House Robber IV

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class Solution:
    def minCapability(self, nums: List[int], k: int) -> int:
        l, r = min(nums), max(nums)
        while l < r:
            m = (l + r) // 2
            last = take = 0
            for n in nums:
                if last:
                    last = 0
                    continue
                if n <= m:
                    take += 1
                    last = 1
            if take >= k:
                r = m
            else:
                l = m + 1
        return l

Identifying Problem Isomorphism

“House Robber IV” has an approximate isomorphism to “Binary Tree Maximum Path Sum”. Both involve a decision-based traversal of a tree (binary tree or general tree) where the task is to maximize the sum of the picked nodes while adhering to certain constraints.

In “House Robber IV”, you’re given a tree where each node has an integer value, and you need to rob nodes in such a way that no two robbed nodes are adjacent, and you aim to maximize the total robbed amount.

In “Binary Tree Maximum Path Sum”, you’re given a binary tree where each node contains an integer value. Your task is to find the maximum possible sum of any path in the tree. A path is defined as any sequence of nodes from starting node to any node in the tree along the parent-child connections.

Both require a dynamic programming approach, as you need to make a decision at each node based on the potential gain from the subtree rooted at that node.

“House Robber IV” is more complex due to the additional constraint of not being able to rob adjacent nodes, requiring a more sophisticated state management than “Binary Tree Maximum Path Sum”.

10 Prerequisite LeetCode Problems

The problem “House Robber IV” involves understanding concepts related to arrays, dynamic programming, binary search, and greedy algorithms.

  1. House Robber (LeetCode #198): This problem is the most basic version of the House Robber series. It involves finding an optimal solution using dynamic programming.

  2. House Robber II (LeetCode #213): This problem is a slight extension of the previous problem where houses are arranged in a circle. It also involves dynamic programming.

  3. House Robber III (LeetCode #337): This problem takes the House Robber problem to trees. The idea of not selecting adjacent elements is the same.

  4. Maximum Subarray (LeetCode #53): This problem teaches you about handling subarrays and using dynamic programming to find an optimal solution.

  5. Partition Equal Subset Sum (LeetCode #416): This problem can help you get a better understanding of the concept of partitioning and dynamic programming.

  6. Jump Game II (LeetCode #45): This is a classic problem that involves dynamic programming and greedy algorithms. It could be helpful to understand the minimum step concept.

  7. Minimum Size Subarray Sum (LeetCode #209): The problem helps you understand how to handle subarrays to achieve a specific condition, similar to the House Robber IV problem where you need to rob a minimum number of houses.

  8. Binary Search (LeetCode #704): This is a simple problem to understand the basic concept of binary search which can be useful in the main problem.

  9. Two Sum II - Input array is sorted (LeetCode #167): This problem helps understand the two-pointer technique which can be quite useful for solving array-related problems.

  10. Find Minimum in Rotated Sorted Array (LeetCode #153): This problem can be useful in understanding the manipulation of sorted arrays and the binary search concept.

Clarification Questions

  1. Is the array of house values sorted in any particular order, or is it random?
  2. Can ‘k’ be equal to the total number of houses, or is it strictly less than half of the total number?
  3. Are negative or zero values possible for the amount of money in each house, or are they all strictly positive integers?
  4. Is it guaranteed that there will always be at least one valid combination of houses to rob?
  5. Is the robber obligated to maximize his total loot, or is he only interested in minimizing his capability?
  6. Can ‘k’ ever be zero, implying the robber doesn’t have to rob any house?
  7. Are there multiple test cases, or is it just one array and one ‘k’ value to consider?
  8. Are duplicate values allowed in the array representing the amount of money in each house?
  9. Is the array’s index 0-based or 1-based?
  10. What should be returned if there are multiple combinations that result in the same minimum capability for the robber?

Asking these questions can help clear any ambiguities and guide the approach to solving the problem.

Problem Analysis and Key Insights

  1. Non-adjacency Constraint: The robber can’t rob adjacent houses. This indicates that a straightforward, greedy approach probably won’t work. It hints towards dynamic programming or recursive solutions.

  2. Minimum Houses to Rob: The robber has to rob at least ‘k’ houses. This adds an additional constraint that has to be factored into any solution. It prevents simply picking the ‘k’ smallest values.

  3. Minimize Capability: The objective is not to maximize the total money stolen but to minimize the robber’s capability, defined as the maximum amount stolen from a single house. This is counter-intuitive and will affect the optimization strategy.

  4. Fixed Array Size and Value Ranges: Constraints on array size (1 <= nums.length <= 105) and individual values (1 <= nums[i] <= 109) suggest that a solution with time complexity better than O(n^2) is likely needed for optimal performance.

  5. Multiple Combinations: Multiple combinations of houses to rob may yield the same minimum capability. This means the solution must be able to handle multiple optimal states.

These insights will guide the approach and the algorithms we may consider for solving the problem.

Problem Boundary

The scope of this problem is focused on finding the optimal robbery plan under specific constraints. Specifically, the problem centers around:

  1. Array of Houses with Money: We’re given an integer array representing the amount of money in each house.

  2. Non-adjacency Constraint: The robber cannot rob adjacent houses, narrowing the selection of possible houses to rob.

  3. Minimum Number of Robberies: The robber must rob at least ‘k’ houses, further restricting the set of valid solutions.

  4. Capability Minimization: The objective is to minimize the maximum amount of money robbed from any single house among all the houses robbed.

  5. Time Complexity: Given the constraints, the solution should aim to have an efficient time complexity.

The scope doesn’t extend to other related problems like finding the actual houses to rob or maximizing the total amount stolen. The sole focus is on minimizing the robber’s capability under the given constraints.

To establish the boundary of this problem, consider the following:

  1. Input Constraints:

    • The array nums has a length of 1 to 10^5.
    • Each element in nums ranges from 1 to 10^9.
    • ‘k’ is between 1 and (nums.length + 1) / 2.

  2. Output Constraints:

    • The output must be a single integer representing the minimum capability of the robber.

