Quantitative finance interviews, especially for roles like Quant Strategist at top firms such as Akuna Capital, often feature problem-solving questions rooted in computer science and mathematics. One such classic is the "House Robber" dynamic programming problem, which tests not only algorithmic thinking but also the ability to adapt familiar paradigms to new variants. In this article, we’ll tackle a nuanced variant of the House Robber problem—specifically, a version where you’re reviewing movies, and you cannot skip more than one movie in a row. We'll unpack the nuances, solve the problem using dynamic programming, and explain every concept in detail to prepare you for high-level quant interviews.

Akuna Capital Quant Strategist Interview Question: Dynamic Programming Variant of House Robber Problem

Problem Statement: The Movie Reviewer’s Dilemma

Imagine you're a movie critic planning to review a lineup of movies. Each movie has a certain "enjoyment value." However, you cannot review two directly adjacent movies (to avoid fatigue and ensure quality), but you also cannot skip more than one movie in a row (to keep your portfolio relevant). The goal is to select movies to maximize the total enjoyment value under these constraints.

• Input: An array reviews where reviews[i] represents the enjoyment value of the i^th movie.
• Constraints:
  • You cannot review two adjacent movies (i.e., cannot select i and i+1).
  • You cannot skip more than one movie in a row (i.e., cannot skip both i+1 and i+2 after reviewing i).
• Output: The maximum total enjoyment value achievable.

Translating to a Dynamic Programming Framework

House Robber Recap

The classic House Robber problem is defined as:

• Given an array of values, maximize the sum of non-adjacent elements.

In Mathjax, the recurrence is:

$$ dp[i] = \max(dp[i-1], dp[i-2] + nums[i]) $$

But in our movie reviewer variant, the "no adjacent" rule remains, but we add a new restriction: you cannot skip more than one movie in a row.

Understanding the New Constraint

Let's rephrase the problem:

• If you skip a movie, you must review at least one of the next two.
• If you review a movie, you must skip the next one, but not necessarily the one after.

This requires us to track not only the choices we make (review or skip) but also the consequences of consecutive skips. This is a classic use case for dynamic programming with state variables.

Defining the DP State

Let’s define our DP state as follows:

• dp[i][k] = Maximum enjoyment value attainable up to movie i, where k indicates how many skips in a row have just occurred (0 or 1).

Thus:

• k = 0: Last movie was reviewed.
• k = 1: Last movie was skipped, but only one skip in a row.

We cannot have k = 2 (cannot skip two in a row).

The DP Recurrence

Let’s analyze transitions:

• If you review movie i, you must have not reviewed i-1 (to avoid adjacent reviews). So, the previous must be a "skip".
• If you skip movie i, you must not have already skipped the previous movie (to avoid two skips in a row).

Let's formalize this:

• Case 1: Review movie i
  You can only review i if you skipped i-1:
  $$ dp[i][0] = dp[i-1][1] + reviews[i] $$
• Case 2: Skip movie i
  You can only skip i if you reviewed i-1 (so that you do not have two skips in a row):
  $$ dp[i][1] = dp[i-1][0] $$

Base Cases:

• At i = 0: If you review the first movie, $dp[0][0] = reviews[0]$.
• At i = 0: If you skip the first movie, $dp[0][1] = 0$.

Step-by-Step DP Table Construction

Let's build the DP table for a sample input to make this concrete.

Explanation:

• dp[0][0] = 4: Review the first movie.
• dp[0][1] = 0: Skip the first movie.
• dp[1][0] = dp[0][1] + reviews[1] = 0 + 2 = 2: Only possible if previous was skipped.
• dp[1][1] = dp[0][0] = 4: Skip now, but previous must have been reviewed.
• Continue in this fashion.

Final Answer: The maximum value is max(dp[n-1][0], dp[n-1][1]), where n is the number of movies.

Python Implementation

def max_enjoyment(reviews):
    n = len(reviews)
    if n == 0:
        return 0
    dp = [[float('-inf')] * 2 for _ in range(n)]
    # Base cases
    dp[0][0] = reviews[0]  # Review first movie
    dp[0][1] = 0           # Skip first movie

    for i in range(1, n):
        # dp[i][0]: review this movie, so previous must be skip
        if dp[i-1][1] != float('-inf'):
            dp[i][0] = dp[i-1][1] + reviews[i]
        # dp[i][1]: skip this movie, so previous must be review
        if dp[i-1][0] != float('-inf'):
            dp[i][1] = dp[i-1][0]

    return max(dp[n-1][0], dp[n-1][1])

Test Case:

reviews = [4, 2, 7, 5, 9, 3]
print(max_enjoyment(reviews))  # Output: 20

Time and Space Complexity Analysis

• Time Complexity: O(n), where n is the number of movies. Each movie is processed once with a constant number of state updates.
• Space Complexity: O(n) for the full DP table. Can be optimized to O(1) by keeping only the last states, since only i-1 is needed at each step.

