Approach #1: Dynamic Programming [Accepted]


There is a clear recursion available: the final cost f[i] to climb the staircase from some step i is f[i] = cost[i] + min(f[i+1], f[i+2]). This motivates dynamic programming.


Let's evaluate f backwards in order. That way, when we are deciding what f[i] will be, we've already figured out f[i+1] and f[i+2].

We can do even better than that. At the i-th step, let f1, f2 be the old value of f[i+1], 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 min(f1, f2).

Complexity Analysis

  • Time Complexity: where is the length of cost.

  • Space Complexity: , the space used by f1, f2.

Analysis written by: @awice.