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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

Thinking

Start at index \(0\) or \(1\), pay the current stair, and climb one or two steps. \(n\le 1000\); plain recursion repeats stairs.

The cost from \(i\) depends only on \(i+1\) and \(i+2\); past the top it is \(0\). Memoization computes each index once.

The answer is \(\min(dfs(0), dfs(1))\). Time and space \(O(n)\).

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 \(\textit{cost}\).

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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))

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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];
    }
}

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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));
    }
};

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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))
}

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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));
}

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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))
    }
}

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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

Thinking

Solution 1 is already linear but recursive. The same relation fills forward.

\(f[i]\) is the min cost to reach index \(i\), from \(i-1\) or \(i-2\) paying that stair. \(f[n]\) is the top.

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 \(\textit{cost}\).

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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]

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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];
    }
}

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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];
    }
};

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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]
}

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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];
}

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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]
    }
}

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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)

Thinking

\(f[i]\) reads only the previous two values, so two scalars suffice. Space \(O(1)\).

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 \(O(1)\).

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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

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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;
    }
}

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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;
    }
};

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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
}

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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;
}

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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
    }
}

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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;
}

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