Min Cost Climbing Stairs

You are given an array of costs to climb a staircase, and you can start from either step 0 or step 1. You can climb one or two steps at a time. The goal is to find the minimum cost to reach the top of the staircase.

A dynamic programming approach can help solve this problem.

Step-by-Step Solution

  1. Initialize a DP Array: Create a DP array of the same length as the cost array to store the minimum cost to reach each step.

  2. Base Cases: You can start from either step 0 or step 1, so the cost for these steps will be the same as the given cost array for those steps.

  3. Bottom-Up Calculation: Iterate through the DP array from step 2 onwards, and for each step, calculate the minimum cost to reach that step by considering both one step and two steps before.

  4. Find the Result: The minimum cost to reach the top will be the minimum of the last two values in the DP array.

Code

 1
 2
 3
 4
 5
 6
 7
 8
 9
10

class Solution:
    def minCostClimbingStairs(self, cost: List[int]) -> int:
        n = len(cost)
        dp = [0] * n
        dp[0], dp[1] = cost[0], cost[1]

        for i in range(2, n):
            dp[i] = cost[i] + min(dp[i - 1], dp[i - 2])

        return min(dp[-1], dp[-2])

Insights

  • Dynamic Programming: The dynamic programming approach builds up the solution from the bottom and avoids redundant calculations, leading to efficient solving.
  • Time Complexity: The time complexity of this solution is (O(n)), where (n) is the length of the cost array.
  • Space Complexity: The space complexity is also (O(n)) due to the additional DP array.

This approach considers both possibilities of climbing one or two steps at each step, and builds up the minimum cost solution. By considering all possibilities and building from the base, it ensures that the final result is the minimum possible cost to reach the top of the staircase.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11

def minCostClimbingStairs(cost):
    n = len(cost)
    first = cost[0]
    second = cost[1]
    if n <= 2:
        return min(first, second)
    for i in range(2, n):
        curr = cost[i] + min(first, second)
        first = second
        second = curr
    return min(first, second)
  1. Implement the recursive solution
  2. Make you have the right recurrence relation
  3. What is the base case?
  4. What is the sub problem
  5. Find the solution for the given problem.

DP can be used in three cases:

  • T / F for something
  • Find min/max of something (optimization)
  • Number of ways

. Operating on string or array (sequence) - Prefixes - Suffixes - Substrings

What is the state in this problem?

Base condition

  • What is the smallest instance of the problem? Trivial case to return the answer

Base Cases

. steps(0) = 0 steps(1) = 0

Recurrence Relation f(n) = f(n-1) + f(n-2) steps(n) = min(steps(n-1) + steps(n-1)) What do we need to compute?

Order of subproblems State Memoization vs Bottom Up

Bottom Up - Smaller test cases to bigger test cases? Recurrence relation is used to break a problem into subproblems

Recursive solution is having overlapping subproblems TLE - Repeated work. Memoization TOP DOWN way to solve this problem

  1. Take a one dimensional array of size equal to cost array
  2. Base cases will initialize the elements at index 0 and 1 to 0.
  3. For the recursive case, I have to apply the recurrence relation to compute the value
  4. Store the answer at that index
  5. Return the last element in the dp array
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19

# @param {Integer[]} cost
# @return {Integer}

def climb_wrapper(cost, step)
   dp = Array.new(cost.size, 0) 
   
   for step in (2..cost.size)
      min_cost_one_step = dp[step-1] + cost[step-1]
      min_cost_two_step = dp[step-2] + cost[step-2]  

      dp[step] = [min_cost_one_step, min_cost_two_step].min 
   end
 
   dp[-1]
end

def min_cost_climbing_stairs(cost)
  climb_wrapper(cost, cost.size)
end

Problem Classification

This problem is an optimization problem in dynamic programming.

What Components:

  1. You are provided with an integer array ‘cost’ which denotes the cost of each step on a staircase.
  2. You can climb either one or two steps at a time.
  3. You can start at either the first step (index 0) or the second step (index 1).
  4. You need to determine the minimum cost to reach the top of the staircase.

The problem can be classified as a Dynamic Programming problem. Because, in the problem, we are asked to find an optimal solution (in this case, the minimum cost to reach the top of the stairs), and we can break down the problem into smaller sub-problems (the minimum cost to reach each individual step).

The concept of choosing between one step or two steps at a time allows for overlapping subproblems, another characteristic feature of dynamic programming problems. The optimal solution at each stage (each step) depends on the solutions of the smaller subproblems (the steps before it).

