746. Min Cost Climbing Stairs

Description

You are given an integer array cost where cost[i] is the cost of i^th 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

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

💡 Hint 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.

💡 Hint 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]

Submitted Code

class Solution(object):
    def minCostClimbingStairs(self, cost):
        """
        :type cost: List[int]
        :rtype: int
        """
        n = len(cost)
        prev2, prev1 = cost[0], cost[1]   # 2계단 뒤, 1계단 뒤

        for i in range(2, n):             # 3번째 계단부터 시작
            curr = cost[i] + min(prev1, prev2)
            prev2, prev1 = prev1, curr
        
        return min(prev1, prev2)    # 마지막 cost[-1], cost[-2] 중 선택

Runtime: 1 ms | Beats 83.11%
Memory: 12.47 MB | Beats 85.11%

점화식과 변수 2개만을 이용하는 최적화된 방법이다. 현재 계단에 이르기 위해 2계단 뒤에서 2칸 올라오거나 1계단 뒤에서 1칸 올라오는 방법 중 더 최소 비용이 드는 쪽을 선택해야 한다.

Other Solutions

1st

class Solution:
    def minCostClimbingStairs(self, cost: List[int]) -> int:
		    if not cost:
            return 0

        dp = [0] * len(cost)
		    dp[0] = cost[0]
		
        if len(cost) >= 2:
            dp[1] = cost[1]
        
        for i in range(2, len(cost)):
            dp[i] = cost[i] + min(dp[i-1], dp[i-2])

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

time complexity: 𝑂(𝑛)
space complexity: 𝑂(𝑛)

i번째 계단에 도달하는 최소 비용을 dp 배열에 저장할 수도 있다.

2nd

class Solution:
    def minCostClimbingStairs(self, cost: List[int]) -> int:
        cost.append(0) # top [10,15,20,0]
        
        for i in range(len(cost)-4, -1, -1):
            cost[i] += min(cost[i+1], cost[i+2])
            
        return min(cost[0], cost[1])

뒤에서부터 앞으로 가는 방법으로, 원본 cost 배열을 바로 수정하기 때문에 공간 낭비를 줄일 수 있다.