On a staircase, the i-th step has some non-negative cost cost[i] assigned (0 indexed).
Once you pay the cost, you can either climb one or two steps.
You need to find minimum cost to reach the top of the floor,
and you can either start from the step with index 0, or the step with index 1.
Example 1
Input: cost = [10, 15, 20]
Output: 15
Explanation: Cheapest is start on cost[1], pay that cost and go to the top.
Example 2
Input: cost = [1, 100, 1, 1, 1, 100, 1, 1, 100, 1]
Output: 6
Explanation: Cheapest is start on cost[0], and only step on 1s, skipping cost[3].
Idea (dynamic programming)
For each starting point, there are two options:
climb one step
climb two step
Minimum cost is the min of these two options.
Solution 1: Recursion
Time: O(2n)
Space: O(1)
class Solution(object):
def minCostClimbingStairs(self, cost):
"""
:type cost: List[int]
:rtype: int
"""
def helper(index, cost):
if index > len(cost)-1:
total = 0
else:
total = cost[index] + min(helper(index+1, cost), helper(index+2, cost))
return total
return min(helper(0, cost), helper(1, cost))
Solution 2: Recursion with memoization
Time: O(n)
Space: O(n)
class Solution(object):
def minCostClimbingStairs(self, cost):
"""
:type cost: List[int]
:rtype: int
"""
hashmap = {}
def helper(index, cost):
if hashmap.get(index) == None:
if index > len(cost)-1:
total = 0
else:
total = cost[index] + min(helper(index+1, cost), helper(index+2, cost))
hashmap[index] = total
return hashmap.get(index)
return min(helper(0, cost), helper(1, cost))
Solution 3: Bottom up
Time: O(n)
Space: O(1)
class Solution(object):
def minCostClimbingStairs(self, cost):
"""
:type cost: List[int]
:rtype: int
"""
f1 = 0
f2 = 0
for x in reversed(cost):
minCostFromThisStep = x + min(f1, f2)
f2 = f1
f1 = minCostFromThisStep
return min(f1, f2)