We have to make a decision at every step, choosing either to go one step or two steps, to minimize the total cost to reach the top. This exhibits the principle of optimality, indicating that the problem can be solved using dynamic programming.

It also has elements of recursion (we solve the problem for the current step by relating it to the solution of previous steps) and memoization (we can remember the minimum cost to reach each step to avoid repeated calculations), which are key techniques used in dynamic programming.

Language Agnostic Coding Drills

  1. Dissect the code and identify each distinct concept it contains:
  • Basic Syntax: This includes defining a function, variable assignment, return statements, using built-in functions like len() and min(), and indexing into a list.
  • Conditional Statements: Here we have an “if” condition to handle cases when there are less than or equal to two steps.
  • Looping: A “for” loop is used to iterate over the range of steps.
  • Basic Arithmetic Operations: Addition and minimum operation used in line calculating curr.
  1. List out the coding concepts or drills in order of increasing difficulty:
  • Basic Syntax: The most fundamental level and easy to understand.
  • Basic Arithmetic Operations: Addition and finding minimum is quite straightforward.
  • Conditional Statements: A step further from the basic syntax. It requires understanding of conditionals but is not as complex as looping.
  • Looping: Understanding loops requires comprehending the flow of control in a program, making it a more complex concept.
  1. Describe the problem-solving approach:
  • The problem here is a classical dynamic programming problem where the solution to a problem depends on solutions to smaller instances of the same problem.
  • Initially, the minimum cost to reach the first and second step is just the cost of the first and second step respectively.
  • For each subsequent step, the minimum cost to reach that step is the cost of that step plus the minimum cost to reach either of the two preceding steps.
  • We use a loop to iterate over all the steps, maintaining the minimum cost to reach the current step and the previous step.
  • Finally, we return the minimum cost between the last two steps, which represents the minimum cost to reach the top of the stairs.

Targeted Drills in Python

  1. Separate Python code for each identified concept:
  • Basic Syntax: Writing a simple function, variable assignment, using built-in functions, and list indexing:
1
2
3
4
5

def basic_syntax():
    numbers = [1, 2, 3, 4, 5]  # list initialization
    first = numbers[0]  # indexing into a list
    print("First element:", first)  # print statement
    return len(numbers)  # using built-in function and return statement
  • Basic Arithmetic Operations: Performing basic addition and minimum operation:
1
2
3
4

def arithmetic_operations(a, b):
    sum_result = a + b  # Addition
    min_result = min(a, b)  # Minimum
    return sum_result, min_result
  • Conditional Statements: Basic if-else conditional statements:
1
2
3
4
5

def conditionals(n):
    if n <= 2:
        return "Small number"
    else:
        return "Large number"
  • Looping: For loop over a range:
1
2
3

def looping(n):
    for i in range(n):
        print(i)
  1. Problem-specific drills:
  • The most significant problem-specific concept in this case is dynamic programming, which allows us to solve each subproblem only once and use the results of these solved subproblems to solve larger subproblems:
1
2
3
4
5
6

def dynamic_programming(cost):
    first, second = cost[0], cost[1]
    for i in range(2, len(cost)):
        curr = cost[i] + min(first, second)
        first, second = second, curr
    return min(first, second)

This drill is essential as it encapsulates the main logic of our problem, the essence of dynamic programming: compute the minimal cost to reach a certain step based on previously computed results.

  1. Integration of the pieces:

  • First, understand the problem’s domain and requirements. In this case, the requirement is to calculate the minimum cost to reach the top of the stairs.

  • Next, we initialize the cost for the first and second step based on the problem statement, which is done using basic syntax and arithmetic operations.

  • We then use a looping construct to iterate over each step, starting from the third. For each step, we calculate the minimum cost to reach it (using our dynamic programming drill).

  • Finally, we return the minimum cost between the last two steps as our result. The combination of these drills should provide the complete solution to our problem.

Dynamic Programming

Stage 1: Recursive Formulation

  1. Can you describe the problem you’re solving, in precise and coherent English?

Here is how I would describe this staircase problem in precise English:

We are given an array of positive integers cost representing the cost to climb each individual step of a staircase. The staircase has n steps numbered 0 to n-1 where n = cost.length.

Starting at either step 0 or step 1, we want to recursively compute the minimum total cost to reach the top step (step n-1) from the initial position.