  3. Functional Constraints:

    • Houses cannot be adjacent.
    • At least ‘k’ houses must be robbed.

  4. Operational Constraints:

    • The solution should be efficient given the input size constraints.

By clearly identifying these constraints and boundaries, you set the stage for focusing on solving the core problem without getting sidetracked.

Problem Classification

The problem belongs to the “Dynamic Programming” domain, which is a subset of the “Algorithms” domain.

‘What’ Components:

  1. Consecutive Houses: A sequence of houses each containing a certain amount of money.
  2. Robber: A character who aims to steal money from these houses.
  3. Non-Adjacent Rule: The robber refuses to rob adjacent houses.
  4. Capability of the Robber: The maximum amount the robber steals from a single house among all houses he robs.
  5. Minimum Number of Houses (k): The robber needs to rob at least ‘k’ houses.
  6. Output: The minimum capability of the robber based on all possible ways to steal from at least ‘k’ houses.
  • Optimization Problem: We are aiming to minimize the robber’s capability while considering the minimum number of houses he must rob.
  • Constraint Satisfaction: The robber must obey the constraint of not robbing adjacent houses and robbing at least ‘k’ houses.
  • Array Manipulation: The problem uses an array to represent the money in each house.

The problem is essentially asking us to find the minimum capability of a robber who has to rob at least ‘k’ non-adjacent houses. We have to consider all possible combinations of non-adjacent houses that satisfy the ‘k’ constraint and identify the one that results in the lowest “capability” for the robber. Dynamic programming could be a suitable approach to efficiently explore these combinations and keep track of the best ones.

Distilling the Problem to Its Core Elements

Fundamental Concept:

The fundamental concept this problem is based upon is “Optimization under Constraints”. You have to optimize (minimize) the robber’s capability while obeying certain rules (not robbing adjacent houses, robbing at least ‘k’ houses).

Simplest Description:

Imagine a robber wants to steal from a row of houses, but he won’t steal from two houses next to each other. Also, he has to steal from at least a certain number of houses. What’s the least amount he can steal from any one house while following these rules?

Core Problem:

The core problem is finding the least amount of money the robber can steal from a single house while robbing at least ‘k’ non-adjacent houses.

Key Components:

  1. List of houses with money.
  2. Rule against robbing adjacent houses.
  3. Minimum number of houses (‘k’) to rob.
  4. Finding the minimum capability of the robber.

Minimal Set of Operations:

  1. Initialize a variable to hold the minimum capability.
  2. Loop through the list of houses to identify valid combinations that follow the non-adjacent rule and the ‘k’ minimum houses rule.
  3. For each valid combination, find the maximum amount the robber can steal from a single house (capability).
  4. Update the minimum capability if the capability from a new valid combination is smaller.
  5. Return the minimum capability.

By following these operations, you can solve the problem efficiently.

Visual Model of the Problem

Visualizing this problem can help in understanding its nuances. Here are some ways to visualize it:

  1. Array Visualization:

    • Imagine the array nums as a straight street with houses.
    • Each house has a value (the money inside), and you can mark these values above each house.
    Street:  | 1 | 2 | 3 | 4 | 5 | 
Money:   | 2 | 7 | 9 | 3 | 1 |

  2. Graphical Plot:

    • Consider plotting the array values on a graph where the x-axis represents the house index and the y-axis represents the money in each house.
    9  -
8  -       *
7  -    *
6  -
5  -
4  -
3  -          *
2  -  * 
1  -               *
0  ----------------------
     1  2  3  4  5  6  7

  3. Color Coding:

    • To make it easier to spot non-adjacent houses, you could color code the houses that can potentially be robbed together.
    Color:  | R | G | R | G | R |  
Money:  | 2 | 7 | 9 | 3 | 1 |

  4. Subsets Visualization:

    • For small-sized arrays, list down all possible subsets of size ‘k’ that can be robbed, observing the non-adjacent rule. Use these subsets to derive the minimum capability.

By visualizing the problem this way, you make it easier to see patterns, which can help in identifying possible strategies for solving it.

Problem Restatement

A robber wants to steal money from a row of houses. Each house contains a specific amount of money, given in an array. The robber can’t rob from two houses next to each other. He also has to rob a minimum number of houses, let’s say ‘k’.

The robber wants to find the approach that requires him to steal the least amount of money from any single house while still meeting the ‘k’ requirement. In other words, out of all the houses he robs, which one has the least amount of money? This is what we call the robber’s “capability.”

Constraints:

  • The array length is between 1 and 105.
  • The amount of money in each house ranges from 1 to 109.
  • The robber has to rob at least ‘k’ houses, where ‘k’ is between 1 and (array length + 1)/2.

Abstract Representation of the Problem

You have a sequence of numerical values (N), and a constraint for selecting a subset (K). You are restricted from selecting adjacent elements from the sequence. The goal is to find the minimal maximum value from all possible non-adjacent subsets of size K.

Key elements:

  • Sequence of numerical values (N)
  • Size constraint for subset selection (K)
  • Restriction on selecting adjacent elements
  • Objective to minimize the maximum value in the chosen subset

Here, the focus is on the structure of the sequence, the subset constraint, and the adjacency restriction, abstracting away the real-world context of robbers and houses.

Terminology

  1. Subset: A collection of elements taken from a larger set. In this problem, we’re interested in subsets of houses to rob.

  2. Non-Adjacent Elements: Elements that are not immediately next to each other. The problem restricts the selection of adjacent houses.

  3. Constraint: A limitation or condition that must be satisfied. In this case, ‘k’ is the constraint, which dictates the minimum number of houses that must be robbed.

  4. Capability: The maximum amount of money that can be stolen from a single house in the chosen subset. The goal is to minimize this capability.

  5. Optimization: Finding the best solution from a set of possible solutions. Here, we want to minimize the robber’s capability.

  6. Dynamic Programming: A method for solving complex problems by breaking them down into simpler overlapping subproblems. This approach could be relevant for solving this problem efficiently.

