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.
Test Cases
Example 1:
| 3 | 1 | 1 |
| 2 | 5 | 1 |
| 1 | 5 | 5 |
| 2 | 1 | 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
Solution
class Solution {
public int cherryPickup(int[][] grid) {
int dir[] = new int[]{-1, 0, 1};
int row = grid.length;
int col = grid[0].length;
int dp[][][] = new int[row][col][col];
for(int i = 0; i < row; i++){
for(int j = 0; j < col; j++){
for(int k = 0; k < col; k++){
dp[i][j][k] = -1;
}
}
}
int col1 = 0;
int col2 = col - 1;
dp[0][col1][col2] = grid[0][col1] + grid[0][col2];
int max = dp[0][col1][col2];
for(int i = 1; i < row; i++){
for(int c1 = 0; c1 < col; c1++){
for(int c2 = 0; c2 < col; c2++){
int prev = dp[i - 1][c1][c2];
if(prev >= 0){
for(int d1: dir){
col1 = d1 + c1;
for(int d2: dir){
col2 = d2 + c2;
if(inRange(col1, col) && inRange(col2, col)){
dp[i][col1][col2] = Math.max(dp[i][col1][col2], prev+(col1 == col2 ? grid[i][col1] : (grid[i][col1] + grid[i][col2])));
max = Math.max(max, dp[i][col1][col2]);
}
}
}
}
}
}
}
return max;
}
public boolean inRange(int val, int limit){
return 0 <= val && val < limit;
}
}