What is the time complexity of Min Cost Climbing Stairs?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Min Cost Climbing Stairs?

This problem uses Array, Dynamic Programming. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Min Cost Climbing Stairs?

Insight 1: Dynamic Programming is well-suited because the optimal solution for reaching a step depends on the optimal solutions for the previous one or two steps. Insight 2: The problem has overlapping subproblems. The minimum cost to reach step `i` is influenced by the minimum costs to reach steps `i-1` and `i-2`.

Which companies ask LeetCode 746. Min Cost Climbing Stairs in interviews?

This problem is frequently asked at Accenture.

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Min Cost Climbing Stairs - LeetCode 746 Solution

Min Cost Climbing Stairs - Complete Solution Guide

Min Cost Climbing Stairs is LeetCode problem 746, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given an integer array cost where cost[i] is the cost of i th step on a staircase. Once you pay the cost, you can either climb one or two steps. You can either start from the step with index 0 , or the step with index 1 . Return the minimum cost to reach the top of the floor . Example 1: Input: cost = [10, 15 ,20] Output: 15 Explanation: You will start at index 1. - Pay 15 and climb two steps to reach the top. The total cost is 15. Example 2: Input: cost = [ 1 ,100, 1 ,1, 1 ,100, 1 , 1 ,

Detailed Explanation

The problem asks us to find the minimum cost to reach the top of a staircase, given an array `cost` where `cost[i]` represents the cost of the i-th step. We can start at either step 0 or step 1. After paying the cost of a step, we can climb one or two steps forward. The goal is to find the path that minimizes the total cost to reach the 'top', which can be thought of as reaching beyond the last step of the `cost` array.

Solution Approach

The provided solution uses dynamic programming to calculate the minimum cost to reach each step of the staircase. A DP array `dp` of size `n+1` is created, where `dp[i]` represents the minimum cost to reach the i-th step (or 'top' if i=n). The base cases are `dp[0] = 0` and `dp[1] = 0`, because we can start from step 0 or step 1 without paying any cost to *reach* those steps. The solution then iterates from step 2 up to step `n` (inclusive), calculating `dp[i]` as the minimum of: (1) the cost to reach the previous step (`dp[i-1]`) plus the cost of the previous step (`cost[i-1]`), and (2) the cost to reach the step two steps before (`dp[i-2]`) plus the cost of that step (`cost[i-2]`). Finally, the solution returns `dp[n]`, which represents the minimum cost to reach the top of the stairs.

Step-by-Step Algorithm

Step 1: Initialize a DP array `dp` of size `n+1` with all elements set to 0, where `n` is the length of the `cost` array. `dp[i]` represents the minimum cost to *reach* step `i`.
Step 2: Set `dp[0] = 0` and `dp[1] = 0`, as it costs nothing to reach the first and second starting points.
Step 3: Iterate from `i = 2` to `n`: `dp[i] = min(dp[i - 1] + cost[i - 1], dp[i - 2] + cost[i - 2])`. This computes the minimum cost to reach the i-th step by either coming from the (i-1)-th step or the (i-2)-th step.
Step 4: Return `dp[n]`, the minimum cost to reach the top of the staircase.

Key Insights

Insight 1: Dynamic Programming is well-suited because the optimal solution for reaching a step depends on the optimal solutions for the previous one or two steps.
Insight 2: The problem has overlapping subproblems. The minimum cost to reach step `i` is influenced by the minimum costs to reach steps `i-1` and `i-2`.
Insight 3: We can define dp[i] as the minimum cost to reach step `i`. The top of the staircase is considered the (n+1)-th step, where n is the size of `cost` array. Hence, we return dp[n].

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Dynamic Programming.

Companies

Asked at: Accenture.

Min Cost Climbing Stairs | LeetCode 746

Easy

Array Dynamic Programming

Time: O(n)Space: O(n)

Solution Code

Available in:

class Solution {
    public int minCostClimbingStairs(int[] cost) {
        int n = cost.length;
        int[] dp = new int[n + 1];
        dp[0] = 0;
        dp[1] = 0;
        for (int i = 2; i <= n; i++) {
            dp[i] = Math.min(dp[i - 1] + cost[i - 1], dp[i - 2] + cost[i - 2]);
        }
        return dp[n];
    }
}

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Problem Info

Description

Problem 746: Min Cost Climbing Stairs - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming

Companies

Accenture

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:746

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

4 steps

1
Step 1: Initialize a DP array `dp` of size `n+1` with all elements set to 0, where `n` is the length of the `cost` array. `dp[i]` represents the minimum cost to *reach* step `i`.
2
Step 2: Set `dp[0] = 0` and `dp[1] = 0`, as it costs nothing to reach the first and second starting points.
3
Step 3: Iterate from `i = 2` to `n`: `dp[i] = min(dp[i - 1] + cost[i - 1], dp[i - 2] + cost[i - 2])`. This computes the minimum cost to reach the i-th step by either coming from the (i-1)-th step or the (i-2)-th step.
4
Step 4: Return `dp[n]`, the minimum cost to reach the top of the staircase.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is O(n) because the for loop iterates n-1 times. The space complexity is O(n) because we store n+1 intermediate results in the `dp` array.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Space Optimized Dynamic Programming

Instead of using an array of size n+1, we can use only two variables to store the minimum cost for the previous two steps. This eliminates the need for the `dp` array.

Time: O(n)Space: O(1)

Trade-offs: The space-optimized version is better when memory usage is a concern. The original DP version might be easier to understand initially for some learners.

✨

Key Insights

1Insight 1: Dynamic Programming is well-suited because the optimal solution for reaching a step depends on the optimal solutions for the previous one or two steps.
2Insight 2: The problem has overlapping subproblems. The minimum cost to reach step `i` is influenced by the minimum costs to reach steps `i-1` and `i-2`.
3Insight 3: We can define dp[i] as the minimum cost to reach step `i`. The top of the staircase is considered the (n+1)-th step, where n is the size of `cost` array. Hence, we return dp[n].

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: `cost` array with only two elements. The loop does not execute, and `dp[2]` is implicitly equal to `min(dp[1]+cost[1], dp[0]+cost[0])`, the logic is handled correctly.
Edge case 2: `cost` array with all elements being 0. The algorithm will still correctly compute the minimum cost as 0.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to initialize `dp[0]` and `dp[1]` correctly. The problem states that you can start at index 0 or index 1. Therefore, the minimum cost to reach step 0 or 1 is 0.

✗Common mistake 2: Incorrectly calculating the index when accessing the `cost` array within the loop. Remember that `cost[i-1]` refers to the cost of the (i-1)-th step.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Resource allocation - minimizing the cost of a project by choosing the cheapest sequence of tasks where each task has a cost and completing a task allows you to perform the next one or the one after it.

🏦Real-world scenario 2: Pathfinding in a graph where nodes have costs associated with them. The goal is to find the cheapest path from a starting node to a destination node.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Don't rush. Explain your thought process clearly. Easy problems are often about communication.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: June 27, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 13

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Min Cost Climbing Stairs"?

For this easy-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?