How to Explain Min Cost Climbing Stairs in a Coding Interview (Step-by-Step)

Min Cost Climbing Stairs is the gateway Dynamic Programming problem — the first place most candidates encounter the idea of subproblems, recurrence relations, and the bottom-up table. Knowing it cold is table stakes; knowing how to explain it — the DP formulation, the O(1) space optimisation, and the “top as a virtual step” framing — is what earns points.

In this post, you’ll learn exactly how to walk an interviewer through LeetCode 746: the recurrence relation from first principles, a full bottom-up DP table trace, the two-variable space optimisation, and every edge case worth naming.

Problem Statement

Examples

Constraints

• 2 <= cost.length <= 1000
• 0 <= cost[i] <= 999

What Is the Problem?

“I’m on a staircase. Each step has a cost to enter it. Once I pay, I can jump 1 or 2 steps forward. I can start for free at step 0 or step 1, and I want to exit at a virtual ‘top’ step one beyond the last index. I need the minimum total cost to reach that top. This is a classic DP problem because the minimum cost to reach any step depends on the minimum cost of reaching the two steps before it.”

The phrase “minimum cost to reach any step depends on the two steps before it” is the entire recurrence relation in plain English. Stating it before writing a formula signals that you understand why DP applies.

Step 1: Compare Approaches Before Jumping to Code

“I’ll present the full bottom-up DP table first — it’s the clearest way to derive and verify the recurrence. Then I’ll show how to compress it to O(1) space using just two rolling variables, since each step only depends on the previous two.”

The Key Insight — The Top is a Virtual Step

“Framing the top as a virtual step at index n unifies all cases — there’s no need to separately handle ‘can I reach the top from the last step or second-to-last step?’ The recurrence handles it automatically.”

The Recurrence Relation

Step 2: Explain the Algorithm in Plain English

DP Table Visualised — cost = [10, 15, 20]

n = 3. The virtual top is at index 3. dp[i] = minimum cost to arrive at step i.

The answer is dp[3] = 15. The optimal path is: start at step 1 (free), pay cost[1]=15, jump 2 steps to the top. Total: 15.

The O(1) Space Optimisation

“Mentioning the connection to Fibonacci is a strong signal to the interviewer — it shows you recognise a pattern class, not just a one-off solution.”

Step 3: Talk Through the Code

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

        # O(1) space — two rolling variables replacing the full DP array
        prev2 = 0   # dp[i-2]: min cost to reach two steps back (base case dp[0] = 0)
        prev1 = 0   # dp[i-1]: min cost to reach one step back  (base case dp[1] = 0)

        # Compute dp[2] through dp[n] iteratively
        for i in range(2, n + 1):
            curr  = min(prev1 + cost[i - 1], prev2 + cost[i - 2])
            prev2 = prev1
            prev1 = curr

        return prev1   # prev1 now holds dp[n] — min cost to reach the top

Three things to narrate: (1) the loop runs from 2 to n inclusive — that’s n−1 iterations, producing dp[2] through dp[n]; (2) cost[i−1] and cost[i−2] access the cost of the step below — because we’re computing the cost to arrive at i, not to leave it; (3) the update order matters — prev2 = prev1 must happen before prev1 = curr, or you’d overwrite the value you still need.

Step 4: Highlight Edge Cases

Step 5: State Time and Space Complexity

“Time is O(n) regardless — one pass, one O(1) operation per step. The DP table approach uses O(n) space; the rolling-variable approach brings it down to O(1). The optimisation is possible because the recurrence only looks back two positions — like Fibonacci. If the recurrence looked back k positions, we’d need O(k) space.”

Interview Checklist for Min Cost Climbing Stairs

• Restate — “minimum cost to reach the virtual top step beyond the last index; can start free at step 0 or 1”
• Explain why naive recursion is exponential — overlapping subproblems
• State the DP definition clearly: dp[i] = min cost to arrive at step i (not including cost[i])
• Write the recurrence: dp[i] = min(dp[i−1] + cost[i−1], dp[i−2] + cost[i−2])
• Explain the base cases: dp[0] = dp[1] = 0 (free start positions)
• Frame the top as a virtual step at index n — unifies the final answer with the recurrence
• Show the full DP table for the example before introducing the space optimisation
• Explain the O(1) space optimisation — two rolling variables, same update-order caveat as Fibonacci
• Mention the Fibonacci connection — recurrence looking back a fixed number of steps collapses to rolling variables
• State O(n) time and O(1) space for the optimised solution

Related Problems to Practice

If you can explain Min Cost Climbing Stairs cleanly, try these next: