At each step, we can take 1 or 2 steps. The total ways to reach step n is the sum of ways to reach step n-1 and step n-2. We use memoization to avoid redundant computation.

1. Base cases: If n == 1, return 1. If n == 2, return 2.
2. Recursive step: Return climb(n-1) + climb(n-2).
3. Use a memo dictionary to cache results.

• Time Complexity: O(N) with memoization.
• Space Complexity: O(N) (memo + recursion stack).

def climb_stairs(n, memo={}):
    if n == 1:
        return 1
    if n == 2:
        return 2
    if n in memo:
        return memo[n]
    
    memo[n] = climb_stairs(n - 1, memo) + climb_stairs(n - 2, memo)
    return memo[n]