At each step, we may advance either one step or two steps further up the staircase. Climbing each step incurs the cost specified in the cost array for that step index.

The solution involves recursively exploring all valid step sequences, computing the total cost at each step, and returning the minimum cost solution that reaches the top.

The key constraints are:

  • n stairs numbered 0 to n-1 where n = cost.length
  • Non-negative integer costs for each step
  • Can start at step 0 or step 1
  • Can climb 1 or 2 steps at each advance
  • Seek minimum total cost to reach top step

Let me know if I’m missing or unclear about any details or requirements in this problem description. Please feel free to suggest any improvements to make the problem statement more precise and coherent.

  1. What are the smaller subproblems that contribute to solving the overall problem?

Here are some of the key subproblems that contribute to solving this overall minimum cost staircase problem:

  • Minimum cost to reach step i starting from step 0

    • Represents cost to reach step i in optimal way from base

  • Minimum cost to reach step i starting from step 1

    • Represents cost to reach step i in optimal way from first step

  • Minimum cost to take 1 step from current position

    • Represents cost of optimal single step from a given position

  • Minimum cost to take 2 steps from current position

    • Represents optimal cost of advancing 2 steps from current position

  • Minimum of taking 1 step vs 2 steps from current position

    • Choosing whether to take 1 or 2 steps at each position based on minimum cost

  • Combining optimal subproblem solutions

    • Building up global solution by combining optimal solutions to subproblems

By finding optimal solutions for these smaller problems of calculating the costs of taking 1 step, 2 steps, or reaching a particular step from the starting positions, we can recursively build up the final minimum cost solution. Solving smaller subproblems is key to solving the overall problem optimally.

  1. Can you write a recursive formula or algorithm that expresses the solution of the problem in terms of its smaller subproblems?

Here is one way to write a recursive formula expressing the solution to this staircase minimum cost problem in terms of smaller subproblems:

Let’s define:

  • cost(i) = cost to take step i
  • dp(i) = minimum cost to reach step i

Then the recursive relationships are:

Base cases:

dp(0) = 0 dp(1) = 0

Recursive formula:

dp(i) = min(dp(i-1) + cost(i-1),
dp(i-2) + cost(i-2))

This expresses the minimum cost to reach step i in terms of the minimum costs of the subproblems of reaching the prior two steps i-1 and i-2, plus their respective step costs.

The overall solution is then:

min(dp(n-1), dp(n-2) + cost(n-2))

Where n is number of steps.

This finds the minimum of the two options of either starting at step 0 and reaching step n-1, or starting at step 1 and reaching step n-2.

So we build up the full solution by combining optimal solutions to the smaller subproblems of reaching each individual step.

Specification

  1. Can you specify the problem in terms of the input and desired output?

Here is one way to specify this staircase problem in terms of input and desired output:

Input:

  • cost - An array of positive integers representing the cost to climb each individual step

  • n - The number of steps, where n = cost.length

Output:

  • An integer representing the minimum total cost to reach the top step (step n-1) starting from either step 0 or step 1 and recursively taking 1 or 2 steps at each point.

More formally:

Given:

cost = [c0, c1, …, cn-1]

where ci >= 0 is the cost to take step i

and n = cost.length

Find:

min(dp(n-1), dp(n-2) + cn-2)

where:

dp(i) = minimum cost to reach step i

Such that:

dp(i) = min(dp(i-1) + ci-1, dp(i-2) + ci-2)

With base cases:

dp(0) = 0 dp(1) = 0

So in summary, given the cost array of step values, we want to return the minimum total cost computed using the optimal substructure of taking 1 or 2 steps recursively. This formal specification captures the core input-output relationship and optimization objective.

  1. What are the boundary conditions or base cases for your recursive formulation?

The boundary conditions or base cases for the recursive formulation of the staircase minimum cost problem are:

dp(0) = 0 dp(1) = 0

Where dp(i) represents the minimum cost to reach step i.

This establishes that:

  • Reaching step 0 has a minimum cost of 0 (the start position)

  • Reaching step 1 from step 0 has a minimum cost of 0

These form the base cases because:

  • Step 0 is the starting position, so no cost to reach it.

  • Step 1 is reachable from step 0 by taking 1 step, which has no associated cost.

  • Any step 2 or higher cannot be directly reached from the starting position in a single step.

So dp(0) and dp(1) represent the base scenarios with defined minimum costs. The recursive formula builds on these cases to calculate dp(i) for i >= 2.

