From 2ada485f46a3bde584363a6904108b7824f33f7b Mon Sep 17 00:00:00 2001
From: pranshugoyal0111 <46605937+pranshugoyal0111@users.noreply.github.com>
Date: Wed, 7 Aug 2019 00:19:49 -0700
Subject: [PATCH 1/2] Create Problem

---
 Problem | 10 ++++++++++
 1 file changed, 10 insertions(+)
 create mode 100644 Problem

diff --git a/Problem b/Problem
new file mode 100644
index 00000000..3cffd6bd
--- /dev/null
+++ b/Problem
@@ -0,0 +1,10 @@
+I currently have difficulty in coding out dp problems. Will update the code after the class
+Howvever for the coin change problem , I have an approach in mind that I think should work.
+
+There are 2 cases for each coin. Either that coin will be included in the sum that we have to achieve or that coin will not be included.
+
+coin(amount , n) -> let this denote the amount that can be achieved using  n coins.
+                = coin(amount- j, n) + 1 if ith coin with value j is included
+                or coin(amount, n-1) if ith coin is not included 
+                
+ 

From 52cc0c1b68738a6fc3439a739aca6aaee062235f Mon Sep 17 00:00:00 2001
From: pranshugoyal0111 <46605937+pranshugoyal0111@users.noreply.github.com>
Date: Wed, 7 Aug 2019 20:49:37 -0700
Subject: [PATCH 2/2] Create Robber.java

---
 Robber.java | 40 ++++++++++++++++++++++++++++++++++++++++
 1 file changed, 40 insertions(+)
 create mode 100644 Robber.java

diff --git a/Robber.java b/Robber.java
new file mode 100644
index 00000000..5e260991
--- /dev/null
+++ b/Robber.java
@@ -0,0 +1,40 @@
+// Time Complexity : O(n)
+// Space Complexity : O(n)
+// Did this code successfully run on Leetcode :yes
+// Any problem you faced while coding this : took help from geeksforgeeks
+
+// First 3 cases handle the scenarios when we have 0 or 1 or 2 houses
+// Then for more than 3 houses, at every house we have to take a decision to rob that house or not. We take this decision
+//   by calculating maximum of ( money robbed till 2 houses before that plus money in that house or money robbed till one house
+//    before it)
+
+
+class Solution {
+    public int rob(int[] nums) {
+        
+        if(nums.length==0 || nums==null)
+            return 0;
+        
+        if(nums.length==1)
+            return nums[0];
+        
+        if(nums.length==2)
+            return Math.max(nums[0],nums[1]);
+        
+    
+        int[] dp = new int[nums.length];
+        
+        //Base Case
+        dp[0] = nums[0];
+        dp[1] = Math.max(nums[0],nums[1]);
+            
+        for(int i =2; i< nums.length ; i++)
+        {
+            dp[i] = Math.max(nums[i] + dp[i-2], dp[i-1]);
+            
+        }
+        
+        return dp[nums.length-1];
+        
+    }
+}