# 746. Min Cost Climbing Stairs

## Description

You are given an integer array cost where cost[i] is the cost of ith step on a staircase. Once you pay the cost, you can either climb one or two steps.

You can either start from the step with index 0, or the step with index 1.

Return the minimum cost to reach the top of the floor.

Example 1:

Input: cost = [10,15,20]
Output: 15
Explanation: You will start at index 1.
- Pay 15 and climb two steps to reach the top.
The total cost is 15.


Example 2:

Input: cost = [1,100,1,1,1,100,1,1,100,1]
Output: 6
Explanation: You will start at index 0.
- Pay 1 and climb two steps to reach index 2.
- Pay 1 and climb two steps to reach index 4.
- Pay 1 and climb two steps to reach index 6.
- Pay 1 and climb one step to reach index 7.
- Pay 1 and climb two steps to reach index 9.
- Pay 1 and climb one step to reach the top.
The total cost is 6.


Constraints:

• 2 <= cost.length <= 1000
• 0 <= cost[i] <= 999

## Solutions

Solution 1: Dynamic Programming

We define $f[i]$ as the minimum cost required to reach the $i$th step, initially $f[0] = f[1] = 0$. The answer is $f[n]$.

When $i \ge 2$, we can directly reach the $i$th step from the $(i - 1)$th step using $1$ step, or reach the $i$th step from the $(i - 2)$th step using $2$ steps. Therefore, we have the state transition equation:

$f[i] = \min(f[i - 1] + cost[i - 1], f[i - 2] + cost[i - 2])$

The final answer is $f[n]$.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the cost array.

We notice that $f[i]$ in the state transition equation is only related to $f[i - 1]$ and $f[i - 2]$. Therefore, we can use two variables $f$ and $g$ to alternately record the values of $f[i - 2]$ and $f[i - 1]$, which optimizes the space complexity to $O(1)$.

• class Solution {
public int minCostClimbingStairs(int[] cost) {
int f = 0, g = 0;
for (int i = 2; i <= cost.length; ++i) {
int gg = Math.min(f + cost[i - 2], g + cost[i - 1]);
f = g;
g = gg;
}
return g;
}
}

• class Solution {
public:
int minCostClimbingStairs(vector<int>& cost) {
int f = 0, g = 0;
for (int i = 2; i <= cost.size(); ++i) {
int gg = min(f + cost[i - 2], g + cost[i - 1]);
f = g;
g = gg;
}
return g;
}
};

• class Solution:
def minCostClimbingStairs(self, cost: List[int]) -> int:
f = g = 0
for i in range(2, len(cost) + 1):
f, g = g, min(f + cost[i - 2], g + cost[i - 1])
return g


• func minCostClimbingStairs(cost []int) int {
var f, g int
for i := 2; i <= n; i++ {
f, g = g, min(f+cost[i-2], g+cost[i-1])
}
return g
}

• function minCostClimbingStairs(cost: number[]): number {
let [f, g] = [0, 0];
for (let i = 2; i <= cost.length; ++i) {
[f, g] = [g, Math.min(f + cost[i - 2], g + cost[i - 1])];
}
return g;
}


• impl Solution {
pub fn min_cost_climbing_stairs(cost: Vec<i32>) -> i32 {
let (mut f, mut g) = (0, 0);
for i in 2..=cost.len() {
let gg = std::cmp::min(f + cost[i - 2], g + cost[i - 1]);
f = g;
g = gg;
}
g
}
}


• function minCostClimbingStairs(cost) {
const n = cost.length;
const f = Array(n).fill(-1);
const dfs = i => {
if (i >= n) {
return 0;
}
if (f[i] < 0) {
f[i] = cost[i] + Math.min(dfs(i + 1), dfs(i + 2));
}
return f[i];
};
return Math.min(dfs(0), dfs(1));
}