  7. Time Complexity: A measure of the amount of computational time taken by an algorithm to run. In terms of this problem, we aim for a solution with reasonable time complexity given the constraints.

  8. Space Complexity: A measure of the amount of working storage an algorithm needs. Like time complexity, we aim for a solution that uses space efficiently.

Understanding these terms is key to grasping both the problem and potential solutions. They serve as the building blocks for forming a strategy to tackle the problem.

Problem Simplification and Explanation

The problem can be boiled down to a thief who wants to steal from a row of houses. The catch is, he can’t steal from two houses that are next to each other. On top of that, he has to rob at least ‘k’ houses. His aim is to minimize the maximum amount he takes from any single house.

Key Concepts

  1. Row of Houses: Imagine a straight street with houses lined up. Each house has some money.

  2. Thief’s Choices: The thief has to make choices about which houses to rob while avoiding adjacent houses.

  3. Minimum Robberies: The thief must rob at least ‘k’ houses; he can’t just rob one and leave.

  4. Capability: The most money taken from a single house among the houses he robbed.

Interaction

  • The thief scans the row of houses and makes strategic choices.
  • His choices are limited by the need to skip adjacent houses and to rob at least ‘k’ houses.
  • Finally, among all his robbery options, he aims to minimize his highest theft from a single house.

Analogy

Imagine you’re apple picking. You have a row of trees, each with a different number of apples. However, you can’t pick from two trees that are next to each other. You also have to pick apples from at least ‘k’ trees. Now, among all the trees you picked from, you want to make sure that the tree from which you picked the most apples has as few apples as possible.

This analogy helps you understand that you’re making a series of constrained choices to reach an optimal outcome, just like the robber.

Constraints

There are several characteristics in the problem statement that can be exploited for an efficient solution:

  1. Array Size Limitation: The maximum size of the array nums is (10^5), and this provides a boundary for the time complexity we should aim for.

  2. Element Value Limitation: Each element in the nums array is at most (10^9). This sets a cap on the individual values, which can be useful in determining algorithmic limits.

  3. Non-Adjacent Constraint: The robber cannot rob adjacent houses, a key constraint that suggests dynamic programming or recursive strategies to optimize choices.

  4. Minimum Number of Houses (k): The robber has to rob at least ‘k’ houses, where (1 \leq k \leq \frac{nums.length + 1}{2}). This constraint could be used to reduce the search space.

  5. Sorted Capability: The goal is to minimize the ‘capability’ (the maximum stolen from a single house), which suggests that sorting or heap-based data structures might be helpful for tracking potential solutions.

By recognizing these characteristics, we set the stage for choosing a suitable algorithmic approach. The constraints particularly guide us toward considering dynamic programming, recursive solutions, or heap-based strategies to solve the problem efficiently.

Analyzing the constraints provides us with several key insights:

  1. Time Complexity: With (10^5) as the upper limit for the array size, algorithms with a time complexity of (O(n \log n)) or (O(n)) are likely needed for efficiency.

  2. Value Constraints: The upper limit of (10^9) for individual elements in nums indicates that arithmetic overflows are unlikely to be an issue, but we still need to handle large numbers efficiently.

  3. Non-Adjacency: The rule that the robber can’t rob adjacent houses is a strong clue that recursion or dynamic programming may be needed for an efficient solution. This constraint significantly reduces the choices the robber has at each step.

  4. Minimum Houses: Knowing that the robber must steal from at least ‘k’ houses provides an opportunity to pre-filter or eliminate some choices right from the beginning, reducing the search space.

  5. Objective to Minimize Capability: The goal is not just to pick houses to rob but to do so in a way that minimizes the maximum value stolen from any single house. This suggests that a greedy approach alone won’t suffice; we need to think more strategically.

These insights from the constraints guide us toward considering approaches that can handle these specific limitations efficiently, such as dynamic programming or recursive strategies with memoization.

Case Analysis

Additional Examples

  1. Single House (Edge Case)

    • Input: nums = [4], k = 1
    • Output: 4
    • Reasoning: Only one house, so the robber has to rob it. The capability is 4.

  2. All Houses the Same (Boundary Case)

    • Input: nums = [5, 5, 5, 5], k = 2
    • Output: 5
    • Reasoning: All houses contain the same amount, so the capability will always be 5 regardless of which ones are robbed.

  3. Descending Order

    • Input: nums = [9, 8, 7, 6], k = 2
    • Output: 7
    • Reasoning: The minimum capability happens when robbing houses with 7 and 9, avoiding the adjacent constraint.

  4. Large Numbers (Boundary Case)

    • Input: nums = [1000000000, 999999999], k = 1
    • Output: 999999999
    • Reasoning: The robber can just rob the house with 999999999, meeting the minimum ‘k’ and the capability is 999999999.

  5. Multiple Optimal Solutions

    • Input: nums = [3, 4, 3, 4], k = 2
    • Output: 3
    • Reasoning: The robber can either rob the first and third houses or the second and fourth houses, either way, the minimum capability is 3.

Edge Cases

  1. Minimum Array Length: When the array length is at the minimum bound, i.e., 1, the problem is trivially solvable.

  2. Maximum Array Length: When the array length is at the maximum bound, i.e., (10^5), the efficiency of the solution will be critically tested.

  3. Minimum and Maximum ‘k’: ‘k’ can be as low as 1 or as high as ((\text{nums.length} + 1)/2), each extreme affecting the number of combinations to check.

  4. Minimum and Maximum House Value: With a range of 1 to (10^9), special attention should be paid to arithmetic operations to avoid overflow or inefficiencies.

By considering these test cases and edge conditions, we can ensure that the solution covers all bases and handles different aspects of the problem effectively.

Visualizing these cases can help to better understand the problem. Here’s a simple way to visualize each of them:

  1. Single House (Edge Case)

    • Visualization: [4]
    • Single house, no choice but to rob it.

  2. All Houses the Same (Boundary Case)

    • Visualization: [5]-[5]-[5]-[5]
    • All houses have the same value. Any two non-adjacent houses can be robbed for the same capability.

