample.java      | 55 ++++++++++++++++++++++++++++++++++++++++++++----
 2 files changed, 92 insertions(+), 4 deletions(-)
 create mode 100644 HouseRobber.java

diff --git a/HouseRobber.java b/HouseRobber.java
new file mode 100644
index 00000000..b068cb9f
--- /dev/null
+++ b/HouseRobber.java
@@ -0,0 +1,41 @@
+// Time Complexity : O(n) where n is the number of houses
+// Space Complexity : O(1) as we are using only two variables
+// Did this code successfully run on Leetcode :yes
+// Any problem you faced while coding this : No
+
+
+// Your code here along with comments explaining your approach
+// As discussed in class, Greedy or selective approach will not work here, 
+// as we might end up with a local optimal solution.
+// We have to go with exhaustive approach, 
+// where we will try all the combinations of houses to make up the amount.
+// But it will lead to exponential time complexity,
+//  so we will use dynamic programming to optimize it.
+// The input variable is nums, which is changing, so we will create a dp array of size n 
+// where n is the number of houses.
+// For every house, if we choose to rob it,
+//  we will take the value of that house and add it to the value of the house two steps
+//  back, as we cannot rob adjacent houses.
+//  If we choose not to rob it, we will take the value from the previous house.
+// Lastly, we will return the value at the last index of the dp array, which will give us the maximum amount we can rob.
+
+
+public class HouseRobber {
+    class Solution {
+    public int rob(int[] nums) {
+        if (nums.length < 2) return nums[0];
+
+        int n = nums.length;
+        int[] dp = new int[n];
+
+        dp[0] = nums[0];
+        dp[1] = Math.max(nums[0], nums[1]);
+
+        for (int i = 2; i < n; i++) {
+            dp[i] = Math.max(nums[i] + dp[i - 2], dp[i - 1]);
+        }
+
+        return dp[n - 1];
+    }
+}
+}
diff --git a/Sample.java b/Sample.java
index 1739a9cb..6e34d119 100644
--- a/Sample.java
+++ b/Sample.java
@@ -1,7 +1,54 @@
-// Time Complexity :
-// Space Complexity :
-// Did this code successfully run on Leetcode :
-// Any problem you faced while coding this :
+// Time Complexity : O(m*n) where m is the number of coins and n is the amount
+// Space Complexity : o(m*n) where m is the number of coins and n is the amount
+// Did this code successfully run on Leetcode :yes
+// Any problem you faced while coding this : not exactly, but I had to be careful 
+// with the initialization of the dp array to avoid overflow issues.
 
 
 // Your code here along with comments explaining your approach
+// As discussed in class, Greedy or selective approach will not work here, 
+// as we might end up with a local optimal solution.
+// We have to go with exhaustive approach, 
+// where we will try all the combinations of coins to make up the amount.
+// But it will lead to exponential time complexity,
+//  so we will use dynamic programming to optimize it.
+// The input variables are coins and amount,which are changing, so we will create a dp array of size (m+1) x (n+1) where m is the number of coins and n is the amount.
+// For every amount, if 0 - not choose we will take the value from the previous row,
+//  if we choose the coin, we will take 1 + value from the same row and column - coin value.
+// Lastly, we will return the value at the bottom right corner of the dp array, if it is still the initialized value, we will return -1,
+//  otherwise we will return that value.
+class Solution {
+    public int coinChange(int[] coins, int amount) {
+
+        int m = coins.length;
+        int n = amount;
+
+        int[][] dp = new int[m + 1][n + 1];
+
+        dp[0][0] = 0;
+
+        // Initialize top row with a large value
+        for (int j = 1; j <= n; j++) {
+            dp[0][j] = Integer.MAX_VALUE - 10; // avoid overflow
+        }
+
+        for (int i = 1; i <= m; i++) {
+            for (int j = 0; j <= n; j++) {
+
+                if (j < coins[i - 1]) {
+                    dp[i][j] = dp[i - 1][j];
+                } else {
+                    dp[i][j] = Math.min(
+                        dp[i - 1][j],
+                        1 + dp[i][j - coins[i - 1]]
+                    );
+                }
+            }
+        }
+
+        int res = dp[m][n];
+
+        if (res == Integer.MAX_VALUE - 10) return -1;
+        return res;
+    }
+}
\ No newline at end of file