# Min Cost Climbing Stairs

​On a staircase, every `ith` step is associated with some non-negative cost `cost[i]`. Once you pay the cost, you can either climb one or two steps. You need to find the 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[] = {2, 5, 3, 1, 7, 3, 4}
Output: 9
Steps: 2->3->1->3```

Example 2:

```Input: cost = {1, 100, 1, 1, 1, 100, 1, 1, 100, 1}
Output: 6
Steps: 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]`.

There is a clear recursion available: the final cost `dp[i]` to climb the staircase from some step `i` is

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

This motivates dynamic programming.

Algorithm:

Let `dp[i]` be the cost to climb the `ith` staircase to from `0th` or `1st` step. Hence:

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

Since `dp[i-1]` and `dp[i-2]` are needed to compute the cost of travelling from `ith` step, a bottom-up approach can be used to solve the problem. The answer will be the minimum cost of reaching `n-1th` stair and `n-2th` stair. Compute the `dp[]` array in a bottom-up manner. Below is the implementation of the above approach.

```public int minCostClimbingStairs(int[] cost) {

if (cost == null || cost.length == 0) {
return 0;
}
int len = cost.length;
if (len == 1) return cost[0];
if (len == 2) return Math.min(cost[0], cost[1]);

int[] dp = new int[len];
dp[0] = cost[0];
dp[1] = cost[1];

for (int i = 2; i < cost.length; i++) {
dp[i] = cost[i] + Math.min(dp[i - 1], dp[i - 2]);
}
return Math.min(dp[len - 2], dp[len - 1]);
}
```

Complexity Analysis

• Time Complexity: O(N) where N is the length of `cost`.
• Space Complexity: O(N), the space used by `dp`.

The problem can also be solved in `O(1)` space without a `dp` array and using some constant variables.

```class Solution {
public int minCostClimbingStairs(int[] cost) {
if (cost == null || cost.length == 0) return 0;
int f1 = cost[0];
int f2 = cost[1];
for (int i = 2; i < cost.length; i++) {
int f3 = cost[i] + Math.min(f1, f2);
f1 = f2;
f2 = f3;
}
return Math.min(f1, f2);
}
}
```

Categories: Dynamic Programming