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Climbing Stairs

Certification: Basic Level Accuracy: 100% Submissions: 2 Points: 5

Write a Java program to solve the climbing stairs problem. You are climbing a staircase. It takes n steps to reach the top. Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?

Example 1
     
  • Input: n = 2
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  • Output: 2
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  • Explanation: 
       
    • Step 1: Identify that this is a dynamic programming problem where each step depends on previous steps.
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    • Step 2: For n = 2, we can either take two 1-step moves or one 2-step move.
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    • Step 3: The two distinct ways are: [1+1] and [2].
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    • Step 4: Return 2 as the total number of ways to climb the staircase.
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    •
Example 2
     
  • Input: n = 3
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  • Output: 3
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  • Explanation: 
       
    • Step 1: Apply the same dynamic programming approach for n = 3.
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    • Step 2: For n = 3, we can take three 1-step moves, or one 1-step and one 2-step move in different orders.
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    • Step 3: The three distinct ways are: [1+1+1], [1+2], and [2+1].
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    • Step 4: Return 3 as the total number of ways to climb the staircase.
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    •
Constraints
     
  • 1 ≤ n ≤ 45
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  • You must solve this efficiently to handle large inputs
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  • Time Complexity: O(n)
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  • Space Complexity: O(1)
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Hint
Solution

Solution Hints

    
  
  • Notice that this problem follows the Fibonacci sequence pattern
    • 
  
  • The number of ways to reach step n is the sum of ways to reach step n-1 and step n-2
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  • Use dynamic programming to avoid redundant calculations
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  • You only need to keep track of the previous two values, not the entire array
    • 
  
  • Initialize base cases for steps 1 and 2, then build up to n
    •

Steps to solve by this approach:

 Step 1: Handle base cases - if n is 1 or 2, return n (1 way for 1 step, 2 ways for 2 steps).
 Step 2: Initialize variables to keep track of the number of ways for the previous two steps: prev1 = 1 (for 1 step) and prev2 = 2 (for 2 steps).
 Step 3: Use a loop to calculate the number of ways for each step from 3 to n.
 Step 4: For each step i, the number of ways is the sum of ways for step i-1 and step i-2.
 Step 5: Update prev1 and prev2 for the next iteration.
 Step 6: After completing the loop, prev2 contains the number of ways to climb n steps.
 Step 7: Time complexity is O(n) and space complexity is O(1) since we only use a constant amount of extra space.

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