# Min Cost Climbing Stairs

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:**

1. `cost`will have a length in the range`[2, 1000]`.
2. Every`cost[i]`will be an integer in the range`[0, 999]`.

## Analysis

**朴素的动态规划DP - Bottom Up**

这里比较明显的状态是用`dp[i]`表示到达第`i`个step时所需要的cost；

递推关系也容易得到，状态转移方程是

```java
dp[i] = Math.min(dp[i - 1], dp[i - 2]) + cost[i]
```

相当于 `i - 1`, `i - 2` 位置的cost加上当前`i`位置的cost

初始状态 dp\[0] = cost\[0], dp\[1] = cost\[1] 分别代表从0，或者1出发

最终结果则是取距离top为1或者2 steps中的较小值 `Math.min(dp[ len - 1], dp[ len - 2])`

## Solution

DP - O(n) space, O(n) time - (13ms 36.68%)

```java
class Solution {
    public int minCostClimbingStairs(int[] cost) {
        int len = cost.length;
        int[] climbCost = new int[len];
        climbCost[0] = cost[0];
        climbCost[1] = cost[1];
        for (int i = 2; i < len; i++) {
            climbCost[i] = Math.min(climbCost[i - 1], climbCost[i - 2]) + cost[i];
        }
        return Math.min(climbCost[len - 1], climbCost[len - 2]);
    }
}
```

DP - O(1) space, O(n) time

```java
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) where N is the length of`cost`.
* Space Complexity:O(1), the space used by`f1, f2`.

## Reference

<https://leetcode.com/problems/min-cost-climbing-stairs/solution/>

<https://leetcode.com/problems/min-cost-climbing-stairs/discuss/110111/Easy-to-understand-C++-using-DP-with-detailed-explanation>