From 46889b116f1f8e8fd7907610446f124abcdd8a20 Mon Sep 17 00:00:00 2001
From: Pranav Sorte <pranavsorte@Pranavs-MacBook-Air.local>
Date: Mon, 24 Nov 2025 21:13:06 -0800
Subject: [PATCH] done

---
 Problem1.java | 42 ++++++++++++++++++++++++++++++++++++++++++
 Problem2.java | 36 ++++++++++++++++++++++++++++++++++++
 2 files changed, 78 insertions(+)
 create mode 100644 Problem1.java
 create mode 100644 Problem2.java

diff --git a/Problem1.java b/Problem1.java
new file mode 100644
index 00000000..4efde39e
--- /dev/null
+++ b/Problem1.java
@@ -0,0 +1,42 @@
+// Time Complexity : O(N) ~ where N is the number of houses to be colored
+// Space Complexity : O(1) 
+// 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
+/*
+Initially took a 1-D array of 3 size where initialzed with the first row of costs.
+Then in each iteration added the current row,col value + the minimum of previous row. Eg - if 0, then previous row 1 and 2 columns.
+Held the previous row red and blue in temp variables since the value will be updated in each iteration
+*/
+
+
+public class Problem1 {
+    public int minCost(int[][] costs) {
+        if(costs.length == 0) {
+            return 0;
+        }
+
+        int m = costs.length;
+        int[] dp = new int[3];
+
+        dp[0] = costs[0][0];
+        dp[1] = costs[0][1];
+        dp[2] = costs[0][2];
+        for(int i = 1;i<m;i++) {
+            int tempR = dp[0];
+            dp[0] = costs[i][0] + Math.min(dp[1], dp[2]);
+            int tempB = dp[1];
+            dp[1] = costs[i][1] + Math.min(tempR, dp[2]);
+            dp[2] = costs[i][2] + Math.min(tempR, tempB);
+        }
+
+        return Math.min(dp[0], Math.min(dp[1], dp[2]));
+    }
+
+    public static void main(String[] args) {
+        Problem1 obj = new Problem1();
+        System.out.println(obj.minCost(new int[][]{{17,2,17},{16,16,5},{14,3,19}}));
+    }
+}
diff --git a/Problem2.java b/Problem2.java
new file mode 100644
index 00000000..30efa9b8
--- /dev/null
+++ b/Problem2.java
@@ -0,0 +1,36 @@
+// Time Complexity : O(M*N)
+// Space Complexity : O(M*N) 
+// 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
+/*
+Initally took the dp array of m+1 and n+1
+Took the amount on the columns whereas took the coins on rows.
+In each iteration, if the denomination of the coin is greater than the current amount, then performed the below calculation
+Added the sum from the row above and from [i][j-coins[i-1]] position
+*/
+
+public class Problem2 {
+    public int change(int amount, int[] coins) {
+        if(coins.length == 0) {
+            return 0;
+        }
+        int m = coins.length;
+        int n = amount;
+        int[][] dp = new int[m+1][n+1];
+        dp[0][0] = 1;
+
+        for(int i = 1;i<=m;i++) {
+            for(int j = 0;j<=n;j++) {
+                if(coins[i-1] > j) {
+                    dp[i][j] = dp[i-1][j];
+                } else {
+                    dp[i][j] = dp[i-1][j] + dp[i][j - coins[i-1]];
+                }
+            }
+        }
+        return dp[m][n];
+    }
+}