Minimum Cost Climbing Stairs

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Steps

setup

Initialize variables for first two steps

We start by storing the cost of the first step in prev2 and the cost of the second step in prev1. These represent the minimum costs to reach steps 0 and 1 respectively.

💡 Initializing these variables sets the base cases for the DP approach, which are essential for building up the solution.

Line:

prev2, prev1 = cost[0], cost[1]

💡 The algorithm treats the first two steps as known minimum costs to build upon.

fill_cells

Start iteration at step 2

We begin the loop at step index 2 to compute the minimum cost to reach this step using the previous two minimum costs.

💡 Starting the iteration here is crucial because the first two steps are base cases already known.

Line:

for i in range(2, n):

💡 The algorithm will now compute costs for steps beyond the base cases.

fill_cells

Calculate current cost at step 2

Calculate curr as cost[2] plus the minimum of prev1 and prev2, i.e., 20 + min(15, 10) = 30.

💡 This step shows how the algorithm uses previously computed minimum costs to find the current minimum cost.

Line:

curr = cost[i] + min(prev1, prev2)

💡 The minimum cost to reach step 2 depends on the cheaper path from the previous two steps.

fill_cells

Update prev2 and prev1 after step 2

Shift prev2 to prev1's value (15), and prev1 to curr's value (30) to prepare for the next iteration.

💡 Updating these variables moves the window forward, maintaining only the last two minimum costs.

Line:

prev2, prev1 = prev1, curr

💡 The algorithm uses a rolling window to save space while keeping track of needed values.

fill_cells

End of loop - no more steps to process

Since the input has only 3 steps, the loop ends here after processing step 2.

💡 The loop covers all steps from 2 to n-1, so no further iterations are needed.

Line:

for i in range(2, n):  # loop ends

💡 The algorithm has computed minimum costs for all steps.

compare

Compute final answer by taking min of last two costs

Return the minimum of prev1 and prev2, i.e., min(30, 15) = 15, representing the minimum cost to reach the top.

💡 The top can be reached from either of the last two steps, so we take the cheaper option.

Line:

return min(prev1, prev2)

💡 The final answer is the minimum cost to reach beyond the last step.

reconstruct

Interpretation of final answer

The minimum cost to reach the top is 15, achieved by starting at step 1 (cost 15) and then climbing two steps to the top.

💡 This step connects the computed value back to the problem context.

💡 The algorithm's output matches the problem's explanation and expected output.

traverse

Summary of computed minimum costs

The minimum costs to reach steps 0, 1, and 2 are 10, 15, and 30 respectively, showing the cumulative cost progression.

💡 Reviewing these values helps understand how the algorithm accumulates costs step-by-step.

💡 The DP approach builds the solution by combining previous results efficiently.

expand

Visualizing the rolling window update

Highlight how prev2 and prev1 move forward after each iteration, representing the last two minimum costs needed for the next calculation.

💡 This rolling update is key to the space optimization of the algorithm.

Line:

prev2, prev1 = prev1, curr

💡 The algorithm only keeps track of two values at a time, reducing memory usage.

finalize

Final confirmation of answer

Confirm the minimum cost to reach the top is 15, which matches the expected output for the input cost array.

💡 This final step reassures the correctness of the algorithm and the trace.

Line:

print(minCostClimbingStairs(cost))  # Output: 15

💡 The algorithm correctly computes the minimum cost climbing stairs problem.

def minCostClimbingStairs(cost):
    n = len(cost)
    prev2, prev1 = cost[0], cost[1]  # STEP 1: Initialize base costs
    for i in range(2, n):  # STEP 2: Iterate from step 2
        curr = cost[i] + min(prev1, prev2)  # STEP 3: Compute current min cost
        prev2, prev1 = prev1, curr  # STEP 4: Update rolling variables
    return min(prev1, prev2)  # STEP 6: Return final answer

if __name__ == '__main__':
    cost = [10, 15, 20]
    print(minCostClimbingStairs(cost))  # STEP 10: Output 15

📊

Minimum Cost Climbing Stairs - Watch the Algorithm Execute, Step by Step

Watching the algorithm step-by-step reveals how dynamic programming builds solutions from smaller subproblems, making it easier to understand the logic behind the minimum cost calculation.

Item 0 - wt:10 val:10

i\w	0	1	2
i=0	10	15	?

prev2 = 10

Item 2 - wt:20 val:20

i\w	0	1	2
i=0	10	15	?

Compute step 2

Item 2 - wt:20 val:30

i\w	0	1	2
i=0	10	15	30

curr = 20 + 10 = 30

Item 2 - wt:20 val:30

i\w	0	1	2
i=0	10	15	30

prev2 updated to 15

Item null - wt:null val:null

i\w	0	1	2
i=0	10	15	30

All steps computed

Item null - wt:null val:15

i\w	0	1	2
i=0	10	15	30

prev2 = 15

Item 1 - wt:15 val:15

i\w	0	1	2
i=0	10	15	30

Chosen step with cost 15

Item null - wt:null val:null

i\w	0	1	2
i=0	10	15	30

Step 0 cost 10

Item 2 - wt:20 val:30

i\w	0	1	2
i=0	10	15	30

prev2 now 15

Item null - wt:null val:15

i\w	0	1	2
i=0	10	15	30

Final answer = 15

Key Takeaways

✓ The algorithm uses a rolling window of two variables to store minimum costs, avoiding the need for a full DP table.

