diff --git a/Problem1.java b/Problem1.java
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+++ b/Problem1.java
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+// Time Complexity : O(MXN) M is the total number of coins and N is the lenght of Array
+// Space Complexity : O(N)
+// Did this code successfully run on Leetcode : Yes
+// Any problem you faced while coding this :Initialise the Array with Max Value when looking for the Minimum
+
+
+// Your code here along with comments explaining your approach
+class Solution{
+    int coinChange(int coins[], int amount){
+        //Edge Case
+        if(amount<1){
+            return 0;
+        }
+        else{
+            // Make an array starting from 0 to the Max amount in the array so that at each position we check the mimum coints to reach that particular value
+            int minCoinsDP[]=new int[amount+1];
+            for(int i=1; i<minCoinsDP.length;i++){
+                minCoinsDP[i]=Integer.MAX_VALUE;
+                for(int coin: coins){
+                    // i Will run from start to the Coing value as with a coin value bigger then i we cannot make the amount. 
+                    // If We are able to make a value with smaller denomation we add that and the previous answer of smaller denomation in the minimum
+                    if(coin<=i && minCoinsDP[i-coin]!=Integer.MAX_VALUE){
+                        minCoinsDP[i]=Math.min(minCoinsDP[i], minCoinsDP[i-coin]+1);
+                    }
+                }
+            }
+            // No Amount was able to make with the coin denomaitions. 
+            if(minCoinsDP[amount]==Integer.MAX_VALUE){
+                return -1;
+            }
+            return minCoinsDP[amount];
+        }
+    }
+}
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diff --git a/Problem2.java b/Problem2.java
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+// Time Complexity : O(N) Each House is Visited
+// Space Complexity :O(Dp Array) We make extra array
+// 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
+class Solution {
+    public int rob(int[] nums) {
+        // If no houses
+        if (nums.length == 0) return 0;
+        // If only one house
+        if (nums.length == 1) return nums[0];
+        // dp[i] = max money we can rob till house i
+        int[] dp = new int[nums.length];
+        // Base case
+        dp[0]=nums[0];
+        dp[1]=Math.max(nums[0], nums[1]);
+        // Fill dp arr
+        for (int i=2;i<nums.length;i++) {
+            dp[i]= Math.max(dp[i-1],nums[i]+dp[i-2]);
+        }
+        // Answer is last value
+        return dp[nums.length - 1];
+    }
+}