Climbing Stairs Problem in DSA Java - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the Climbing Stairs Problem?

It is a problem where you count how many distinct ways you can climb to the top of a staircase with n steps, taking either 1 or 2 steps at a time.

beginner

What is the base case for the Climbing Stairs Problem?

For 1 step, there is 1 way to climb. For 2 steps, there are 2 ways: (1+1) or (2).

intermediate

What is the relation between ways to climb n steps and smaller steps?

The ways to climb n steps = ways to climb (n-1) steps + ways to climb (n-2) steps, because you can reach step n from step n-1 or n-2.

intermediate

Why is the Climbing Stairs Problem similar to Fibonacci numbers?

Because the number of ways to climb n steps follows the same pattern as Fibonacci sequence where each number is the sum of the two before it.

intermediate

What is the time complexity of the dynamic programming solution for the Climbing Stairs Problem?

The time complexity is O(n) because we compute the number of ways for each step from 1 to n once.

How many ways are there to climb 3 steps if you can take 1 or 2 steps at a time?

A 3

B 2

C 4

D 5

The ways are: (1+1+1), (1+2), (2+1), so total 3 ways.

What is the recurrence relation used in the Climbing Stairs Problem?

A ways(n) = ways(n-1) + ways(n-2)

B ways(n) = ways(n-1) * ways(n-2)

C ways(n) = ways(n-1) - ways(n-2)

D ways(n) = ways(n-1) / ways(n-2)

The number of ways to reach step n is sum of ways to reach n-1 and n-2.

What is the number of ways to climb 1 step?

A 3

B 0

C 2

D 1

There is only 1 way to climb 1 step: a single step.

Which algorithmic technique is best suited for solving the Climbing Stairs Problem efficiently?

A Brute force recursion

B Dynamic programming

C Greedy algorithm

D Divide and conquer

Dynamic programming avoids repeated calculations by storing intermediate results.

What is the space complexity of the optimized dynamic programming solution using two variables?

A O(n)

B O(log n)

C O(1)

D O(n^2)

Using two variables to store previous results reduces space to constant O(1).

Explain how to solve the Climbing Stairs Problem using dynamic programming.

Think about how you can build the solution step by step from smaller problems.

Describe why the Climbing Stairs Problem is related to the Fibonacci sequence.

Focus on the pattern of how the number of ways grows.

Practice

1. What does the Climbing Stairs problem ask you to find?
Given n steps, how many distinct ways can you climb to the top if you can take 1 or 2 steps at a time?

easy

A. The maximum number of steps you can take at once

B. The minimum number of steps to reach the top

C. The total number of distinct ways to reach the top

D. The number of steps you skip

Solution

1. Step 1: Understand the problem statement

   The problem asks for the count of different ways to climb to the top using 1 or 2 steps at a time.

2. Step 2: Identify what is being counted

   We count all possible sequences of 1-step and 2-step moves that sum to n steps.

3. Final Answer:

   The total number of distinct ways to reach the top -> Option C

4. Quick Check:

   Counting ways = total distinct sequences [OK]

Hint: Count sequences of 1 or 2 steps summing to n [OK]

Common Mistakes:

• Confusing number of ways with minimum steps
• Thinking about maximum steps allowed
• Counting skipped steps instead of ways

2. Which of the following Java method signatures correctly represents a function to solve the Climbing Stairs problem for n steps?

easy

A. public int climbStairs(int n)

B. public void climbStairs(int n)

C. public int climbStairs()

D. public int climbStairs(String n)

Solution

1. Step 1: Check method return type and parameters

   The method should return an integer count and accept an integer input for steps.

2. Step 2: Validate each option

   public int climbStairs(int n) returns int and takes int n, which fits the problem. Others either have wrong return type or parameter type.

3. Final Answer:

   public int climbStairs(int n) -> Option A

4. Quick Check:

   Return int count, input int steps [OK]

Hint: Return int, input int n for steps [OK]

Common Mistakes:

• Using void return type instead of int
• Missing input parameter
• Using wrong parameter type like String

3. What is the output of the following Java code snippet?

public class Solution {
  public static int climbStairs(int n) {
    if (n <= 2) return n;
    int a = 1, b = 2, c = 0;
    for (int i = 3; i <= n; i++) {
      c = a + b;
      a = b;
      b = c;
    }
    return c;
  }
  public static void main(String[] args) {
    System.out.println(climbStairs(4));
  }
}

medium

A. 3

B. 7

C. 4

D. 5

Solution

1. Step 1: Trace the loop for n=4

   Initial a=1, b=2. Loop runs for i=3 and i=4.
   i=3: c=1+2=3, a=2, b=3
   i=4: c=2+3=5, a=3, b=5

2. Step 2: Return value after loop

   Return c=5, which is the number of ways to climb 4 steps.

3. Final Answer:

   5 -> Option D

4. Quick Check:

   Ways(4) = 5 [OK]

Hint: Use Fibonacci-like addition for steps > 2 [OK]

Common Mistakes:

• Returning b instead of c
• Off-by-one errors in loop
• Misunderstanding base cases

4. Identify the error in the following Java code for climbing stairs:

public int climbStairs(int n) {
  if (n == 1) return 1;
  int[] dp = new int[n];
  dp[0] = 1;
  dp[1] = 2;
  for (int i = 2; i <= n; i++) {
    dp[i] = dp[i-1] + dp[i-2];
  }
  return dp[n];
}

medium

A. Wrong return statement should be dp[n-1]

B. Array index out of bounds due to dp size and loop range

C. Incorrect base case for n == 1

D. Missing initialization of dp[2]

Solution

1. Step 1: Check array size and indexing

   Array dp size is n, but loop runs i from 2 to n inclusive, accessing dp[n], which is out of bounds.

2. Step 2: Identify correct loop range and return

   Loop should run i < n, and return dp[n-1] to avoid index error.

3. Final Answer:

   Array index out of bounds due to dp size and loop range -> Option B

4. Quick Check:

   Loop index exceeds array size [OK]

Hint: Array size n means max index n-1 [OK]

Common Mistakes:

• Looping to i <= n instead of i < n
• Returning dp[n] instead of dp[n-1]
• Incorrect base case handling

5. You want to find the number of ways to climb 10 steps but now you can take 1, 2, or 3 steps at a time. Which approach correctly adapts the Climbing Stairs problem?

hard

A. Use dp[i] = dp[i-1] + dp[i-2] + dp[i-3] with base dp[0]=1, dp[1]=1, dp[2]=2

B. Use dp[i] = dp[i-1] * dp[i-2] * dp[i-3] with base dp[0]=1

C. Use dp[i] = dp[i-1] + dp[i-2] only, ignoring 3 steps

D. Use recursion without memoization for all steps

Solution

1. Step 1: Understand the new step options

   Since you can take 1, 2, or 3 steps, ways to reach step i depend on previous 3 steps.

2. Step 2: Formulate dp relation and base cases

   dp[i] = dp[i-1] + dp[i-2] + dp[i-3]. Base cases: dp[0]=1 (one way to stand still), dp[1]=1, dp[2]=2.

3. Final Answer:

   Use dp[i] = dp[i-1] + dp[i-2] + dp[i-3] with base dp[0]=1, dp[1]=1, dp[2]=2 -> Option A

4. Quick Check:

   Include all step sizes in dp sum [OK]

Hint: Sum dp of last 3 steps for 1,2,3 step moves [OK]

Common Mistakes:

• Ignoring 3-step moves in dp formula
• Multiplying dp values instead of adding
• Using recursion without caching causing slow run