readme.md

746. Min Cost Climbing Stairs

You are given an integer array cost where cost[i] is the cost of ith step on a staircase. Once you pay the cost, you can either climb one or two steps.

You can either start from the step with index 0, or the step with index 1.

Return the minimum cost to reach the top of the floor.

Example 1:

Input: cost = [10,15,20] Output: 15 Explanation: You will start at index 1.

• Pay 15 and climb two steps to reach the top. The total cost is 15.

Example 2:

Input: cost = [1,100,1,1,1,100,1,1,100,1] Output: 6 Explanation: You will start at index 0.

• Pay 1 and climb two steps to reach index 2.
• Pay 1 and climb two steps to reach index 4.
• Pay 1 and climb two steps to reach index 6.
• Pay 1 and climb one step to reach index 7.
• Pay 1 and climb two steps to reach index 9.
• Pay 1 and climb one step to reach the top. The total cost is 6.

Constraints:

• 2 <= cost.length <= 1000
• 0 <= cost[i] <= 999

Solution

class Solution:
    def minCostClimbingStairs(self, cost: List[int]) -> int:
        n = len(cost)
        dp = [0] * n
        dp[0], dp[1] = cost[0], cost[1]

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

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

Thoughts

Time Complexity

The time complexity is O(n), where n is the length of the cost array, as we need to iterate through the array once to fill the dp array.

Space Complexity

The space complexity is O(n) for the dp array. This can be reduced to O(1) by using two variables to store the costs of the last two steps instead of using an array.