2742. Painting the Walls

Description

You are given two 0-indexed integer arrays, cost and time, of size n representing the costs and the time taken to paint n different walls respectively. There are two painters available:

A paid painter that paints the i^th wall in time[i] units of time and takes cost[i] units of money.
A free painter that paints any wall in 1 unit of time at a cost of 0. But the free painter can only be used if the paid painter is already occupied.

Return the minimum amount of money required to paint the n walls.

Example 1:

Input: cost = [1,2,3,2], time = [1,2,3,2]
Output: 3
Explanation: The walls at index 0 and 1 will be painted by the paid painter, and it will take 3 units of time; meanwhile, the free painter will paint the walls at index 2 and 3, free of cost in 2 units of time. Thus, the total cost is 1 + 2 = 3.

Example 2:

Input: cost = [2,3,4,2], time = [1,1,1,1]
Output: 4
Explanation: The walls at index 0 and 3 will be painted by the paid painter, and it will take 2 units of time; meanwhile, the free painter will paint the walls at index 1 and 2, free of cost in 2 units of time. Thus, the total cost is 2 + 2 = 4.

Constraints:

1 <= cost.length <= 500
cost.length == time.length
1 <= cost[i] <= 10⁶
1 <= time[i] <= 500

Solutions

Solution 1: Memorization

We can consider whether each wall is painted by a paid painter or a free painter. Design a function $dfs(i, j)$, which means that from the $i$th wall, and the current remaining free painter working time is $j$, the minimum cost of painting all the remaining walls. Then the answer is $dfs(0, 0)$.

The calculation process of function $dfs(i, j)$ is as follows:

If $n - i \le j$, it means that there are no more walls than the free painter's working time, so the remaining walls are painted by the free painter, and the cost is $0$;
If $i \ge n$, return $+\infty$;
Otherwise, if the $i$th wall is painted by a paid painter, the cost is $cost[i]$, then $dfs(i, j) = dfs(i + 1, j + time[i]) + cost[i]$; if the $i$th wall is painted by a free painter, the cost is $0$, then $dfs(i, j) = dfs(i + 1, j - 1)$.

Note that the parameter $j$ may be less than $0$. Therefore, in the actual coding process, except for the $Python$ language, we add an offset $n$ to $j$ so that the range of $j$ is $[0, 2n]$.

Time complexity $O(n^2)$, space complexity $O(n^2)$. Where $n$ is the length of the array.

Python3JavaC++GoRust

class Solution:
    def paintWalls(self, cost: List[int], time: List[int]) -> int:
        @cache
        def dfs(i: int, j: int) -> int:
            if n - i <= j:
                return 0
            if i >= n:
                return inf
            return min(dfs(i + 1, j + time[i]) + cost[i], dfs(i + 1, j - 1))

        n = len(cost)
        return dfs(0, 0)

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

    public int paintWalls(int[] cost, int[] time) {
        n = cost.length;
        this.cost = cost;
        this.time = time;
        f = new Integer[n][n << 1 | 1];
        return dfs(0, n);
    }

    private int dfs(int i, int j) {
        if (n - i <= j - n) {
            return 0;
        }
        if (i >= n) {
            return 1 << 30;
        }
        if (f[i][j] == null) {
            f[i][j] = Math.min(dfs(i + 1, j + time[i]) + cost[i], dfs(i + 1, j - 1));
        }
        return f[i][j];
    }
}

class Solution {
public:
    int paintWalls(vector<int>& cost, vector<int>& time) {
        int n = cost.size();
        int f[n][n << 1 | 1];
        memset(f, -1, sizeof(f));
        function<int(int, int)> dfs = [&](int i, int j) -> int {
            if (n - i <= j - n) {
                return 0;
            }
            if (i >= n) {
                return 1 << 30;
            }
            if (f[i][j] == -1) {
                f[i][j] = min(dfs(i + 1, j + time[i]) + cost[i], dfs(i + 1, j - 1));
            }
            return f[i][j];
        };
        return dfs(0, n);
    }
};

func paintWalls(cost []int, time []int) int {
    n := len(cost)
    f := make([][]int, n)
    for i := range f {
        f[i] = make([]int, n<<1|1)
        for j := range f[i] {
            f[i][j] = -1
        }
    }
    var dfs func(i, j int) int
    dfs = func(i, j int) int {
        if n-i <= j-n {
            return 0
        }
        if i >= n {
            return 1 << 30
        }
        if f[i][j] == -1 {
            f[i][j] = min(dfs(i+1, j+time[i])+cost[i], dfs(i+1, j-1))
        }
        return f[i][j]
    }
    return dfs(0, n)
}

impl Solution {
    #[allow(dead_code)]
    pub fn paint_walls(cost: Vec<i32>, time: Vec<i32>) -> i32 {
        let n = cost.len();
        let mut record_vec: Vec<Vec<i32>> = vec![vec![-1; n << 1 | 1]; n];
        Self::dfs(&mut record_vec, 0, n as i32, n as i32, &time, &cost)
    }

    #[allow(dead_code)]
    fn dfs(
        record_vec: &mut Vec<Vec<i32>>,
        i: i32,
        j: i32,
        n: i32,
        time: &Vec<i32>,
        cost: &Vec<i32>,
    ) -> i32 {
        if n - i <= j - n {
            // All the remaining walls can be printed at no cost
            // Just return 0
            return 0;
        }
        if i >= n {
            // No way this case can be achieved
            // Just return +INF
            return 1 << 30;
        }
        if record_vec[i as usize][j as usize] == -1 {
            // This record hasn't been written
            record_vec[i as usize][j as usize] = std::cmp::min(
                Self::dfs(record_vec, i + 1, j + time[i as usize], n, time, cost)
                    + cost[i as usize],
                Self::dfs(record_vec, i + 1, j - 1, n, time, cost),
            );
        }
        record_vec[i as usize][j as usize]
    }
}

2742. Painting the Walls

Description

Solutions

Solution 1: Memorization

Comments