Question

1463. Cherry Pickup II

中文文档

Description

You are given a rows x cols matrix grid representing a field of cherries where grid[i][j] represents the number of cherries that you can collect from the (i, j) cell.

You have two robots that can collect cherries for you:

• Robot #1 is located at the top-left corner (0, 0), and
• Robot #2 is located at the top-right corner (0, cols - 1).

Return _the maximum number of cherries collection using both robots by following the rules below_:

• From a cell (i, j), robots can move to cell (i + 1, j - 1), (i + 1, j), or (i + 1, j + 1).
• When any robot passes through a cell, It picks up all cherries, and the cell becomes an empty cell.
• When both robots stay in the same cell, only one takes the cherries.
• Both robots cannot move outside of the grid at any moment.
• Both robots should reach the bottom row in grid.

 

Example 1:

Input: grid = [[3,1,1],[2,5,1],[1,5,5],[2,1,1]]
Output: 24
Explanation: Path of robot #1 and #2 are described in color green and blue respectively.
Cherries taken by Robot #1, (3 + 2 + 5 + 2) = 12.
Cherries taken by Robot #2, (1 + 5 + 5 + 1) = 12.
Total of cherries: 12 + 12 = 24.

Example 2:

Input: grid = [[1,0,0,0,0,0,1],[2,0,0,0,0,3,0],[2,0,9,0,0,0,0],[0,3,0,5,4,0,0],[1,0,2,3,0,0,6]]
Output: 28
Explanation: Path of robot #1 and #2 are described in color green and blue respectively.
Cherries taken by Robot #1, (1 + 9 + 5 + 2) = 17.
Cherries taken by Robot #2, (1 + 3 + 4 + 3) = 11.
Total of cherries: 17 + 11 = 28.

 

Constraints:

• rows == grid.length
• cols == grid[i].length
• 2 <= rows, cols <= 70
• 0 <= grid[i][j] <= 100

Solutions

Solution 1

class Solution:
  def cherryPickup(self, grid: List[List[int]]) -> int:
    m, n = len(grid), len(grid[0])
    f = [[[-1] * n for _ in range(n)] for _ in range(m)]
    f[0][0][n - 1] = grid[0][0] + grid[0][n - 1]
    for i in range(1, m):
      for j1 in range(n):
        for j2 in range(n):
          x = grid[i][j1] + (0 if j1 == j2 else grid[i][j2])
          for y1 in range(j1 - 1, j1 + 2):
            for y2 in range(j2 - 1, j2 + 2):
              if 0 <= y1 < n and 0 <= y2 < n and f[i - 1][y1][y2] != -1:
                f[i][j1][j2] = max(f[i][j1][j2], f[i - 1][y1][y2] + x)
    return max(f[-1][j1][j2] for j1, j2 in product(range(n), range(n)))

class Solution {
  public int cherryPickup(int[][] grid) {
    int m = grid.length, n = grid[0].length;
    int[][][] f = new int[m][n][n];
    for (var g : f) {
      for (var h : g) {
        Arrays.fill(h, -1);
      }
    }
    f[0][0][n - 1] = grid[0][0] + grid[0][n - 1];
    for (int i = 1; i < m; ++i) {
      for (int j1 = 0; j1 < n; ++j1) {
        for (int j2 = 0; j2 < n; ++j2) {
          int x = grid[i][j1] + (j1 == j2 ? 0 : grid[i][j2]);
          for (int y1 = j1 - 1; y1 <= j1 + 1; ++y1) {
            for (int y2 = j2 - 1; y2 <= j2 + 1; ++y2) {
              if (y1 >= 0 && y1 < n && y2 >= 0 && y2 < n && f[i - 1][y1][y2] != -1) {
                f[i][j1][j2] = Math.max(f[i][j1][j2], f[i - 1][y1][y2] + x);
              }
            }
          }
        }
      }
    }
    int ans = 0;
    for (int j1 = 0; j1 < n; ++j1) {
      for (int j2 = 0; j2 < n; ++j2) {
        ans = Math.max(ans, f[m - 1][j1][j2]);
      }
    }
    return ans;
  }
}

