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].
Note:
cost will have a length in the range [2, 1000].cost[i] will be an integer in the range [0, 999].Source
def min_cost_climbing_stairs(cost):
"""Recursive solution with memoization."""
def dp(n):
nonlocal memo
# base cases
if n < 2:
return cost[n]
if n in memo:
return memo[n]
# recursive relation
cost_amount = cost[n] + min(dp(n-1), dp(n-2))
memo[n] = cost_amount
return cost_amount
memo = {}
return min(dp(len(cost)-1), dp(len(cost)-2))
cost = [10, 15, 20]
min_cost_climbing_stairs(cost)
15
cost = [1, 100, 1, 1, 1, 100, 1, 1, 100, 1]
min_cost_climbing_stairs(cost)
6
Solve it both recursively and iteratively.
def min_cost_climbing_stairs(cost):
"""Dynamic Programming version with O(n) space complexity."""
if not cost:
return 0
n = len(cost)
if n < 3:
return min(cost)
dp = [0] * n
dp[0] = cost[0]
dp[1] = cost[1]
for i in range(2, n):
dp[i] = cost[i] + min(dp[i-2], dp[i-1])
return min(dp[-2], dp[-1])
Solve it with O(1) (i.e. constant) memory.
def min_cost_climbing_stairs(cost):
"""Dynamic Programming version with O(1) space complexity."""
f1 = f2 = 0
for step in cost:
f1, f2 = step + min(f1, f2), f1
return min(f1, f2)