This space optimization is hard to see from code alone but becomes clear when watching the step-by-step updates.

✓ Each step's minimum cost depends only on the minimum of the previous two steps plus the current step's cost, showing a Fibonacci-like dependency.

Visualizing this dependency clarifies why the algorithm only needs two previous values.

✓ The final answer is the minimum of the last two computed costs, reflecting that the top can be reached from either of the last two steps.

This decision point is subtle in code but explicit in the visualization, helping understand the problem's boundary condition.

Practice

1. Consider the following Python code implementing the House Robber problem using space-optimized DP. What is the value of prev1 after the third iteration of the loop when input is [2, 1, 4]?

def rob(nums):
    prev1, prev2 = 0, 0
    for num in nums:
        temp = prev1
        prev1 = max(prev1, prev2 + num)
        prev2 = temp
    return prev1

easy

A. 6

B. 4

C. 5

D. 7

Solution

1. Step 1: Trace iterations

   Iteration 1 (num=2): prev1= max(0,0+2)=2, prev2=0 Iteration 2 (num=1): prev1= max(2,0+1)=2, prev2=2 Iteration 3 (num=4): prev1= max(2,2+4)=6, prev2=2

2. Step 2: Identify value after third iteration

   After third iteration, prev1=6

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Manual calculation confirms prev1=6 after third iteration [OK]

Hint: Track prev1 and prev2 updates carefully each iteration [OK]

Common Mistakes:

• Confusing prev1 and prev2 updates
• Off-by-one iteration counting

2. Consider this buggy code snippet for maxProfit with at most two transactions. Which line contains the subtle bug causing incorrect profit calculation?

medium

A. Line 10: Inner loop iterating forwards instead of backwards

B. Line 7: Initialization of maxDiff array

C. Line 9: Outer loop iterating from 1 to n

D. Line 6: Initialization of profit array

Solution

1. Step 1: Identify iteration order in inner loop

   The inner loop iterates k from 1 to 2 (forwards). This causes profit[k-1] to be updated before profit[k], leading to incorrect maxDiff updates.

2. Step 2: Correct iteration order

   Iterating k backwards (from 2 down to 1) ensures profit[k-1] is from previous iteration, preserving correct state transitions.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Backward iteration prevents premature state overwriting [OK]

Hint: Inner loop must run backwards to avoid overwriting states [OK]

Common Mistakes:

• Iterating inner loop forwards
• Incorrect maxDiff updates

3. What is the time complexity of the space-optimized dynamic programming solution for the climbing stairs problem, and why might some candidates incorrectly think it is higher?

medium

A. O(2^n) due to recursive calls without memoization

B. O(n^2) because of nested loops to compute each step

C. O(n) because each step is computed once in a single loop

D. O(n) but with additional O(n) space for memoization arrays

Solution

1. Step 1: Identify loop structure

   The space-optimized DP uses a single loop from 2 to n, computing each step once.

2. Step 2: Understand why time is O(n)

   Each iteration does constant work, so total time is linear in n.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Single loop with constant work per iteration -> O(n) [OK]

Hint: Single loop with constant work means O(n) time [OK]

Common Mistakes:

• Confusing recursion without memoization with DP
• Assuming nested loops cause O(n^2)

4. What is the time complexity of the optimal dynamic programming solution for the House Robber II problem using the space-optimized approach shown below?

def rob_linear(nums, start, end):
    prev, curr = 0, 0
    for i in range(start, end + 1):
        prev, curr = curr, max(curr, prev + nums[i])
    return curr

def rob(nums):
    n = len(nums)
    if n == 1:
        return nums[0]
    return max(rob_linear(nums, 0, n - 2), rob_linear(nums, 1, n - 1))

medium

A. O(n) with constant extra space

B. O(n²) due to nested loops

C. O(2^n) because of recursion without memoization

D. O(n) because each house is processed twice in two linear scans

Solution

1. Step 1: Identify loops in rob_linear

   rob_linear runs a single loop from start to end, which is O(n) in worst case.

2. Step 2: Count calls in rob

   rob calls rob_linear twice, each over roughly n/2 elements, so total O(n) time.

3. Step 3: Consider space usage

   The DP uses only constant extra space (variables prev and curr) during computation.

4. Final Answer:

   Option A -> Option A

5. Quick Check:

   Two linear scans over n elements with constant space -> O(n) time and O(1) space [OK]

Hint: Two linear scans sum to O(n) time with constant space [OK]

Common Mistakes:

• Confusing recursion with DP
• Assuming nested loops cause O(n²)
• Ignoring that calls are sequential, not nested

5. Suppose the climbing stairs problem is modified so you can climb 1, 2, or 3 steps at a time. Which of the following changes to the space-optimized DP solution correctly computes the number of ways to reach the top?

hard

A. Keep two variables and update current as sum of the previous two only

B. Use three variables to store ways for last three steps and update current as their sum

C. Use recursion without memoization to handle the extra step size

D. Use a greedy approach to always take 3 steps when possible

Solution

1. Step 1: Understand the new step sizes

   With steps of size 1, 2, or 3, the number of ways to reach step i depends on the sum of ways to reach i-1, i-2, and i-3.

2. Step 2: Adjust space-optimized DP variables

   We need to track the last three computed values, so three variables are required to update current as their sum.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Summing last three states matches the new recurrence [OK]

Hint: Number of variables equals max step size for space optimization [OK]

Common Mistakes:

• Using only two variables for three-step jumps
• Recursion without memoization causes inefficiency