Added Solutions to both DP-2 Problesms

Problem1.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 : No
+class Solution {
+    public int minCost(int[][] costs) {
+        // If nohouses
+        if (costs==null|| costs.length==0) return 0;
+        int n = costs.length;
+        int[][]dp=new int[n][3];
+        // Base casefirst house cost stays same
+        dp[0][0] = costs[0][0]; //red
+        dp[0][1] = costs[0][1]; //blue
+        dp[0][2] = costs[0][2]; //green
+        // Fill dp table
+        for (int i = 1; i < n; i++) {
+            //Dp[i] the number of ways to make amout i
+            dp[i][0]=costs[i][0]+Math.min(dp[i-1][1],dp[i-1][2]);
+            dp[i][1]=costs[i][1]+Math.min(dp[i-1][0],dp[i-1][2]);
+            dp[i][2]=costs[i][2]+Math.min(dp[i-1][0],dp[i-1][1]);
+        }
+        // Minimum of last house
+        return Math.min(dp[n - 1][0], 
+               Math.min(dp[n - 1][1], dp[n - 1][2]));
+    }
+}

Problem2.java

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+
+// Time Complexity : O(amount * number of coins)
+// Space Complexity : O(amount)
+// Did this code successfully run on Leetcode : Yes
+// Any problem you faced while coding this : No
+class Solution {
+    public int change(int amount, int[] coins) {
+        // dp[i] = number of ways to make amount i
+        int[]dp=new int[amount+1];
+        // Base case
+        dp[0]=1;
+        // For each coin
+        for (int coin : coins) {
+            for (int i=coin; i<= amount; i++) {
+                dp[i]=dp[i]+dp[i-coin];
+            }
+        }
+        return dp[amount];
+    }
+}