746. Min Cost Climbing Stairs

1. Description

On a staircase, the i-th step has some non-negative cost cost[i] assigned (0 indexed).
Once you pay the cost, you can either climb one or two steps. You need to find minimum cost to reach the top of the floor, and you can either start from the step with index 0, or the step with index 1.

2. Example

Example 1:
Input: cost = [10, 15, 20]
Output: 15
Explanation: Cheapest is start on cost[1], pay that cost and go to the top.

Example 2:
Input: cost = [1, 100, 1, 1, 1, 100, 1, 1, 100, 1]
Output: 6
Explanation: Cheapest is start on cost[0], and only step on 1s, skipping cost[3].

3. Note

• cost will have a length in the range [2, 1000].
• Every cost[i] will be an integer in the range [0, 999].

4. Solutions

My Accepted Solution

n = i_cost.size()
Time complexity: O(n)
Space complexity: O(1)

previousTwoStairsCost[i] means the cost to climb the last i-th stair
dp = previousTwoStairsCost
$$ \begin{aligned} &dp[i] = \begin{cases} 0, & \text{i == 0, 1} \hspace{100cm}\\
min(dp[i - 2] + i\_cost[i - 2], dp[i - 1] + i\_cost[i - 1]), & \text{i > 1} \\
\end{cases} \end{aligned} $$

// we could only jump one step or two steps, so we just need the last two dp values
class Solution {
public:
    // int minCostClimbingStairs(vector<int>& cost)
    int minCostClimbingStairs(const vector<int>& i_cost) {
        vector<int> previousTwoStairsCost{0, 0};
        for (int i = 2; i <= i_cost.size(); ++i) {
            int cost =
                    min(previousTwoStairsCost[0] + i_cost[i - 2],
                        previousTwoStairsCost[1] + i_cost[i - 1]);

            previousTwoStairsCost[0] = previousTwoStairsCost[1];
            previousTwoStairsCost[1] = cost;
        }

        return previousTwoStairsCost.back();
    }
};