746. Min Cost Climbing Stairs

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.

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].

Note:

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

Solution

class Solution:
    def minCostClimbingStairs(self, cost: List[int]) -> int:
        table = [0] * len(cost)

        table[0] = cost[0]
        table[1] = cost[1]

        for i in range(2, len(cost)):
            table[i] = cost[i] + min(table[i - 1], table[i - 2])

        return min(table[-1], table[-2])

This problem might originally seem complicated, or like it requires a method like recursive backtracking. In actuality, you can solve the problem very simply with dynamic programming. We use a bottom up approach, starting at the beginning of the “stairs” and working our way up.

We take the first two values and manually put them into our table. Then, to populate the rest of the table, we follow this algorithm:

- For each "stair" in the staircase
    - Add the cost from the table
    - Compare the costs of the stairs 1 and 2 positions below current
        - We want the lowest possible score, so add whichever stair has a lower value
- Finally, we compare the costs of the last two steps.
- Whichever one of those is cheaper is the one that we return.

Try it on Leetcode