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.
Input: cost = [10, 15, 20]
Output: 15
Explanation: Cheapest is start on cost[1], pay that cost and go to the top.
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].
cost will have a length in the range [2, 1000].
Every cost[i] will be an integer in the range [0, 999].
For the DP approach we create an array of size n where n is length of cost array. We first initialize the first two elements of the array with the first two elements of the cost array. Then, the third element of the DP array is the sum of the third element of the cost array plus the minimum value of the two previous values of the DP array. Last, we return the minimum of the last two values of the DP array. Therefore the time and space complexity should be an order of O(n).
def minCostClimbingStairs(cost):
"""
:type cost: List[int]
:rtype: int
:Time Complexity: O(N)
:Space Complexity: O(N)
"""
N = len(cost)
dp = [0] * N
dp[0] = cost[0]
dp[1] = cost[1]
for i in range(2, N):
dp[i] = min(dp[i-1], dp[i-2]) + cost[i]
return min(dp[N-1], dp[N-2])