Count ways to N'th Stair(Order does not matter)

Slug: count-ways-to-nth-stairorder-does-not-matter5639 Difficulty: Medium Id: 704431 Topic Tags: Dynamic Programming, Mathematical, Algorithms Company Tags: Accolite, Amazon

Summary

We are asked to find the number of ways to climb N stairs, where each step can be either one or two units. The order of climbing does not matter. This problem is a classic example of dynamic programming and mathematical optimization.

Detailed Explanation

The key insight here is that the number of ways to climb N stairs depends on the last step taken. If we take one step, then the remaining steps are equivalent to climbing N-1 stairs. If we take two steps, then the remaining steps are equivalent to climbing N-2 stairs. This recursive relationship allows us to build a dynamic programming solution.

Here's a step-by-step breakdown of the solution:

• Create an array dp of size N+1, where dp[i] represents the number of ways to climb i stairs.
• Initialize dp[0] = 1, since there is one way to climb 0 stairs (i.e., not climbing at all).
• For each step from 1 to N, calculate dp[i] as follows:
  • If we take one step to reach the ith stair, then the remaining steps are equivalent to climbing i-1 stairs. Therefore, add dp[i-1] to the current value.
  • If we take two steps to reach the ith stair, then the remaining steps are equivalent to climbing i-2 stairs. Therefore, add dp[i-2] to the current value.
• The final answer is dp[N], which represents the number of ways to climb N stairs.

Here's a diagram illustrating this process:

  0: 1
  1: 1 + dp[0]
  2: dp[1] + dp[0]
  ...
  N: dp[N-1] + dp[N-2]

Time complexity: O(N) Space complexity: O(N)

Optimized Solutions

Java