Space Optimized Version:

def max_enjoyment_optimized(reviews):
    n = len(reviews)
    if n == 0:
        return 0
    prev_review = reviews[0]
    prev_skip = 0

    for i in range(1, n):
        curr_review = prev_skip + reviews[i]   # Review now, previous must be skip
        curr_skip = prev_review                # Skip now, previous must be review
        prev_review, prev_skip = curr_review, curr_skip

    return max(prev_review, prev_skip)

Edge Cases and Validation

• All reviews have zero value: Returns 0.
• Only one movie: Returns its value.
• Two movies: Returns the maximum of their values (since you can't review both).
• Long runs of low-value movies with occasional spikes: The algorithm will always select spikes without violating constraints.

Visualizing the DP Transitions

Let’s visualize the choices as a decision tree for a small input, say reviews = [3, 2, 5, 10]:

• At index 0: Can review (3) or skip (0).
• At index 1: If previous reviewed, must skip; if previous skipped, can review.
• Continue pattern, noting that two skips in a row are not allowed.

The DP ensures that only valid paths are considered and picks the one with the maximum sum.

Why Is This Variant Used in Quant Interviews?

• Tests abstraction: Candidates must recognize the base dynamic programming formulation and adapt it to new constraints.
• Models real-world trading: In quantitative strategies, constraints are everywhere (liquidity, regulations, risk budgets); the ability to model and optimize under constraints is crucial.
• Demonstrates coding rigor: The interviewer observes if you handle edge cases, explain state transitions, and optimize space/time.

Generalizing the Problem: Skipping K In a Row

Suppose the rule changes: "You cannot skip more than K movies in a row."

• State space increases: dp[i][k] for k = 0 to K (number of consecutive skips).
• Transitions:
  • If reviewing: previous must have been skipped (any skip count from 1 to K).
  • If skipping: previous skip count must be less than K.

Recurrence:

$$ dp[i][0] = \max_{j=1}^{K} dp[i-1][j] + reviews[i] $$ $$ dp[i][k] = dp[i-1][k-1], \quad 1 \leq k \leq K $$

The method scales, but state management becomes more complex.

Common Pitfalls and How to Avoid Them

• Forgetting to initialize base cases: Always set up dp[0] states according to the problem description.
• Allowing illegal transitions: Ensure no two skips in a row are possible.
• Off-by-one errors: Be careful with indices, especially at the start.
• Not optimizing space when possible: In interviews, mention how you’d optimize for space/time if the array is large.

Practice Question for Readers

Given reviews = [5, 1, 1, 5, 10, 2, 4], apply the above method and compute the answer by hand or code. What is the maximum enjoyment value?

reviews = [5, 1, 1, 5, 10, 2, 4]
print(max_enjoyment(reviews))

Conclusion

The "Dynamic Programming Variant of House Robber Problem" as posed in Akuna Capital Quant Strategist interviews is a fantastic test of both algorithmic fluency and the ability to handle nuanced constraints. The key is to:

• Frame the state variables precisely.
• Write clear, correct recurrences.
• Implement with attention to base cases and edge conditions.
• Optimize where possible, and always explain your thought process.

Mastering this variant not only prepares you for interviews but provides a toolkit for tackling a wide range of real-worldproblems in quantitative research, algorithmic trading, and operations research.

Deep Dive: Mathematical Justification of the DP Approach

Let’s briefly justify why dynamic programming is the right tool for this problem, and why greedy or brute-force approaches would fail.

Why Not Greedy?

A naive greedy strategy would be to always select the movie with the highest available review value, but this doesn’t account for future constraints. For instance, taking a high-value movie now might force you to skip two valuable movies later (because you’d be forced into two skips, which is not allowed).

Dynamic programming, in contrast, keeps track of all possible states (i.e., whether you skipped or reviewed the previous movie) and ensures that you never violate the constraints, while globally maximizing the enjoyment value.

State Representation and Optimal Substructure

The problem exhibits optimal substructure: the optimal solution to the whole problem is composed of optimal solutions to its subproblems (the enjoyment value up to the previous movies, under valid skip/review patterns).

Overlapping subproblems also exist, as for each movie, we only need to know the result of certain states from the previous step, which can be reused multiple times as we build up the solution.

