DP-1 completed

CoinChange.java

@@ -0,0 +1,31 @@
+// Time Complexity : O(amount * n) where n is number of coins
+// Space Complexity : O(amount * n) where n is number of coins
+// Did this code successfully run on Leetcode : No, time exceeded
+// Any problem you faced while coding this : 
+
+// Your code here along with comments explaining your approach
+//Bruce force - recursive solution
+class Solution {
+    public int coinChange(int[] coins, int amount) {
+        int res= helper(coins, amount, 0);
+        if (res>=Integer.MAX_VALUE-10) {
+            return -1;
+        }
+        return res;
+    }
+        private int helper(int[] coins, int amount, int idx){
+            //base 
+            if (amount ==0 ) return 0;
+            if (idx== coins.length || amount <0 ) return Integer.MAX_VALUE - 10;
+            // just so it infinity wont overflow when we add the coins addition.
+
+            //logic
+            // did not choose
+            int case1 = helper (coins, amount, idx +1);
+            // choose
+            int case2 =1 + helper(coins, amount- coins[idx], idx);
+            int res= Math.min(case1, case2);
+            return res;
+
+        }
+}

Sample.java

@@ -1,7 +1,30 @@
-// Time Complexity :
-// Space Complexity :
-// Did this code successfully run on Leetcode :
-// Any problem you faced while coding this :
-
+// Time Complexity :O(n)
+// Space Complexity : O(1)
+// Did this code successfully run on Leetcode : Yes
+// Any problem you faced while coding this : 
 
 // Your code here along with comments explaining your approach
+/* 
+At every house, you have 2 choices, either rob it or skip it.
+	1.	Skip the current house → keep previous maximum.
+	2.	Rob the current house → add its money to the best value from two houses back.
+*/ 
+
+class Solution {
+    public int rob(int[] nums) {
+        int n = nums.length;
+        //base 
+        if (n==0) return 0;
+        if (n==1) return nums[0];
+
+        int[] dp = new int[n];
+        int prev=nums[0];
+        int curr= Math.max(nums[0], nums[1]);
+        for (int i = 2; i<n; i++){
+            int temp = curr;
+            curr= Math.max(curr, nums[i] + prev);
+            prev=temp;
+        }
+        return curr;
+    }
+}