Create Q-04: House Robber.java

Author: py3-coder

DP Fibonacci || 1D-DP/Q-04: House Robber.java

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+/*
+
+198. House Robber
+
+You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security systems connected and it will automatically 
+contact the police if two adjacent houses were broken into on the same night.
+Given an integer array nums representing the amount of money of each house, return the maximum amount of money you can rob tonight without alerting the police.
+
+Example 1:
+Input: nums = [1,2,3,1]
+Output: 4
+Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3).
+Total amount you can rob = 1 + 3 = 4.
+
+Example 2:
+Input: nums = [2,7,9,3,1]
+Output: 12
+Explanation: Rob house 1 (money = 2), rob house 3 (money = 9) and rob house 5 (money = 1).
+Total amount you can rob = 2 + 9 + 1 = 12.
+ 
+Constraints:
+
+1 <= nums.length <= 100
+0 <= nums[i] <= 400
+*/
+
+class Solution {
+    static int[] memo = new int[101];
+    public int rob(int[] nums) {
+        int n=nums.length;
+        //Recursion::
+        //return solveRec(nums,n-1);
+        Arrays.fill(memo,-1);
+        return solveMemo(nums,n-1);
+        
+    }
+    public static int solveRec(int[] arr,int n){
+        //Base Case ::
+        if(n<0) return 0;
+        //Choices we have ::
+        int rob = arr[n] +solveRec(arr,n-2);
+        int notrob =solveRec(arr,n-1);
+
+        return memo[n]=Math.max(rob,notrob);
+    }
+    public static int solveMemo(int[] arr,int n){
+        //Base Case ::
+        if(n<0) return 0;
+        // precamputed value::
+        if(memo[n]!=-1){
+            return  memo[n];
+        }
+        //Choices we have ::
+        int rob = arr[n] +solveMemo(arr,n-2);
+        int notrob =solveMemo(arr,n-1);
+
+        return memo[n]=Math.max(rob,notrob);
+    }
+   
+}