Question

2742. Painting the Walls

中文文档

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 ith 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.

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]
  }
}