  3. Descending Order

    • Visualization: [9]-[8]-[7]-[6]
    • Rob the first and third houses for minimum capability.

  4. Large Numbers (Boundary Case)

    • Visualization: [1000000000]-[999999999]
    • Only two houses, but with very large values.

  5. Multiple Optimal Solutions

    • Visualization: [3]-[4]-[3]-[4]
    • Multiple ways to achieve the same minimum capability.

You can think of each house as a node and the connections between them imply adjacency, which means adjacent houses cannot be robbed together. The number inside each node is the amount of money in that house.

Understanding these visual representations can give insights into the inherent structure of the problem and possible solutions.

Analyzing different cases provides several key insights:

  1. Edge Cases Matter: When you have just one house, the solution is straightforward. But as the problem scales, the complexity grows, making it important to account for edge cases.

  2. Equal Values: If all houses have the same value, any set of non-adjacent houses can be robbed without affecting the outcome. This can simplify the problem under certain conditions.

  3. Descending Order: When the values are in descending order, robbing the first and third houses will generally yield the minimum capability, indicating that sometimes the arrangement of values can point to a quick solution.

  4. Large Numbers: Handling very large numbers might require careful coding to avoid overflow or memory issues, although Python is generally good with large integers. But the concept remains important for languages with limitations on integer sizes.

  5. Multiple Optimal Solutions: There could be multiple ways to achieve the same minimum capability, indicating that the problem might have multiple valid solutions.

Understanding these aspects can help in devising a more efficient and robust algorithm.

Identification of Applicable Theoretical Concepts

Certainly. Several mathematical and algorithmic concepts could be useful for tackling this problem:

  1. Dynamic Programming: This problem has overlapping subproblems, and we could use dynamic programming to store already computed results for reuse.

  2. Recursion: The problem has a recursive nature, as the decision to rob a particular house depends on the previous choices.

  3. Greedy Approach: While a greedy approach of always choosing the lowest value might not work directly due to the ‘k’ constraint, a modified greedy algorithm could potentially offer some efficiency.

  4. Graph Theory: This problem could be represented as a graph where nodes are houses and edges indicate adjacency. Dynamic programming could then be used to find the optimal path through the graph that respects the ‘k’ constraint.

  5. Sorting Algorithms: Sorting the array of houses based on the amount of money could provide a quicker way to identify the minimum capability required.

  6. Combinatorial Mathematics: The constraint that we have to rob ‘k’ houses could lead us to combinatorial ways to select these ‘k’ from ’n’ houses, taking adjacency into account.

  7. Min-Max Theory: The goal is to find the minimum capability out of the maximum amounts stolen in various scenarios, which aligns with min-max optimization problems.

By recognizing these mathematical and algorithmic properties, we can take advantage of existing theories and methodologies to come up with a more efficient solution.

Simple Explanation

Imagine a neighborhood with a row of houses, and each house has a piggy bank with some money in it. A robber wants to steal piggy banks, but he has two rules. First, he won’t steal from houses next to each other because that’s too risky. Second, he wants to steal from at least a certain number of houses, let’s say 2 or 3.

The robber wants to be as sneaky as possible, so he tries to take piggy banks that have the least amount of money. The goal is to find out the minimum amount of money the robber will end up with in his biggest piggy bank after robbing the required number of houses without getting caught.

So, think of it like a game where the robber tries to score as low as possible, but still has to follow his own rules. It’s like playing hopscotch, but you’re only allowed to land on certain numbers, and you want to get the lowest score possible.

Problem Breakdown and Solution Methodology

Approach to Solving the Problem

  1. List Possible Combinations:

    • Start by listing all the combinations of houses the robber can target, while respecting the two rules: not robbing adjacent houses and robbing at least ‘k’ houses.
    • Metaphor: Imagine playing a game of hopscotch where you can only land on non-adjacent squares and have to land a minimum number of times.

  2. Calculate Capabilities:

    • For each combination, find the house with the most money. This represents the robber’s “capability” for that particular combination.
    • Metaphor: In a fruit basket, pick the biggest apple from each set of non-adjacent apples.

  3. Find the Minimum Capability:

    • After calculating the capabilities for all combinations, find the minimum one.
    • Analogy: In a race, you time several paths and pick the one with the shortest time.

  4. Return the Result:

    • The minimum capability is the answer.

Effects of Parameter Changes:

  • Increasing ‘k’:

    • If the minimum number of houses (‘k’) to be robbed increases, the combinations decrease. This might make the calculation easier but could also increase the minimum capability.

  • Increasing House Money:

    • If the money in any house increases, the minimum capability might also increase, depending on where that house is in the optimal sequence.

  • Increasing Number of Houses:

    • More houses would increase the number of possible combinations, making the problem computationally more intense.

Example Cases:

  • Example 1:

    • Houses: [2,3,5,9], k = 2
    • Combinations: [(2,5), (2,9), (3,9)]
    • Capabilities: [5, 9, 9]
    • Minimum Capability: 5

  • Example 2:

    • Houses: [2,7,9,3,1], k = 2
    • Combinations: [(2,9), (2,3), (2,1), (7,3), (7,1), (9,1)]
    • Capabilities: [9, 3, 2, 7, 7, 9]
    • Minimum Capability: 2

This approach helps the robber pick the least risky set of houses to rob while adhering to his self-imposed rules. Like assembling the least expensive basket from a fruit stand with specific picking rules.

Inference of Problem-Solving Approach from the Problem Statement

  1. Consecutive Houses:

    • This term tells us that the houses are adjacent to one another. The strategy influenced by this term is to skip over adjacent houses when generating combinations.

  2. Money Inside:

    • Each house contains a specific amount of money, represented as integers. This forms the basis of calculating the robber’s capability for each combination of houses.

  3. Robber’s Capability:

    • The term refers to the maximum amount of money robbed from a single house in a given combination. We use this to evaluate which combination of houses to rob.

  4. Minimum Number of Houses (k):

    • The robber has to rob at least ‘k’ houses. This constraint eliminates certain combinations and affects our approach to generating valid combinations of houses to rob.

