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Easy
Array
Dynamic Programming

746. Min Cost Climbing Stairs

中文文档

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: Memoization Search

We design a function $\textit{dfs}(i)$, which represents the minimum cost required to climb the stairs starting from the $i$-th step. Therefore, the answer is $\min(\textit{dfs}(0), \textit{dfs}(1))$.

The execution process of the function $\textit{dfs}(i)$ is as follows:

  • If $i \ge \textit{len(cost)}$, it means the current position has exceeded the top of the stairs, and there is no need to climb further, so return $0$;
  • Otherwise, we can choose to climb $1$ step with a cost of $\textit{cost}[i]$, then recursively call $\textit{dfs}(i + 1)$; or we can choose to climb $2$ steps with a cost of $\textit{cost}[i]$, then recursively call $\textit{dfs}(i + 2)$;
  • Return the minimum cost between these two options.

To avoid repeated calculations, we use memoization search, saving the results that have already been calculated in an array or hash table.

The time complexity is $O(n)$, and the space complexity is $O(n)$, where $n$ is the length of the array $\textit{cost}$.

Python3

class Solution:
    def minCostClimbingStairs(self, cost: List[int]) -> int:
        @cache
        def dfs(i: int) -> int:
            if i >= len(cost):
                return 0
            return cost[i] + min(dfs(i + 1), dfs(i + 2))

        return min(dfs(0), dfs(1))

Java

class Solution {
    private Integer[] f;
    private int[] cost;

    public int minCostClimbingStairs(int[] cost) {
        this.cost = cost;
        f = new Integer[cost.length];
        return Math.min(dfs(0), dfs(1));
    }

    private int dfs(int i) {
        if (i >= cost.length) {
            return 0;
        }
        if (f[i] == null) {
            f[i] = cost[i] + Math.min(dfs(i + 1), dfs(i + 2));
        }
        return f[i];
    }
}

C++

class Solution {
public:
    int minCostClimbingStairs(vector<int>& cost) {
        int n = cost.size();
        int f[n];
        memset(f, -1, sizeof(f));
        auto dfs = [&](this auto&& dfs, int i) -> int {
            if (i >= n) {
                return 0;
            }
            if (f[i] < 0) {
                f[i] = cost[i] + min(dfs(i + 1), dfs(i + 2));
            }
            return f[i];
        };
        return min(dfs(0), dfs(1));
    }
};

Go

func minCostClimbingStairs(cost []int) int {
	n := len(cost)
	f := make([]int, n)
	for i := range f {
		f[i] = -1
	}
	var dfs func(int) int
	dfs = func(i int) int {
		if i >= n {
			return 0
		}
		if f[i] < 0 {
			f[i] = cost[i] + min(dfs(i+1), dfs(i+2))
		}
		return f[i]
	}
	return min(dfs(0), dfs(1))
}

TypeScript

function minCostClimbingStairs(cost: number[]): number {
    const n = cost.length;
    const f: number[] = Array(n).fill(-1);
    const dfs = (i: number): number => {
        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));
}

Rust

impl Solution {
    pub fn min_cost_climbing_stairs(cost: Vec<i32>) -> i32 {
        let n = cost.len();
        let mut f = vec![-1; n];

        fn dfs(i: usize, cost: &Vec<i32>, f: &mut Vec<i32>, n: usize) -> i32 {
            if i >= n {
                return 0;
            }
            if f[i] < 0 {
                let next1 = dfs(i + 1, cost, f, n);
                let next2 = dfs(i + 2, cost, f, n);
                f[i] = cost[i] + next1.min(next2);
            }
            f[i]
        }

        dfs(0, &cost, &mut f, n).min(dfs(1, &cost, &mut f, n))
    }
}

JavaScript

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));
}

Solution 2: Dynamic Programming

We define $f[i]$ as the minimum cost needed to reach the $i$-th stair. Initially, $f[0] = f[1] = 0$, and the answer is $f[n]$.

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

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

The final answer is $f[n]$.

The time complexity is $O(n)$, and the space complexity is $O(n)$, where $n$ is the length of the array $\textit{cost}$.

Python3

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

Java

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

C++

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

Go

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

TypeScript

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

Rust

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

JavaScript

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

Solution 3: Dynamic Programming (Space Optimization)

We notice that the state transition equation for $f[i]$ only depends on $f[i - 1]$ and $f[i - 2]$. Therefore, we can use two variables $f$ and $g$ to alternately record the values of $f[i - 2]$ and $f[i - 1]$, thus optimizing the space complexity to $O(1)$.

Python3

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

Java

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;
    }
}

C++

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;
    }
};

Go

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
}

TypeScript

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

Rust

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
    }
}

JavaScript

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