How can Dynamic Programming be used to solve the House Robber problem?

Dynamic Programming can be used to solve the House Robber problem by breaking it down into smaller subproblems and storing the solutions to these subproblems in a table.

The House Robber problem involves a row of houses, each containing a certain amount of money. The robber wants to maximize the amount of money he can steal without robbing adjacent houses.

To solve this problem using Dynamic Programming, we can define a function, let's call it "rob", that takes an array of house values as input and returns the maximum amount of money that can be stolen.

We start by defining the base cases. If there are no houses, the maximum amount that can be stolen is 0. If there is only one house, the robber can steal the money from that house.

Next, we define the recursive formula for the "rob" function. At each house, the robber has two choices: either rob the current house and skip the next house, or skip the current house and move on to the next one.

If the robber chooses to rob the current house, the maximum amount that can be stolen is the value of the current house plus the maximum amount that can be stolen from the house two positions before.

If the robber chooses to skip the current house, the maximum amount that can be stolen is the maximum amount that can be stolen from the previous house.

We can use a table to store the solutions to the subproblems. The table will have a size equal to the number of houses. We initialize the table with the base cases and then fill it up iteratively using the recursive formula.

Finally, the maximum amount that can be stolen is the value stored in the last position of the table.

By using Dynamic Programming, we avoid redundant calculations and improve the efficiency of the solution. The time complexity of this approach is O(n), where n is the number of houses.