Step by Step – The DP Way Up!

Climbing Stairs – One Step at a Time

Today’s challenge seems simple but teaches powerful fundamentals of dynamic programming.

Problem:
You are climbing a staircase. It takes n steps to reach the top.
Each time you can climb either 1 or 2 steps.
How many unique ways can you reach the top?

Let’s explore from brute force to highly optimized solutions!

Best Data Structure for Solving It

No complex data structures required!

  • In memoization, we use a HashMap to cache computed results and avoid recalculating.

  • In tabulation, only two variables are needed — super efficient in terms of space.

Different Approaches – Brute Force to Optimized Solutions

1. Brute Force (Recursive without Memoization)

Try all possible combinations recursively. Simple but inefficient.


public int climbStairsBrute(int n) {
    if (n == 0 || n == 1) return 1;
    return climbStairsBrute(n - 1) + climbStairsBrute(n - 2);
}

2. Memoization (Top-Down Dynamic Programming)

Store the results of subproblems using a HashMap. Much faster!


public int climbStairs(int n) {
    Map<Integer, Integer> memo = new HashMap<>();
    return climb(n, memo);
}

private int climb(int n, Map<Integer, Integer> memo) {
    if (n == 0 || n == 1) return 1;
    if (!memo.containsKey(n)) {
        memo.put(n, climb(n - 1, memo) + climb(n - 2, memo));
    }
    return memo.get(n);
}

3. Tabulation (Bottom-Up Dynamic Programming)

Iteratively build the answer using just two variables.


public int climbStairsTab(int n) {
    if (n == 0 || n == 1) return 1;
    int first = 1, second = 1;
    for (int i = 2; i <= n; i++) {
        int temp = first + second;
        first = second;
        second = temp;
    }
    return second;
}

⚠️ Common Mistakes & Pitfalls

  • Forgetting to handle base cases n = 0 or n = 1.

  • Using brute force for large n, which leads to slow or crashing code.

  • Wasting extra space when an O(1) solution works just fine.

📊 Time & Space Complexity Summary

  • Brute Force: Time → O(2^n), Space → O(n)-Stack

  • Memoization: Time → O(n), Space → O(n)-HashMap

  • Tabulation: Time → O(n), Space → O(1)

🌍 Real-World Applications

  • Pathfinding problems in robotics and video games

  • Solving similar step or coin change problems in combinatorics

  • Strengthening fundamentals in recursion, DP, and space optimization — vital in tech interviews and real systems

Even simple problems like this can help build rock-solid intuition for dynamic programming.

I hope you found this article insightful! Keep exploring, keep learning, and don’t forget to check back daily for more exciting problem breakdowns. If you know someone who would benefit from this, share it with them and help them grow too! That’s it for today—see you in the next post!

                                                                                                                                         Signing off!! 

                                                                                                Master DSA One Problem at a Time :)



Comments

Post a Comment

Popular posts from this blog

Trailing Zeros in Factorial – How Many and Why?

Trailing Zeros – Counting Zeros in Factorial Factorial-related problems are a fundamental part of Data Structures and Algorithms (DSA). One such classic problem involves counting the number of trailing zeros in N! . In this article, we will explore different approaches to solving this problem, optimize our solution, and examine its real-world applications. Optimal Data Structure for Solving the Problem For this problem, no additional data structures are required. Since factorials grow exponentially, storing the actual factorial value is impractical. Instead, we leverage mathematical properties and division operations to efficiently determine the number of trailing zeros. This approach ensures an optimal solution in terms of both time and space complexity. Approach – From Brute Force to an Optimized Solution Brute Force Approach – Compute and Count A naive approach involves computing the factorial of N and counting the number of trailing zeros. However, this method is highly ineff...
Read more

Divisible Drama — Who’s In, Who’s Out?

 Divisible Drama — Who’s In, Who’s Out? Problem Statement You’re given two positive integers n and m . num1 = sum of all numbers from 1 to n not divisible by m num2 = sum of all numbers from 1 to n divisible by m Your mission, should you choose to accept it: Return num1 - num2 Example Input: n = 10 , m = 3 Output: 19 Breakdown: Not divisible by 3 → [ 1 , 2 , 4 , 5 , 7 , 8 , 10 ] → num1 = 37 Divisible by 3 → [ 3 , 6 , 9 ] → num2 = 18 Answer = 37 - 18 = 19 Best Data Structure for Solving the Problem This one’s a math.  Why overthink it when Gauss already gave us the formula for sum of 1 to n? Different Approaches (from Brute Force to Optimized) Approach 1: Brute Force (aka List & Sum Vibes) Loop from 1 to n , split the nums based on whether they’re divisible by m , and sum them up. class Solution { public int differenceOfSums ( int n, int m) { int num1 = 0 , num2 = 0 ; for ( int i = 1 ; i <= n; i++)...
Read more

Zero Array After Queries

Zero Array After Queries You are given an integer array nums and a list of queries, where each query is a pair [li, ri] representing a range. Each query allows you to decrement any subset of elements in the range [li, ri] by 1. Your task is to determine if it is possible to make every element in nums equal to zero after applying all queries in order . Example Input: nums = [ 1 , 0 , 1 ] queries = [[0, 2]] Output: true Explanation: We can select indices 0 and 2 and decrement both by 1 to obtain [0, 0, 0] . Best Data Structure for the Job We are working with multiple range updates and need to evaluate coverage over indices. The best choice here is the difference array combined with a prefix sum . It allows efficient range additions and helps track how many times each index is affected by the queries. Different Approaches Approach 1: Brute Force This naive method iterates over every index for every query and performs the decrement operations directly. class Solution {...
Read more