class Solution {
public:
  int cherryPickup(vector<vector<int>>& grid) {
    int m = grid.size(), n = grid[0].size();
    int f[m][n][n];
    memset(f, -1, sizeof(f));
    f[0][0][n - 1] = grid[0][0] + grid[0][n - 1];
    for (int i = 1; i < m; ++i) {
      for (int j1 = 0; j1 < n; ++j1) {
        for (int j2 = 0; j2 < n; ++j2) {
          int x = grid[i][j1] + (j1 == j2 ? 0 : grid[i][j2]);
          for (int y1 = j1 - 1; y1 <= j1 + 1; ++y1) {
            for (int y2 = j2 - 1; y2 <= j2 + 1; ++y2) {
              if (y1 >= 0 && y1 < n && y2 >= 0 && y2 < n && f[i - 1][y1][y2] != -1) {
                f[i][j1][j2] = max(f[i][j1][j2], f[i - 1][y1][y2] + x);
              }
            }
          }
        }
      }
    }
    int ans = 0;
    for (int j1 = 0; j1 < n; ++j1) {
      for (int j2 = 0; j2 < n; ++j2) {
        ans = max(ans, f[m - 1][j1][j2]);
      }
    }
    return ans;
  }
};

func cherryPickup(grid [][]int) int {
  m, n := len(grid), len(grid[0])
  f := make([][][]int, m)
  for i := range f {
    f[i] = make([][]int, n)
    for j := range f[i] {
      f[i][j] = make([]int, n)
      for k := range f[i][j] {
        f[i][j][k] = -1
      }
    }
  }
  f[0][0][n-1] = grid[0][0] + grid[0][n-1]
  for i := 1; i < m; i++ {
    for j1 := 0; j1 < n; j1++ {
      for j2 := 0; j2 < n; j2++ {
        x := grid[i][j1]
        if j1 != j2 {
          x += grid[i][j2]
        }
        for y1 := j1 - 1; y1 <= j1+1; y1++ {
          for y2 := j2 - 1; y2 <= j2+1; y2++ {
            if y1 >= 0 && y1 < n && y2 >= 0 && y2 < n && f[i-1][y1][y2] != -1 {
              f[i][j1][j2] = max(f[i][j1][j2], f[i-1][y1][y2]+x)
            }
          }
        }
      }
    }
  }
  ans := 0
  for j1 := 0; j1 < n; j1++ {
    for j2 := 0; j2 < n; j2++ {
      ans = max(ans, f[m-1][j1][j2])
    }
  }
  return ans
}

function cherryPickup(grid: number[][]): number {
  const m = grid.length;
  const n = grid[0].length;
  const f: number[][][] = new Array(m)
    .fill(0)
    .map(() => new Array(n).fill(0).map(() => new Array(n).fill(-1)));
  f[0][0][n - 1] = grid[0][0] + grid[0][n - 1];
  for (let i = 1; i < m; ++i) {
    for (let j1 = 0; j1 < n; ++j1) {
      for (let j2 = 0; j2 < n; ++j2) {
        const x = grid[i][j1] + (j1 === j2 ? 0 : grid[i][j2]);
        for (let y1 = j1 - 1; y1 <= j1 + 1; ++y1) {
          for (let y2 = j2 - 1; y2 <= j2 + 1; ++y2) {
            if (y1 >= 0 && y1 < n && y2 >= 0 && y2 < n && f[i - 1][y1][y2] !== -1) {
              f[i][j1][j2] = Math.max(f[i][j1][j2], f[i - 1][y1][y2] + x);
            }
          }
        }
      }
    }
  }
  let ans = 0;
  for (let j1 = 0; j1 < n; ++j1) {
    for (let j2 = 0; j2 < n; ++j2) {
      ans = Math.max(ans, f[m - 1][j1][j2]);
    }
  }
  return ans;
}

Solution 2

class Solution:
  def cherryPickup(self, grid: List[List[int]]) -> int:
    m, n = len(grid), len(grid[0])
    f = [[-1] * n for _ in range(n)]
    g = [[-1] * n for _ in range(n)]
    f[0][n - 1] = grid[0][0] + grid[0][n - 1]
    for i in range(1, m):
      for j1 in range(n):
        for j2 in range(n):
          x = grid[i][j1] + (0 if j1 == j2 else grid[i][j2])
          for y1 in range(j1 - 1, j1 + 2):
            for y2 in range(j2 - 1, j2 + 2):
              if 0 <= y1 < n and 0 <= y2 < n and f[y1][y2] != -1:
                g[j1][j2] = max(g[j1][j2], f[y1][y2] + x)
      f, g = g, f
    return max(f[j1][j2] for j1, j2 in product(range(n), range(n)))

