[Using DP to solve rob and coin change]

Sample.java

@@ -5,3 +5,41 @@
 
 
 // Your code here along with comments explaining your approach
+class Sample {
+    // Time Complexity : O(m * n) where m is amount and n is number of coins
+    // Space Complexity : O(amount)
+    // Did this code successfully run on Leetcode : Yes
+    // Use DP to solve this problem, create dp array to store minimum number of coins needed for amount
+    public int coinChange(int[] coins, int amount) {
+        int[] dp = new int[amount + 1];
+        for (int i = 1; i < amount + 1; i++) {
+            dp[i] = Integer.MAX_VALUE - 1;
+        }
+
+        for (int i = 0; i < coins.length; i++) {
+            for (int j = 0; j < amount + 1; j++) {
+                if (j - coins[i] >= 0) {
+                    dp[j] = dp[j - coins[i]] + 1 < dp[j] ? dp[j - coins[i]] + 1 : dp[j];
+                }
+            }
+        }
+        return dp[amount] == Integer.MAX_VALUE - 1 ? -1 : dp[amount];
+
+    }
+
+    // Time Complexity : O(n)
+    // Space Complexity : O(1)
+    // Did this code successfully run on Leetcode : yes
+    public int rob(int[] nums) {
+        int max = 0;
+        for (int j = 0; j < nums.length; j++) {
+            if (j - 2 == 0) { // this when we have reached the first second house we can rob
+                nums[j] = nums[j - 2] + nums[j];// make changes in the array
+            } else if (j - 2 > 0) { // compare the 2 houses before and choose the maximum
+                nums[j] = nums[j - 2] > nums[j - 3] ? nums[j - 2] + nums[j] : nums[j - 3] + nums[j];
+            }
+            max = Math.max(nums[j], max);
+        }
+        return max;
+    }
+}