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, 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, and only step on 1s, skipping cost.
costwill have a length in the range
cost[i]will be an integer in the range
There is a clear recursion available: the final cost
f[i] to climb the staircase from some step
f[i] = cost[i] + min(f[i+1], f[i+2]). This motivates dynamic programming.
f backwards in order. That way, when we are deciding what
f[i] will be, we\'ve already figured out
We can do even better than that. At the
i-th step, let
f1, f2 be the old value of
f[i+2], and update them to be the new values
f[i], f[i+1]. We keep these updated as we iterate through
i backwards. At the end, we want
Time Complexity: where is the length of
Space Complexity: , the space used by
Analysis written by: @awice.\n