class Solution {
  public int cherryPickup(int[][] grid) {
    int m = grid.length, n = grid[0].length;
    int[][] f = new int[n][n];
    int[][] g = new int[n][n];
    for (int i = 0; i < n; ++i) {
      Arrays.fill(f[i], -1);
      Arrays.fill(g[i], -1);
    }
    f[0][n - 1] = grid[0][0] + grid[0][n - 1];
    for (int i = 1; i < m; ++i) {
      for (int j1 = 0; j1 < n; ++j1) {
        for (int j2 = 0; j2 < n; ++j2) {
          int x = grid[i][j1] + (j1 == j2 ? 0 : grid[i][j2]);
          for (int y1 = j1 - 1; y1 <= j1 + 1; ++y1) {
            for (int y2 = j2 - 1; y2 <= j2 + 1; ++y2) {
              if (y1 >= 0 && y1 < n && y2 >= 0 && y2 < n && f[y1][y2] != -1) {
                g[j1][j2] = Math.max(g[j1][j2], f[y1][y2] + x);
              }
            }
          }
        }
      }
      int[][] t = f;
      f = g;
      g = t;
    }
    int ans = 0;
    for (int j1 = 0; j1 < n; ++j1) {
      for (int j2 = 0; j2 < n; ++j2) {
        ans = Math.max(ans, f[j1][j2]);
      }
    }
    return ans;
  }
}

class Solution {
public:
  int cherryPickup(vector<vector<int>>& grid) {
    int m = grid.size(), n = grid[0].size();
    vector<vector<int>> f(n, vector<int>(n, -1));
    vector<vector<int>> g(n, vector<int>(n, -1));
    f[0][n - 1] = grid[0][0] + grid[0][n - 1];
    for (int i = 1; i < m; ++i) {
      for (int j1 = 0; j1 < n; ++j1) {
        for (int j2 = 0; j2 < n; ++j2) {
          int x = grid[i][j1] + (j1 == j2 ? 0 : grid[i][j2]);
          for (int y1 = j1 - 1; y1 <= j1 + 1; ++y1) {
            for (int y2 = j2 - 1; y2 <= j2 + 1; ++y2) {
              if (y1 >= 0 && y1 < n && y2 >= 0 && y2 < n && f[y1][y2] != -1) {
                g[j1][j2] = max(g[j1][j2], f[y1][y2] + x);
              }
            }
          }
        }
      }
      swap(f, g);
    }
    int ans = 0;
    for (int j1 = 0; j1 < n; ++j1) {
      for (int j2 = 0; j2 < n; ++j2) {
        ans = max(ans, f[j1][j2]);
      }
    }
    return ans;
  }
};

func cherryPickup(grid [][]int) int {
  m, n := len(grid), len(grid[0])
  f := make([][]int, n)
  g := make([][]int, n)
  for i := range f {
    f[i] = make([]int, n)
    g[i] = make([]int, n)
    for j := range f[i] {
      f[i][j] = -1
      g[i][j] = -1
    }
  }
  f[0][n-1] = grid[0][0] + grid[0][n-1]
  for i := 1; i < m; i++ {
    for j1 := 0; j1 < n; j1++ {
      for j2 := 0; j2 < n; j2++ {
        x := grid[i][j1]
        if j1 != j2 {
          x += grid[i][j2]
        }
        for y1 := j1 - 1; y1 <= j1+1; y1++ {
          for y2 := j2 - 1; y2 <= j2+1; y2++ {
            if y1 >= 0 && y1 < n && y2 >= 0 && y2 < n && f[y1][y2] != -1 {
              g[j1][j2] = max(g[j1][j2], f[y1][y2]+x)
            }
          }
        }
      }
    }
    f, g = g, f
  }
  ans := 0
  for j1 := 0; j1 < n; j1++ {
    for j2 := 0; j2 < n; j2++ {
      ans = max(ans, f[j1][j2])
    }
  }
  return ans
}

function cherryPickup(grid: number[][]): number {
  const m = grid.length;
  const n = grid[0].length;
  let f: number[][] = new Array(n).fill(0).map(() => new Array(n).fill(-1));
  let g: number[][] = new Array(n).fill(0).map(() => new Array(n).fill(-1));
  f[0][n - 1] = grid[0][0] + grid[0][n - 1];
  for (let i = 1; i < m; ++i) {
    for (let j1 = 0; j1 < n; ++j1) {
      for (let j2 = 0; j2 < n; ++j2) {
        const x = grid[i][j1] + (j1 === j2 ? 0 : grid[i][j2]);
        for (let y1 = j1 - 1; y1 <= j1 + 1; ++y1) {
          for (let y2 = j2 - 1; y2 <= j2 + 1; ++y2) {
            if (y1 >= 0 && y1 < n && y2 >= 0 && y2 < n && f[y1][y2] !== -1) {
              g[j1][j2] = Math.max(g[j1][j2], f[y1][y2] + x);
            }
          }
        }
      }
    }
    [f, g] = [g, f];
  }
  let ans = 0;
  for (let j1 = 0; j1 < n; ++j1) {
    for (let j2 = 0; j2 < n; ++j2) {
      ans = Math.max(ans, f[j1][j2]);
    }
  }
  return ans;
}