How do you solve Climbing Stairs II on LeetCode?

Climbing Stairs II (LeetCode #3693) 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 reduces redundant calculations.

What is the time complexity of Climbing Stairs II?

The optimal solution for Climbing Stairs II runs in O(n) time and uses O(n) space. We compute each step's cost once, leading to linear time complexity.

What is the brute force solution for Climbing Stairs II?

The brute force approach for Climbing Stairs II is Brute Force, with O(n²) time complexity and O(1) space complexity. We can explore all possible paths to reach the last step by recursively calculating costs for each jump. This method is straightforward but inefficient due to overlapping subproblems. The optimal Optimal Solution approach improves this to O(n).

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

Climbing Stairs II uses the following concepts: Array, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Greedy Algorithms.

What are the interview tips for Climbing Stairs II?

Explain your thought process clearly. Discuss edge cases upfront. Practice similar dynamic programming problems.

What are common mistakes when solving Climbing Stairs II?

Not considering all jump options. Forgetting to initialize dp array correctly.

#3693

Climbing Stairs II

Medium

ArrayDynamic ProgrammingDynamic ProgrammingGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

Using dynamic programming, we can store the minimum cost to reach each step, building on previously computed results. This avoids redundant calculations and significantly reduces time complexity.

⚙️

Algorithm

3 steps

1Step 1: Initialize a dp array of size n+1 with dp[0] = 0 and others as infinity.
2Step 2: Iterate through each step i from 0 to n, updating dp[j] for j = i+1, i+2, i+3 based on the cost formula.
3Step 3: Return dp[n] as the minimum cost to reach the last step.

solution.py8 lines

1def minCost(n, costs):
2    dp = [float('inf')] * (n + 1)
3    dp[0] = 0
4    for i in range(n):
5        for j in range(1, 4):
6            if i + j <= n:
7                dp[i + j] = min(dp[i + j], costs[i] + j * j + dp[i])
8    return dp[n]

ℹ

Complexity note: We compute each step's cost once, leading to linear time complexity.

1Dynamic programming reduces redundant calculations.
2Understanding cost transitions is crucial.

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