The boundary conditions anchor the recurrence relation since costs for all higher steps depend on costs of their preceding steps. Defining the base cases avoids infinite recursion.

Solution

  1. What does the recursive formula look like for solving the whole problem?

The full recursive formula for solving this staircase minimum cost problem is:

dp(i) = minimum cost to reach step i

Base cases:

dp(0) = 0 dp(1) = 0

Recursive formula:

dp(i) = min(dp(i-1) + cost(i-1), dp(i-2) + cost(i-2))

Overall solution:

min(dp(n-1), dp(n-2) + cost(n-2))

Where:

n = number of steps

cost(i) = cost to take step i

To summarize:

  • The base cases establish the minimum costs to reach the starting steps 0 and 1.

  • The recursive formula expresses the cost to reach step i in terms of the optimal costs of reaching the prior two steps i-1 and i-2 plus their step costs.

  • The overall solution combines the optimal cost to reach the top steps n-1 and n-2 (from starting positions 0 and 1) and takes the minimum.

By leveraging the optimal substructure and building up the minimum costs incrementally from the base cases, this recurrence relation computes the final lowest total cost recursively.

  1. Can the recursive formula be broken down into simpler sub-problems?

Yes, the recursive formula for this staircase problem can be broken down into simpler sub-problems:

The full recurrence is:

dp(i) = min(dp(i-1) + cost(i-1), dp(i-2) + cost(i-2))

This can be decomposed into two simpler sub-problems:

  1. Cost to take 1 step from current position:

dp1(i) = dp(i-1) + cost(i-1)

  1. Cost to take 2 steps from current position:

dp2(i) = dp(i-2) + cost(i-2)

The overall solution combines these:

dp(i) = min(dp1(i), dp2(i))

So we break down the decision at each step into:

  1. What is the minimum cost if we take 1 step from here?

  2. What is the minimum cost if we take 2 steps from here?

  3. Which of those two options has the smaller cost?

By decomposing into these atomic single-step and double-step subproblems, we simplify the recurrence into more modular pieces.

Stage 2: Bottom-up Solution

  1. Can you identify all the subproblems your recursive algorithm can call?

Based on the recursive formula, here are all the possible subproblems that would be called by this algorithm to solve the staircase minimum cost problem:

  • dp(0) - Minimum cost to reach step 0

    • Base case, pre-computed

  • dp(1) - Minimum cost to reach step 1

    • Base case, pre-computed

  • dp(2) - Minimum cost to reach step 2

    • Depends on dp(0) and dp(1)

  • dp(3) - Minimum cost to reach step 3

    • Depends on dp(1) and dp(2)

  • …

  • dp(i) - Minimum cost to reach step i

    • Depends on dp(i-1) and dp(i-2)

  • …

  • dp(n-2) - Minimum cost to reach step n-2

    • Depends on dp(n-3) and dp(n-4)

  • dp(n-1) - Minimum cost to reach step n-1

    • Depends on dp(n-2) and dp(n-3)

So in summary, the possible subproblems are the set of minimum cost values dp(0) to dp(n-1) for each individual step. Each dp(i) subproblem depends on the two preceding subproblems dp(i-1) and dp(i-2), with base cases dp(0) and dp(1) pre-computed.

  1. What kind of data structure would be suitable for memoization? Is it an array, a hashmap, or something else?

For memoizing and storing the results of the subproblems in this staircase problem, the most suitable data structure would be a 1D array.

We can create an array dp of length n, where n is the number of steps:

dp = [0] * n

Then dp[i] will store the minimum cost to reach step i.

An array is a good choice because:

  • The subproblems have a natural 1D ordering by step number that maps cleanly to array indices.

  • Looking up dp[i] takes O(1) constant time.

  • We can iterate through the steps linearly to populate the array.

  • Arrays use less memory than dictionaries for sequential data.

A hashmap/dictionary could also work, but provides no benefit over a linear array for this problem and would use more memory.

So in summary, an array dp of size n, indexed by step number, provides an efficient way to cache and lookup already computed subproblem solutions.

Identify Dependencies

  1. Can you draw a dependency graph to identify which subproblems rely on other subproblems?

Here is a dependency graph showing how the subproblems in this staircase problem rely on one another:

          dp(0)
            |
            |
dp(1) -- dp(2) -- dp(3) -- ... -- dp(n-2) -- dp(n-1)
            |          |            |           |
            |          |            |           |
         dp(3)      dp(4)        dp(n-3)     dp(n-2)

Where:

  • dp(i) - Minimum cost to reach step i

  • An arrow from dp(i) to dp(j) indicates dp(j) depends directly on dp(i).

