70. Climbing Stairs

You are climbing a stair case. It takes n steps to reach to the top.

Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?

Note: Given n will be a positive integer.

Example 1:

Input: 2
Output: 2
Explanation: There are two ways to climb to the top.
1. 1 step + 1 step
2. 2 steps

Example 2:

Input: 3
Output: 3
Explanation: There are three ways to climb to the top.
1. 1 step + 1 step + 1 step
2. 1 step + 2 steps
3. 2 steps + 1 step

// Top-Down DP with Memoization
int helper(int n, vector<int>& memo) {
    if (n == 0) return 1;
    if (n <= 2) return n;
    if (memo[n] != -1) return memo[n];
    memo[n] = helper(n - 1, memo) + helper(n - 2, memo);
    return memo[n];
}
int climbStairs(int n) { // time: O(n); space: O(n)
    vector<int> memo(n + 1, -1);
    return helper(n, memo);
}

// Dynamic Programming
int climbStairs(int n) { // time: O(n); space: O(n)
    if (n <= 1) return n;
    vector<int> dp(n, 0);
    dp[0] = 1;
    dp[1] = 2;
    for (int i = 2; i < n; ++i) {
        dp[i] = dp[i - 1] + dp[i - 2];
    }
    return dp.back();
}

// Space Optimized Dynamic Programming
int climbStairs(int n) { // time: O(n); space: O(1)
    if (n <= 1) return n;
    int two_step = 1, one_step = 2, res = 2;
    for (int i = 2; i < n; ++i) {
        res = two_step + one_step;
        two_step = one_step;
        one_step = res;
    }
    return res;
}