Further Enhancements: Traceback to Find Which Movies to Review

In interviews or real-world applications, you might be asked not only for the maximum value, but also which movies to select. This can be done by keeping a "parent" pointer or reconstructing the choices from the DP table.

def max_enjoyment_with_path(reviews):
    n = len(reviews)
    if n == 0:
        return 0, []
    dp = [[float('-inf')] * 2 for _ in range(n)]
    parent = [[-1] * 2 for _ in range(n)]
    dp[0][0] = reviews[0]
    dp[0][1] = 0

    for i in range(1, n):
        # Review
        if dp[i-1][1] + reviews[i] > dp[i][0]:
            dp[i][0] = dp[i-1][1] + reviews[i]
            parent[i][0] = 1  # Came from skip

        # Skip
        if dp[i-1][0] > dp[i][1]:
            dp[i][1] = dp[i-1][0]
            parent[i][1] = 0  # Came from review

    # Reconstruct path
    result = []
    i = n-1
    state = 0 if dp[n-1][0] >= dp[n-1][1] else 1

    while i >= 0:
        if state == 0:  # Reviewed
            result.append(i)
            state = parent[i][0]
        else:           # Skipped
            state = parent[i][1]
        i -= 1

    return max(dp[n-1][0], dp[n-1][1]), result[::-1]

This function returns both the maximum enjoyment and the list of movie indices you should review.

Real-World Analogies and Applications

The pattern in this problem appears in many areas beyond movie reviews or house robbing:

• Portfolio construction: Picking assets where regulatory or risk limits prevent too many consecutive omissions or inclusions.
• Workforce scheduling: Assigning shifts to employees such that no one works (or skips) too many shifts consecutively.
• Energy management: Operating equipment with mandatory rest periods, but also constraints on consecutive inactivity.
• Sports and training plans: Designing rest and activity days with interval constraints.

The methodology—tracking a state variable that encodes recent history—is a powerful framework for optimization under sequencing constraints.

Interview Tips: How to Tackle Such Questions

1. Clarify the Constraints:
   • Ask questions to make sure you understand all rules—what does “no more than one skip in a row” mean? What about the first or last item?
2. Draw State Diagrams:
   • Sketch out possible transitions between states—this helps you avoid missing edge cases.
3. Define the State Precisely:
   • State should be minimal yet sufficient to enforce all constraints.
4. Write the Recurrence and Base Cases:
   • Don’t rush to code; first, write equations and small tables by hand.
5. Code Cleanly, Mention Optimizations:
   • In interviews, mention how you’d reduce space/time if needed.
6. Test on Small and Edge Cases:
   • Check single-element, two-element, and all-zero/negative scenarios.
7. Explain Your Thought Process:
   • Discuss trade-offs and possible extensions (like the K-skip generalization).

Frequently Asked Questions

• Q: What if negative values are allowed in reviews?
  A: The DP still works. Skipping negative-value movies is optimal; the state machine ensures you never violate constraints.
• Q: Can you solve this problem recursively with memoization?
  A: Yes. You can write a recursive function with memoization (using @lru_cache in Python) that tracks the index and the number of skips in a row.

  
  from functools import lru_cache
  def max_enjoyment_recursive(reviews):
      n = len(reviews)
      @lru_cache(None)
      def dfs(i, skip):
          if i >= n:
              return 0
          res = float('-inf')
          # Review this movie if previous was skipped
          if skip == 1:
              res = max(res, dfs(i+1, 0) + reviews[i])
          # Skip this movie if previous was reviewed
          if skip == 0:
              res = max(res, dfs(i+1, 1))
          return res
      return max(dfs(0, 0), dfs(0, 1))
  

Summary Table: DP Transitions

Cheat Sheet: Steps to Solve

• Initialize dp[0][0] = reviews[0], dp[0][1] = 0
• For each movie i from 1 to n-1:
  • dp[i][0] = dp[i-1][1] + reviews[i]
  • dp[i][1] = dp[i-1][0]
• Answer is max(dp[n-1][0], dp[n-1][1])

References and Further Reading

• Leetcode House Robber (classic variant)
• Wikipedia: Dynamic Programming
• Akuna Capital Careers
• CP Algorithms: Dynamic Programming

Final Thoughts

The dynamic programming variant of the house robber problem, as seen in Akuna Capital quant interviews, is an excellent demonstration of how algorithmic thinking and careful state management enable us to solve real-world constrained optimization problems. Mastering this pattern will not only help you ace interviews but also bolster your problem-solving skills for a variety of practical applications in quantitative finance and beyond.

We encourage you to practice with different variants, challenge yourself with modifications (like K skips), and always strive to understand the underlying principles. Good luck with your interviews and happy coding!