Why Climbing Stairs Problem in DSA Java?

The Big Idea

What if you could instantly know all the ways to climb a huge staircase without trying each path?

The Scenario

Imagine you want to climb a staircase with many steps, but you can only take one or two steps at a time. Counting all the ways to reach the top by hand is like trying to list every possible path without missing any.

The Problem

Manually counting each possible way is slow and confusing. It's easy to forget some paths or count some twice. As the number of steps grows, the task becomes impossible to do quickly or correctly.

The Solution

The Climbing Stairs Problem uses a simple rule: the ways to reach step n equals the ways to reach step n-1 plus the ways to reach step n-2. This rule helps us build the answer step by step without missing or repeating any paths.

Before vs After

✗ Before

int countWays(int steps) {
  // Try to list all paths manually
  // Very long and error-prone
  return 0;
}

✓ After

int countWays(int steps) {
  if (steps <= 2) return steps;
  int first = 1, second = 2;
  for (int i = 3; i <= steps; i++) {
    int current = first + second;
    first = second;
    second = current;
  }
  return second;
}

What It Enables

This problem's solution lets us quickly find how many ways to climb any number of steps, even very large staircases, without listing all paths.

Real Life Example

Planning different ways to climb stairs in a building or designing a game where a character can jump one or two steps at a time.

Key Takeaways

Manual counting is slow and error-prone for many steps.

The problem uses a simple addition rule to count ways efficiently.

This approach scales well for large numbers of steps.

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