  5. Integer Array (nums):

    • An array that represents the money inside each house. This is our primary dataset that we’ll manipulate and evaluate.

  6. Maximum and Minimum Constraints:

    • The problem defines maximum and minimum amounts and lengths, guiding us toward optimized algorithms that can handle the range of possible inputs.

Each keyword or concept informs a specific part of the problem-solving strategy, from how to generate valid combinations of houses to rob (considering “Consecutive Houses” and “Minimum Number of Houses”), to how to evaluate each combination (“Robber’s Capability” and “Money Inside”), and finally to how to handle different scales of the problem (“Maximum and Minimum Constraints”).

  1. Consecutive Houses and Skipping Adjacent Ones:

    • You could draw a number line or a row of houses and use marks or colors to show which houses can and cannot be robbed together.

  2. Money Inside:

    • You can add the monetary value above each house in the row or number line to visually represent the money inside each house.

  3. Robber’s Capability:

    • For each valid combination of houses, you can highlight or circle the house with the maximum value. This will show the robber’s capability for that specific combination.

  4. Minimum Number of Houses (k):

    • You could add a different kind of mark or annotation next to the combinations that meet the minimum ‘k’ houses to emphasize them.

  5. Integer Array (nums):

    • This could be a simple table with each row representing a house and columns for house index and amount of money.

  6. Constraints:

    • You can annotate the minimum and maximum values in your table, perhaps with special markers or color-coding.

By combining these elements, you can create a composite visual representation. For example, take a row of houses with money values above them. Then, for each combination of houses that meets the ‘k’ minimum, circle the house with the maximum value and annotate it. This way, you can easily compare different combinations to find the one with the minimum robber capability.

How did you infer from the problem statement that this problem can be solved using ?

Simple Explanation of the Proof

I’m having trouble understanding the proof of this algorithm. Could you explain it in a way that’s easy to understand?

Stepwise Refinement

  1. Stepwise Refinement of Approach

    • Step 1: Understand Constraints

      • Recognize the problem’s parameters and limitations.

    • Step 2: List All Combinations

      • Generate all combinations of houses to rob, respecting the constraint of not robbing adjacent houses and robbing at least ‘k’ houses.

    • Step 3: Calculate Capability for Each Combination

      • For each valid combination, find the maximum amount that could be robbed (Capability).

    • Step 4: Find Minimum Capability

      • Out of all the capabilities, find the minimum one.

  2. Granular, Actionable Steps

    • Step 1:

      • Validate the input against the constraints.

    • Step 2:

      • Start from the first house, then iterate to find all non-adjacent combinations that include at least ‘k’ houses. This might involve recursive function calls or dynamic programming.

    • Step 3:

      • For each valid combination, iterate through the houses and find the maximum monetary value. Store this value.

    • Step 4:

      • Use a sorting function or simply iterate through all stored maximum values to find the minimum one.

  3. Independent Parts

    • The task of generating all valid combinations of houses to rob can be separated from the task of calculating the capability for each combination.

    • Calculating the maximum value in each combination is also an independent task that can be parallelized.

  4. Repeatable Patterns

    • The pattern of skipping adjacent houses and picking the next is repetitive.

    • The process of finding the maximum value in a given combination is also a repetitive task that applies to each combination generated.

By identifying these elements, we can organize our approach into well-defined steps and modular components that can either be solved independently or used repetitively to simplify the overall problem-solving process.

Solution Approach and Analysis

Detailed Approach to Solving the Problem

1. Input Validation

Start by checking if the input array and ‘k’ value meet the given constraints.

2. Initialize Variables

Create an empty list to store the “capabilities” (maximum amounts robbed from each possible combination of houses).

3. Generate House Combinations

Imagine the houses as nodes in a tree. The root is an empty node, and each child represents the decision of whether or not to rob a particular house. Navigate this tree recursively, avoiding adjacent houses. You’ll only consider those paths that include at least ‘k’ houses.

4. Calculate Capability

For each valid combination, find the maximum amount of money that can be robbed (the “capability”). Think of this as finding the richest house in each group.

5. Store Minimum Capability

As you calculate each capability, compare it with the current minimum capability. If it’s lower, update the minimum.

6. Return Result

Return the minimum capability.

Changes in Parameters and Their Effects

  • Increase in ‘k’: The higher the value of ‘k’, the fewer combinations you’ll have, but each combination will include more houses. This could make the problem easier or harder, depending on other parameters.

  • Array Size: A bigger array will increase the number of combinations exponentially, making the problem computationally harder.

  • Array Values: Higher values in the array will skew the capabilities towards those higher values, potentially making the minimum capability higher.

Example Case:

Let’s use the first example from the problem statement: nums = [2,3,5,9], k = 2.

  1. Input Validation: nums has 4 elements, each between 1 and 10^9. ‘k’ is 2, which is valid. All good.

  2. Initialize Variables: Initialize min_capability to inf (infinity).

  3. Generate House Combinations:

    • [2, 5]
    • [2, 9]
    • [3, 9]

  4. Calculate Capability:

    • For [2, 5], max is 5.
    • For [2, 9], max is 9.
    • For [3, 9], max is 9.

  5. Store Minimum Capability:

    • For [2, 5], min_capability becomes 5.
    • For [2, 9] and [3, 9], min_capability remains 5.

  6. Return Result:

    • The result is 5, which is the minimum capability.

By breaking down this complex problem into smaller, more manageable tasks, we can solve each piece independently and combine them to find the solution.

Identify Invariant

The invariant in this problem is that the robber cannot rob adjacent houses. This constraint remains true throughout the problem and is crucial for generating valid combinations of houses to rob. By ensuring that this invariant holds, we can confidently proceed with other parts of our solution approach, such as calculating capabilities based on different combinations.

Identify Loop Invariant

In the context of algorithm design, a loop invariant is a condition that holds before and after each iteration of a loop. For this problem, if we were to use a loop to explore possible combinations of houses to rob, a potential loop invariant could be that the set of houses considered so far does not contain any adjacent houses. This would ensure that as we iterate through different combinations, we’re always adhering to the primary rule of not robbing adjacent houses. Keeping this loop invariant true allows us to maintain the integrity of the solution as we explore the problem space.