Key observations:

  • Each subproblem dp(i) directly depends on the two preceding subproblems dp(i-1) and dp(i-2).

  • The base cases dp(0) and dp(1) have no dependencies.

  • All other subproblems form a linear chain of dependencies going backwards two steps.

  • The final solution depends directly on dp(n-1) and dp(n-2).

This dependency graph helps visualize the relationships between subproblems and how costs iteratively build up from the base cases. We can use it to guide memoization and tabular solution building.

  1. Can you formalize these dependencies? For example, does solving subproblem i depend on solving subproblem i-1?

Yes, we can formalize the dependencies between subproblems dp(i) in this staircase problem as:

  • Base cases:

    • dp(0) has no dependencies
    • dp(1) only depends on dp(0)

  • Recursive case:

    • dp(i) depends on:
      • dp(i-1)
      • dp(i-2)

More formally:

Let’s define DEPENDS(A, B) to mean “Problem A depends on problem B”.

Then the dependencies are:

  • DEPENDS(dp(1), dp(0))

  • DEPENDS(dp(i), dp(i-1)) for i >= 2

  • DEPENDS(dp(i), dp(i-2)) for i >= 2

  • NOT(DEPENDS(dp(0), dp(other))) for any other subproblem

So in summary, each subproblem dp(i) depends directly on the prior two subproblems dp(i-1) and dp(i-2), except the base cases dp(0) and dp(1) which have no dependencies.

This formalization captures the key insight that solving each subproblem requires first solving the two preceding subproblems it directly depends on.

Evaluation Order

  1. Can you find a good order to evaluate the subproblems so that each subproblem is solved only after its dependencies?

Based on the dependency relationships, here is a good order to evaluate the subproblems for the staircase problem such that each subproblem dp(i) is solved after its dependencies dp(i-1) and dp(i-2):

  1. dp(0)

  2. dp(1)

  3. dp(2)

  4. dp(3)

…

i. dp(i)

…

n-2. dp(n-2)

n-1. dp(n-1)

In more formal order notation:

[dp(0), dp(1), dp(2), …, dp(i), …, dp(n-2), dp(n-1)]

This order ensures we solve the subproblems incrementally in ascending step order. Each subproblem dp(i) is evaluated after its dependencies dp(i-1) and dp(i-2) have already been computed in a prior iteration.

The base cases dp(0) and dp(1) come first since they have no dependencies. Then we build up the solutions iteratively.

This ordering guarantees that when solving dp(i), we will have already solved dp(i-1) and dp(i-2), satisfying the dependency constraints.

  1. Are there any potential pitfalls in the order of subproblem evaluation?

Here are some potential pitfalls to watch out for when ordering the subproblem evaluations in this staircase minimum cost problem:

  • Attempting to evaluate dp(i) before dp(i-1) or dp(i-2) - This violates the dependencies and recurses infinitely.

  • Solving subproblems in completely random order - Without considering dependencies, a vital subproblem result may not be ready when needed.

  • Solving subproblems in descending step order - For example dp(n-1), dp(n-2), etc. This reverses the dependencies.

  • Solving subproblems breadth-first - Evaluating all length 2 subproblems before length 3. Dependency gaps can occur.

  • Caching subproblem results out of order - If result for dp(i-1) is cached after dp(i), stale dependency used.

  • Assuming base cases dp(0) and dp(1) are zero - Important to set them explicitly first.

  • Hardcoding step costs based on indexes - Need to parameterize costs as separate input.

  • Solving overlapping subproblems repeatedly - Fail to memoize and reuse prior results.

Overall, adhering to the ascending step order, explicitly initializing base cases, and caching incremental results while properly parameterizing costs are keys to avoiding pitfalls around order and dependencies in evaluating the subproblems.

Space and Time Analysis

  1. What is the space complexity of your solution? How many distinct subproblems do you need to solve?

The space complexity of this dynamic programming solution is O(n), where n is the number of steps in the staircase.

This is because we need to store the results for up to n distinct subproblems in the memoization array dp.

Specifically, we must solve these subproblems:

  • dp(0)
  • dp(1)
  • dp(2)
  • …
  • dp(n-2)
  • dp(n-1)

There are n total staircase steps, so n distinct subproblems dp(0) through dp(n-1) must be solved and memoized.

