Skip to content

746. Min Cost Climbing Stairs

Easy

Description

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

Solutions

Approach: Functional Relationship

Time complexity: \(O(n)\)

Space complexity: \(O(n)\)

Way 1:

Top Down DP via inbuilt library

We define f(i) as the minimum cost required to reach the i-th step,

Initially f(0) = f(1) = 0. The answer is f(n).

When i ≥ 2, we can directly reach the ith step from the (i - 1)th step using 1 step, or reach the ith step from the (i - 2)th step using 2 steps. Therefore, we have the state transition equation: f(i) = min(f(i - 1) + cost[i - 1], f(i - 2) + cost[i - 2])

The final answer is f(n).

The time complexity is O(n), and the space complexity is O(n). Here, n is the length of the cost array.

class Solution:
    def minCostClimbingStairs(self, cost: List[int]) -> int:
        n = len(cost)
        @cache
        def f(i):
            if i==0:
                return 0
            if i==1:
                return 0
            return min(cost[i-1]+f(i-1), cost[i-2]+f(i-2))
        return f(n)
Top Down DP(Recursion+Memoization) 
class Solution:
    def minCostClimbingStairs(self, cost: List[int]) -> int:
        n = len(cost)
        def f(i,nb={}):
            if i==0:
                return 0
            if i==1:
                return 0

            if i in nb:
                return nb[i]

            nb[i] = min(cost[i-1]+f(i-1), cost[i-2]+f(i-2))

            return nb[i]

        return f(n)

Way 2:

Time complexity: O(n)

Space complexity: O(1)

No-memory DP

  • We notice that f(i) in the state transition equation is only related to f(i — 1) and f(i - 2)
  • Therefore, we can use two variables f0 and f1 to alternately record the values of f(i - 2) and f(i - 1), which optimizes the space complexity to O(1).
class Solution:
    def minCostClimbingStairs(self, cost: List[int]) -> int:
        #Inital states
        n = len(cost)
        f0 , f1 = 0 , 0
        for i in range(2,n+1):
            fi = min(f1+cost[i-1], f0+cost[i-2])
            f0 , f1 = f1 , fi
        return fi
Bottom Up DP- Tabulation 
class Solution:
    def minCostClimbingStairs(self, cost: List[int]) -> int:
        #Inital states
        n = len(cost)
        dp = [0,0]+[0]*(n-1)

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

        return dp[n]