In this problem, the terms “invariant” and “loop invariant” may serve different roles but could potentially be closely related.

An “invariant” would be a condition that remains unchanged throughout the execution of the entire algorithm. For instance, an invariant could be that no two adjacent houses are ever robbed, aligning with the problem’s constraint.

A “loop invariant,” on the other hand, specifically applies to the condition or set of conditions that hold true before and after each iteration of a loop within the algorithm.

In this problem, the rule about not robbing adjacent houses could serve as both an invariant for the entire algorithm and a loop invariant if we use a loop to iterate through combinations. So, in this specific scenario, the invariant and loop invariant could be the same.

Thought Process

Thought Process

  1. Understand Constraints: The first cue is that the robber cannot rob adjacent houses. This rules out simple approaches like sorting.

  2. Minimum Number of Houses: We have a lower limit (k) on the number of houses the robber must rob. This suggests we can’t just pick the ‘k’ smallest values.

  3. Minimum Capability: The problem asks for the minimum capability of the robber among all valid ways to rob ‘k’ houses. This tells us we’re looking for a specific type of optimization.

  4. Dynamic Programming Insight: The requirements strongly hint that this problem could be solved using dynamic programming.

  5. Initialization: We will start with an array to store the minimum capability required if the robber robs up to house ‘i’.

  6. State Transition: For each house ‘i’, the robber has two options - rob it or skip it. Robbing it would add a certain amount to the minimum capability, and skipping it would also have its implications.

  7. Final Answer: Our final answer will be the minimum value in our dynamic programming array that corresponds to robbing at least ‘k’ houses.

Code in Python3

Here’s how you could implement this in Python:

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def min_capability(nums, k):
    n = len(nums)
    dp = [float('inf')] * n
    
    for i in range(n):
        # Initialize 'current_min' to nums[i], representing robbing only this house
        current_min = nums[i]
        
        for j in range(i, max(-1, i - k) - 1, -1):
            current_min = min(current_min, nums[j])
            
            if i - j >= k - 1:
                prev_min = dp[j - 1] if j > 0 else 0
                dp[i] = min(dp[i], max(prev_min, current_min))
                
    return min(dp[-k:])

# Test the function
print(min_capability([2, 3, 5, 9], 2))  # Output should be 5
print(min_capability([2, 7, 9, 3, 1], 2))  # Output should be 2

Key Insights

  • The problem is optimization-focused, specifically looking for a ‘minimum’ capability.
  • The constraint of not robbing adjacent houses and the minimum number of houses that must be robbed suggests a dynamic programming approach.
  • The dynamic programming array tracks the minimum capability needed for robbing up to ‘i’ houses, making the problem easier to solve.
  • We iterate through each house and consider robbing it or skipping it, updating our dp array based on these choices.
  • The problem can be solved in O(n*k) time complexity.

By breaking down the problem like this, you not only solve it but also gain a deeper understanding of dynamic programming and optimization strategies.

Establishing Preconditions and Postconditions

Parameters:

  1. Inputs: The inputs to the method are an integer array nums and an integer k.
  2. Types: nums is a list of integers and k is an integer.
  3. Representation: nums represents the amount of money in each house. k represents the minimum number of houses the robber must rob.

Preconditions:

  1. State: No specific state requirements for the program before calling the method.
  2. Constraints:
  • 1 <= nums.length <= 10^5
  • 1 <= nums[i] <= 10^9
  • 1 <= k <= (nums.length + 1)/2
  1. Program State: No special program state is required.

Method Functionality:

  1. Expected: The method is expected to return the minimum capability of the robber among all valid ways to rob at least k houses.
  2. Interaction: The method solely relies on its input parameters nums and k to compute the result.

Postconditions:

  1. State: The program state is unchanged as the method does not have any side effects.
  2. Return Value: The return value is an integer that represents the minimum capability of the robber based on the rules given.
  3. Side Effects: There are no side effects.

Error Handling:

  1. Preconditions Not Met: If the preconditions related to the constraints on nums and k are not met, the behavior is undefined. Ideally, you’d include a check at the beginning of the method to handle such cases.
  2. Response: The method could throw an exception or return a special value like -1 to indicate that the input parameters are not within the allowed range. However, according to the problem statement, it is always possible to steal from at least k houses, so no special error handling is strictly necessary for the stated problem.

Problem Decomposition

Problem Understanding:

  1. In My Own Words: The problem requires finding the minimum capability of a robber who can rob from k out of n houses, each containing different amounts of money. The key components are the list of amounts in each house (nums) and the minimum number of houses to rob (k).

Initial Breakdown:

  1. Major Parts:
  • Sort the houses by the amount of money.
  • Find combinations of k houses to rob.
  • Calculate the robber’s capability for each combination.
  • Identify the minimum capability among these.

Subproblem Refinement:

  1. Sorting: Implement a sorting function or use a built-in one.
  2. Combination Finding: Write a function to find all valid combinations of k houses.
  3. Capability Calculation: Implement logic to calculate the capability of the robber for each combination.
  4. Minimum Capability: Store and update the minimum capability as you calculate for each combination.

Task Identification:

  1. Sorting: This is a unique task.
  2. Combination Finding and Capability Calculation: These could be combined into a single, generalized task as they’re closely related.
  3. Minimum Capability: This task is updated within the “Capability Calculation” task.

Task Abstraction:

  1. Sorting: Clear and reusable.
  2. Combination Finding and Capability Calculation: Could be abstracted into a function that takes the sorted list and k as parameters, and returns the minimum capability.
  3. Minimum Capability: Its purpose is clear, though it’s part of the capability calculation.

Method Naming:

  1. Sorting: sortHouses
  2. Combination Finding and Capability Calculation: findMinimumCapability
  3. Minimum Capability: Included within findMinimumCapability, so no separate name.

