Problem 1

Problem

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+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 
+
+

Robber.java

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+// 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];
+
+    }
+}