Even though each subproblem dp(i) depends on prior subproblems, we only need to store the most recent results. The dp array acts as a sliding window of size 2.

So in summary, the number of distinct subproblems is linear in the size of the input n. We need O(n) space to store the results for later lookup.

  1. Can you estimate the time complexity by considering how much work each subproblem requires?

We can estimate the time complexity for this dynamic programming staircase solution by looking at the work required for each subproblem:

  • Each dp(i) subproblem depends on two prior subproblems dp(i-1) and dp(i-2).

  • To compute dp(i), we need to:

    1. Look up dp(i-1) and dp(i-2) results (constant time)

    2. Compute cost of taking 1 step (constant time)

    3. Compute cost of taking 2 steps (constant time)

    4. Take minimum of two costs (constant time)

  • So each subproblem takes constant O(1) work.

  • There are n total subproblems from dp(0) to dp(n-1).

Therefore, the overall time complexity is O(n) since we do constant work n times.

Even though each subproblem recursively depends on prior subproblems, we memoize results so each subproblem is only solved once.

So the work done is linear in the number of steps n, giving an overall time complexity of O(n).

Implementation

  1. Can you write down the algorithm considering the evaluation order and the memoization data structure?

Here is one way to write the full algorithm for this staircase problem using the evaluation order and memoization:

// Memoization array
dp = new int[n]

// Base cases  
dp[0] = 0
dp[1] = 0

for i = 2 to n-1:
  
  // Recurrence relation
  dp[i] = min(dp[i-1] + cost[i-1], 
              dp[i-2] + cost[i-2])
              
// Final result  
result = min(dp[n-1], dp[n-2] + cost[n-2])

return result

Where:

  • dp[] memoizes the minimum cost to reach each step

  • We evaluate dp[i] in ascending order after dependencies

  • Base cases are set explicitly

  • We apply the recurrence relation to compute each dp[i]

  • Final result combines dp[n-1] and dp[n-2]

This order of evaluation along with memoization ensures each subproblem is solved only once using previously cached results. We build up the optimal costs incrementally.

Here is one way to implement the minimum cost staircase climbing problem in Python:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19

class Solution:

  def minCostClimbingStairs(self, cost):

    # Number of steps
    n = len(cost)

    # Memoization array
    dp = [0] * (n+1)

    # Base cases
    dp[0] = dp[1] = 0

    # Build up dp[] array
    for i in range(2, n+1):
      dp[i] = min(dp[i-1] + cost[i-1], dp[i-2] + cost[i-2])

    # Return final result
    return min(dp[n], dp[n-1])

The key steps are:

  • Initialize size n+1 memoization array dp[]
  • Prepopulate base cases dp[0] and dp[1]
  • Build up dp using min cost recurrence
  • Return final min of last two elements

This iteratively calculates the optimal cost for each step using results from prior steps stored in the dp table.

Wrong Answer Runtime: 76 ms Case 1 Case 2 Input cost = [10,15,20] Output 10 Expected 15

Claude generates buggy code.

My correct answer:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10

class Solution:
    def minCostClimbingStairs(self, cost: List[int]) -> int:
        n = len(cost)
        dp = [0] * n
        dp[0], dp[1] = cost[0], cost[1]

        for i in range(2, n):
            dp[i] = cost[i] + min(dp[i - 1], dp[i - 2])

        return min(dp[-1], dp[-2])

ChatGPT generates correct answer:

To solve this problem, you can use dynamic programming. You’ll create an array to store the minimum cost to reach each step, and then iterate through the steps to calculate the minimum cost at each step. Here’s the code to implement this solution:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18

class Solution:
    def minCostClimbingStairs(self, cost: List[int]) -> int:
        n = len(cost)
        
        # Create an array to store the minimum cost to reach each step
        dp = [0] * n
        
        # Initialize the first two steps with their respective costs
        dp[0] = cost[0]
        dp[1] = cost[1]
        
        # Iterate through the steps starting from the third step
        for i in range(2, n):
            # Calculate the minimum cost to reach the current step
            dp[i] = cost[i] + min(dp[i-1], dp[i-2])
        
        # The minimum cost to reach the top floor will be either the last step or the second-to-last step
        return min(dp[-1], dp[-2])

You can use this minCostClimbingStairs function to find the minimum cost to reach the top floor given the cost array as input.