Climbing Stairs

Clarify the problem:

• The problem requires finding the number of distinct ways to climb to the top of a staircase with n steps.
• Each time, we can either climb one or two steps.
• We need to implement a function that takes an integer n as input and returns the number of distinct ways to climb to the top.

Analyze the problem:

• Input: An integer n representing the number of steps in the staircase.
• Output: An integer representing the number of distinct ways to climb to the top.
• Constraints:
  • n is a non-negative integer.
  • The problem asks for the number of distinct ways, so we need to count all possible combinations.

Design an algorithm:

• We can solve this problem using dynamic programming.
• We initialize an array dp of size n+1 to store the number of distinct ways to reach each step.
• We set dp[0] = 1 to handle the base case where there are no steps to climb.
• We set dp[1] = 1 to handle the case where there is only one step to climb.
• For each step i from 2 to n, we calculate the number of distinct ways to reach step i by summing up the number of ways to reach the previous two steps: dp[i] = dp[i-1] + dp[i-2].
• Finally, we return dp[n], which represents the number of distinct ways to climb to the top.

Explain your approach:

• We will implement a function called climbStairs that takes an integer n as input.
• We initialize an array dp of size n+1 and set dp[0] = 1 and dp[1] = 1.
• We iterate over the range from 2 to n and calculate dp[i] = dp[i-1] + dp[i-2].
• Finally, we return dp[n] as the number of distinct ways to climb to the top.

Write clean and readable code:

python

def climbStairs(n):
    dp = [0] * (n + 1)
    dp[0] = 1
    dp[1] = 1

    for i in range(2, n + 1):
        dp[i] = dp[i - 1] + dp[i - 2]

    return dp[n]

Test your code:

python

# Test case 1: Example case
n = 2
# There are two ways to climb to the top: [1, 1] and [2].
print(climbStairs(n))  # Output: 2

# Test case 2: Small input
n = 3
# There are three ways to climb to the top: [1, 1, 1], [1, 2], and [2, 1].
print(climbStairs(n))  # Output: 3

# Test case 3: Large input
n = 10
# There are 89 ways to climb to the top.
print(climbStairs(n))  # Output: 89

# Test case 4: Base case
n = 0
# There is one way to climb to the top, which is not to take any steps.
print(climbStairs(n))  # Output: 1

# Test case 5: Edge case
n = 1
# There is one way to climb to the top, which is to take one step.
print(climbStairs(n))  # Output: 1

Optimize if necessary:

• The dynamic programming solution has a time complexity of O(n) since we iterate from 2 to n once.
• The space complexity is O(n) since we use an array of size n+1 to store the intermediate results.

Handle error cases:

• We need to consider the case where the input n is negative. In such cases, we can return 0 or raise an error, depending on the problem's requirements.

Discuss complexity analysis:

• Let n be the input number of steps.
• The time complexity of the solution is O(n) since we iterate from 2 to n once.
• The space complexity is O(n) since we use an array of size n+1 to store the intermediate results.