This is essentially the Fibonacci sequence. dp[i] represents the number of ways to reach step i. We use bottom-up tabulation with O(1) space by only keeping the last two values.

1. Base cases: dp[1] = 1, dp[2] = 2.
2. For each step from 3 to n: dp[i] = dp[i-1] + dp[i-2].
3. Optimize space by keeping only two variables.

• Time Complexity: O(N).
• Space Complexity: O(1).

def climb_stairs(n):
    if n <= 2:
        return n
    prev2, prev1 = 1, 2
    for i in range(3, n + 1):
        current = prev1 + prev2
        prev2 = prev1
        prev1 = current
    return prev1