746 Min Cost Climbing Stairs

On a staircase, thei-th step has some non-negative costcost[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].

Approach #1: Dynamic Programming [Accepted]

Intuition

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

Algorithm

Let's evaluatefbackwards in order. That way, when we are deciding whatf[i]will be, we've already figured outf[i+1]andf[i+2].

We can do even better than that. At thei-th step, letf1, f2be the old value off[i+1],f[i+2], and update them to be the new valuesf[i], f[i+1]. We keep these updated as we iterate throughibackwards. At the end, we wantmin(f1, f2).

class Solution {
    public int minCostClimbingStairs(int[] cost) {
        int f1 = 0, f2 = 0;
        for (int i = cost.length - 1; i >= 0; --i) {
            int f0 = cost[i] + Math.min(f1, f2);
            f2 = f1;
            f1 = f0;
        }
        return Math.min(f1, f2);
    }
}

Time Complexity:O(N)O(N)whereNNis the length ofcost.

• Space Complexity:O(1)O(1), the space used byf1, f2.