746.Min_Cost_Climbing_Stairs(Easy).md

746. Min Cost Climbing Stairs (Easy)

Date and Time: Aug 22, 2024, 0:22 (EST)

Link: https://leetcode.com/problems/min-cost-climbing-stairs/

Question:

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

Walk-through:

Follow the hints:

1. Build an array dp where dp[i] is the minimum cost to climb to the top starting from the ith staircase.

2. Assuming we have n staircase labeled from 0 to n - 1 and assuming the top is n, then dp[n] = 0, marking that if you are at the top, the cost is 0.

3. Now, looping from n - 1 to 0, the dp[i] = cost[i] + min(dp[i + 1], dp[i + 2]). The answer will be the minimum of dp[0] and dp[1]

Python Solution:

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

Time Complexity: $O(n)$
Space Complexity: $O(n)$