[NeetCode 150] Min Cost Climbing Stairs

Min Cost Climbing Stairs

You are given an array of integers cost where cost[i] is the cost of taking a step from the ith floor of a staircase. After paying the cost, you can step to either the (i + 1)th floor or the (i + 2)th floor.

You may choose to start at the index 0 or the index 1 floor.

Return the minimum cost to reach the top of the staircase, i.e. just past the last index in cost.

Example 1:

Input: cost = [1,2,3]

Output: 2

Explanation: We can start at index = 1 and pay the cost of cost[1] = 2 and take two steps to reach the top. The total cost is 2.

Example 2:

Input: cost = [1,2,1,2,1,1,1]

Output: 4

Explanation: Start at index = 0.

Pay the cost of cost[0] = 1 and take two steps to reach index = 2.
Pay the cost of cost[2] = 1 and take two steps to reach index = 4.
Pay the cost of cost[4] = 1 and take two steps to reach index = 6.
Pay the cost of cost[6] = 1 and take one step to reach the top.
The total cost is 4.

Constraints:

2 <= cost.length <= 100
0 <= cost[i] <= 100

Solution

A classic dp problem. Let $dp[i]$ represents the optimal cost spent to climb to stair $i$, the transfer function is
$$dp[i]=\min (dp[j-1]+cost[j-1],dp[j-2]+cost[j-2])$$
Notice that we should initialize a list with $cost.length+1$ length, where the last value represents the optimal cost we spent to stand on the top.

Code

class Solution:
    def minCostClimbingStairs(self, cost: List[int]) -> int:
        dp = [0]*(len(cost)+1)
        for i in range(2, len(dp)):
            dp[i] = min(dp[i-1]+cost[i-1], dp[i-2]+cost[i-2])
        return dp[-1]