How do you solve Min Cost Climbing Stairs on LeetCode?

Min Cost Climbing Stairs (LeetCode #746) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Dynamic programming is a powerful technique for optimizing recursive solutions by storing intermediate results.

What is the brute force solution for Min Cost Climbing Stairs?

The brute force approach for Min Cost Climbing Stairs is Brute Force, with O(n²) time complexity and O(n) space complexity. The brute force approach involves recursively calculating the minimum cost to reach the top from each step. This means exploring all possible paths, which can be inefficient but is straightforward to understand. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Min Cost Climbing Stairs?

Min Cost Climbing Stairs uses the following concepts: Array, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Array.

What are the interview tips for Min Cost Climbing Stairs?

Always clarify the problem constraints and edge cases before diving into coding. Explain your thought process and approach before writing code to ensure clarity. Practice dry-running your code with sample inputs to catch potential errors early.

What are common mistakes when solving Min Cost Climbing Stairs?

Forgetting to handle the base cases in recursive solutions. Not considering the minimum cost from both possible previous steps.

#746

Min Cost Climbing Stairs

Easy

ArrayDynamic ProgrammingDynamic ProgrammingArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(n)	O(n)

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Intuition

Time O(n)Space O(n)

The optimal solution uses dynamic programming to build up the minimum costs from the bottom of the stairs to the top, ensuring that we only calculate each step once. This is efficient and avoids redundant calculations.

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Algorithm

4 steps

1Step 1: Create a dp array where dp[i] represents the minimum cost to reach step i.
2Step 2: Initialize dp[0] and dp[1] with the costs of the first two steps.
3Step 3: Iterate from step 2 to the last step, calculating dp[i] as the cost of step i plus the minimum of dp[i-1] and dp[i-2].
4Step 4: The result is the minimum of dp[n-1] and dp[n-2], where n is the length of the cost array.

solution.py9 lines

1# Full working Python code
2
3def minCost(cost):
4    n = len(cost)
5    dp = [0] * (n + 1)
6    dp[0], dp[1] = cost[0], cost[1]
7    for i in range(2, n):
8        dp[i] = cost[i] + min(dp[i - 1], dp[i - 2])
9    return min(dp[n - 1], dp[n - 2])

ℹ

Complexity note: The time complexity is O(n) because we only loop through the cost array once. The space complexity is O(n) due to the dp array storing the minimum costs.

1Dynamic programming is a powerful technique for optimizing recursive solutions by storing intermediate results.
2Understanding the problem as a pathfinding issue helps in visualizing the solution.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.