Subproblem Interactions:

  1. Order:
    • First, sort the houses.
    • Then, find combinations and calculate capabilities.
    • Finally, determine the minimum capability.
  2. Dependencies:
    • Sorting must happen before finding combinations.
    • Finding combinations and calculating capabilities happen together.
    • Minimum capability is updated as part of the capability calculations.

From Brute Force to Optimal Solution

Brute Force Solution:

  1. Sort the houses by their amounts:

    • First, sort the array nums in ascending order.
    • Time Complexity: (O(n \log n))
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    nums.sort()

  2. Find all combinations of k houses to rob:

    • Generate all possible combinations of robbing k houses out of n sorted houses.
    • Time Complexity: (O(\binom{n}{k}))
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    from itertools import combinations
all_combinations = list(combinations(nums, k))

  3. Calculate Capability for each combination:

    • For each combination, sum up the amounts to calculate the robber’s capability.
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    min_capability = float('inf')
for comb in all_combinations:
    capability = sum(comb)
    min_capability = min(min_capability, capability)

Inefficiencies:

  1. Time Complexity:

    • (O(n \log n) + O(\binom{n}{k}) \times k)
    • Generating all combinations is particularly expensive.

  2. Space Complexity:

    • (O(\binom{n}{k}))
    • Storing all combinations of k houses can consume significant memory.

Optimization Steps:

  1. No need to sort:

    • Sorting doesn’t aid in solving this problem more efficiently. Remove it.
    • Time Complexity saved: (O(n \log n))

  2. Dynamic Programming to find minimum capability:

    • Instead of generating all combinations, use dynamic programming.
    • Start with the smallest possible capability and build upon it.
    • This approach has a Time Complexity of (O(n \times k))

Here’s the code snippet:

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def find_min_capability(nums, k):
    dp = [float('inf')] * (k + 1)
    dp[0] = 0
    
    for num in nums:
        for i in range(k, num - 1, -1):
            dp[i] = min(dp[i], dp[i - num] + num)
            
    return dp[k]

nums = [3, 1, 4, 2, 2, 1]
k = 3
print(find_min_capability(nums, k))

Optimized Time and Space Complexity:

  1. Time Complexity: (O(n \times k))
  2. Space Complexity: (O(k))

As we can see, the optimized solution significantly improves both time and space complexity.

PENDING TEST

Code Explanation and Design Decisions

1. Initial Parameters:

  • nums: This is the array of integers representing the amount in each house.
  • k: The number of houses to rob.

These parameters define the universe of our problem. nums provides the data we are to manipulate, and k sets the constraint on how many houses can be robbed.

2. Primary Loop:

The primary loop iterates over each amount in nums. Each iteration signifies examining a new house to make a decision on whether robbing it contributes to achieving the minimum capability.

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for num in nums:

3. Conditions within the Loop:

Inside the primary loop, we have a secondary loop:

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for i in range(k, num - 1, -1):

This loop iterates from k down to num. The condition i >= num ensures that we are only considering combinations that include the current num. It signifies that you can’t rob a house if it has more amount than the current capability under consideration.

4. Updates within the Loop:

Within the nested loop, we update our dp array:

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dp[i] = min(dp[i], dp[i - num] + num)

Here, dp[i] is being updated to reflect the minimum capability needed to rob i amount. This change is crucial as it builds up the solution incrementally and ensures that the constraint of robbing only k houses is met.

5. Invariant:

The invariant here is the dp array. At dp[i], it always contains the minimum capability needed to rob i amount from any k houses. This invariant ensures that as we proceed through the array, our solution remains aligned with the problem’s objective of finding the minimum capability.

6. Final Output:

The final output is dp[k], which represents the minimum capability needed to rob k houses. It satisfies the problem’s requirements by providing the least amount of capability needed while respecting the constraints set by nums and k.

Coding Constructs

1. High-Level Strategies:

The code uses Dynamic Programming (DP) to solve the problem incrementally. The strategy is to build a solution based on smaller sub-problems.

2. Purpose to a Non-Programmer:

Imagine you’re a thief planning to rob a neighborhood. You want to know the minimum skill you need to rob a certain number of houses without getting caught. This code helps figure that out by evaluating different scenarios in a smart way.

3. Logical Elements:

  • Iteration: Looping through lists of numbers.
  • Conditionals: Checking whether certain conditions are met.
  • State: Maintaining a state (dp array) to remember previous calculations.

4. Algorithmic Approach in Plain English:

The code starts by assuming you need an impossibly high skill level to rob any number of houses. It then looks at each house one by one and updates the required skill level based on what it finds. It does this in a way that ensures it always knows the minimum skill needed to rob any number of houses, up to your target.

5. Key Steps:

  • Initialization: Set up an array to hold the minimum skill levels needed.
  • Iteration: Loop through each amount in each house.
  • Update: Modify the array to record the minimum skill level needed so far.
  • Check: Keep track of the minimum skill needed to meet your target number of houses.

Each step is crucial for building up to the final answer in an efficient manner.

6. Algorithmic Patterns:

  • Dynamic Programming: Building up a solution by solving smaller sub-problems.
  • Loop Invariants: Maintaining a property (dp array) that holds throughout the algorithm’s execution.
  • Nested Loops: Using one loop inside another to examine combinations.

Language Agnostic Coding Drills

1. Dissect the Code:

Here are some language-agnostic coding concepts present in the code:

  1. Variable Initialization: Declaring variables to hold state or intermediate results.
  2. Arrays: Setting up and using an array to store multiple values.
  3. Basic Loops: Iterating through an array or list.
  4. Nested Loops: Having a loop inside another loop.
  5. Conditionals: Using if, else or similar constructs to control flow.
  6. Dynamic Programming (DP): Using an array to save computational state.
  7. Loop Invariants: Maintaining a property that holds throughout the algorithm’s execution.
  8. Array Updates: Updating array elements conditionally based on some criteria.

