Find maximum possible stolen value - House Robber Problem

Find Maximum Possible Stolen Value from Houses

The problem of finding the maximum possible stolen value from houses is a variation of the "House Robber Problem," which is a well-known dynamic programming problem. The challenge is to determine the maximum amount of money a thief can steal without triggering alarms by robbing adjacent houses.

Problem Statement

Given a list of non-negative integers representing the amount of money stored in each house, you need to determine the maximum amount of money you can steal without stealing from two consecutive houses.

Example Problem

Suppose you have the following list of house values:

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House values: [6, 7, 1, 30, 8, 2, 4]

In this case, the optimal solution is to rob houses with values 6, 30, and 4, resulting in a total of 40.

Key Concepts Covered

1. Dynamic Programming Approach: Breaking down the problem into overlapping subproblems and solving them using a bottom-up approach.
2. Optimal Substructure: The problem can be solved by combining solutions of subproblems, which makes it a good candidate for dynamic programming.
3. Memoization and Tabulation: Storing the results of subproblems to avoid redundant calculations.

Steps to Solve the Problem

Identify the Subproblems:

• Let dp[i] represent the maximum amount of money that can be stolen up to house i.
• The value at dp[i] is the maximum of either:
  • Stealing from the current house and skipping the previous one: house[i] + dp[i-2]
  • Skipping the current house: dp[i-1]

Recursive Formula:

• The recursive relation is:
• For the first house, the maximum stolen value is simply the value of that house: dp[0] = house[0].
• For the second house, the maximum stolen value is the maximum between the first and second house: dp[1] = max(house[0], house[1]).

Dynamic Programming Table:

• Use a 1D array dp[] to store the results of subproblems.
• Build the table in a bottom-up manner by iterating over the houses and filling in the dp[] array according to the recursive formula.

Implementation:

• Initialize the dp[] array with the base cases.
• Loop through the remaining houses and fill in the table using the recursive relation.
• The answer will be stored in dp[n-1], where n is the total number of houses.

Example Solution

For the given list of house values:

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House values: [6, 7, 1, 30, 8, 2, 4]

The dynamic programming array will be filled as follows:

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dp[0] = 6 dp[1] = 7 dp[2] = max(1 + 6, 7) = 7 dp[3] = max(30 + 7, 7) = 37 dp[4] = max(8 + 7, 37) = 37 dp[5] = max(2 + 37, 37) = 39 dp[6] = max(4 + 37, 39) = 41

The maximum amount that can be stolen is 41.

Applications and Use Cases

• Security Planning: Similar logic can be applied in scenarios where certain assets cannot be protected simultaneously due to adjacency or other constraints.
• Resource Allocation: Used in optimizing the allocation of resources or assets where adjacent resources cannot be chosen simultaneously.

Challenges

• Handling Edge Cases: Handling cases where there are very few houses or all houses have the same value.
• Space Optimization: The solution can be optimized to use constant space by only keeping track of the last two computed values.

Conclusion

The problem of finding the maximum possible stolen value from houses is an excellent example of dynamic programming. It showcases the importance of optimal substructure and overlapping subproblems, making it a great exercise for mastering dynamic programming techniques.

For a detailed step-by-step guide, check out the full article: https://www.geeksforgeeks.org/find-maximum-possible-stolen-value-houses/.