2. Concepts in Order of Increasing Difficulty:

  1. Variable Initialization: Easiest to understand. Just setting up a place to store values.
  2. Arrays: A step up, understanding that you can store multiple values in a single data structure.
  3. Basic Loops: Iterating through arrays is usually one of the first loops one learns.
  4. Conditionals: Adds a layer of complexity by requiring decision-making.
  5. Array Updates: A little more challenging as you now manipulate array data based on conditions.
  6. Nested Loops: Requires understanding of how loops interact, which adds complexity.
  7. Loop Invariants: Slightly advanced, understanding a property that remains unchanged in a loop.
  8. Dynamic Programming (DP): Most challenging, combining all prior elements to solve problems efficiently.

3. Problem-Solving Approach:

  1. Understand the Problem: Read and comprehend what’s required. This sets up the need for variables and a DP array.

  2. Variable Initialization: Set up variables like min_skill, which will ultimately store your answer.

  3. Arrays and Dynamic Programming: Initialize a DP array to save state. This is the core data structure you will work with.

  4. Basic Loop for House Iteration: Start by looping through each house. This is where you’ll begin to populate your DP array.

  5. Nested Loop for Amount Iteration: Within the house loop, iterate through possible amounts to rob. This will serve to update your DP array.

  6. Conditionals and Array Updates: Within these loops, you’ll have conditions checking whether it’s better to rob this house based on the skill levels. Update the DP array accordingly.

  7. Loop Invariants: Throughout the loops, the DP array serves as a loop invariant, always containing the minimum skill level needed to rob up to i houses.

  8. Final Evaluation: At the end of the loops, your DP array will have the minimum skill needed for any number of houses. Extract your answer from here.

Each of these “drills” serves to build the next. You start simple with variables and arrays, then get into the loops where the real logic happens, underpinned by your use of Dynamic Programming and Loop Invariants.

Targeted Drills in Python

1. Python Coding Drills for General Concepts

Variable Initialization

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# Initializing a variable
x = 10

Arrays

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# Initializing an array
arr = [1, 2, 3, 4]

Basic Loops

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# Looping through an array
for i in arr:
    print(i)

Conditionals

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# Using if-else statements
if x > 5:
    print("Greater")
else:
    print("Smaller")

Array Updates

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# Updating an array element
arr[0] = 5

Nested Loops

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# Nested loop
for i in range(3):
    for j in range(3):
        print(i, j)

Loop Invariants

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# Loop with invariant
invariant = 0
for i in range(5):
    invariant += i
    assert invariant == i*(i+1)//2  # This condition remains true during each iteration

Dynamic Programming

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# Simple DP example: Fibonacci sequence
dp = [0, 1]
for i in range(2, 10):
    dp.append(dp[i-1] + dp[i-2])

2. Problem-Specific Drills

For our problem, understanding how to use a Dynamic Programming (DP) array to store the minimum skill levels is critical.

Dynamic Programming for Minimum Skill Level

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# Initialize a DP array with a large number
dp = [float('inf')] * 10  # Assuming 10 houses
dp[0] = 0  # Base case: 0 skill needed for 0 houses

This DP drill is essential because it forms the backbone of the problem-solving strategy. The DP array will store the minimum skill needed for each house, which we will update using our loops and conditionals.

3. Integrating the Drills

  1. Initialize Variables and Arrays: Start by declaring any variables you’ll need, as well as your DP array.

  2. Start Looping: Use a loop to go through each house, updating your DP array as you go.

  3. Nested Loop: Within the house loop, have another loop to consider different amounts to rob, updating the DP array.

  4. Conditional Statements and Array Updates: Within the nested loops, include conditionals to update the DP array based on whether it’s better to rob a particular house or not.

  5. Loop Invariant: The DP array serves as a loop invariant, providing us with the minimum skill levels as we progress through the houses.

  6. Final Evaluation: After all loops are done, the DP array should contain your answer. Extract and return it.

In summary, each drill teaches a basic Python programming concept and how it can be applied to the problem at hand. Assembling all these drills in the right sequence should give you a complete and optimized solution to the problem.

Q&A

Similar Problems

Here are 10 LeetCode problems that use similar problem-solving strategies or underlying concepts:

  1. Climbing Stairs

    • Why: This problem also uses Dynamic Programming to find the number of ways to climb a stair, which is conceptually similar to finding the minimum skill needed for each house in our problem.

  2. Coin Change

    • Why: The problem employs Dynamic Programming to find the minimum number of coins needed to make a certain amount, similar to how we found the minimum skill level in our problem.

  3. House Robber II

    • Why: This is an extension of the House Robber problem, adding the constraint that houses are arranged in a circle. The problem-solving strategy would largely remain the same, with minor tweaks.

  4. Maximum Product Subarray

    • Why: This problem requires optimizing a value (max product) based on previous sub-problems, akin to our Dynamic Programming approach.

  5. Unique Paths

    • Why: The problem is solved using Dynamic Programming to find the number of unique paths from one corner of a grid to another, which involves a similar breakdown of the problem into smaller subproblems.

  6. Longest Increasing Subsequence

    • Why: This problem requires finding the longest increasing subsequence in an array, employing Dynamic Programming to do so. This is related to our problem’s need to optimize a certain value based on previous values.

  7. Minimum Path Sum

    • Why: The problem requires finding a path from the top-left to the bottom-right corner of a grid while minimizing the sum of the numbers along the path. It employs Dynamic Programming, similar to our problem.

  8. Best Time to Buy and Sell Stock

    • Why: Though not a typical Dynamic Programming problem, it does involve optimizing a value (profit) based on the state of an array, which is similar to our DP approach.

  9. Longest Common Subsequence

    • Why: This problem requires finding the longest common subsequence between two strings. It’s similar to our problem in that it breaks down a large problem into smaller subproblems using Dynamic Programming.

  10. Partition Equal Subset Sum

    • Why: The problem uses Dynamic Programming to find if an array can be partitioned into two subsets with equal sums. The dynamic programming array is used to keep track of possible sums, much like how we kept track of minimum skill levels.

Each of these problems involves using similar steps or techniques, mostly centered around Dynamic Programming, to break down a larger problem into smaller, more manageable subproblems.