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.lengthcols == grid[i].length2 <= rows, cols <= 700 <= grid[i][j] <= 100
Solutions
Solution 1
Intuition
store max cherries in each row
Code
class Solution {
private static final int[] DIRS = { -1, 0, 1 };
public int cherryPickup(int[][] grid) {
int rows = grid.length, cols = grid[0].length;
int[][][] dp = new int[rows][cols][cols];
dp[0][0][cols - 1] = grid[0][0] + grid[0][cols - 1];
int res = 0;
// from row 1 to bottom
for (int row = 1; row < rows; row++) {
// robot#1 from left top \
for (int col1 = 0; col1 < cols && col1 <= row; col1++) {
// robot#2 form right top /
for (int col2 = cols - 1; col2 > col1 && col2 >= (cols - 1 - row); col2--) {
int prevCherries = 0;
for (int dir1 : DIRS) {
for (int dir2 : DIRS) {
int prevRow = row - 1, prevCol1 = col1 + dir1, prevCol2 = col2 + dir2;
if (prevCol1 >= 0 && prevCol1 < cols && prevCol2 >= 0 && prevCol2 < cols
&& prevCol1 != prevCol2) {
prevCherries = Math.max(prevCherries, dp[prevRow][prevCol1][prevCol2]);
}
}
}
int cherries = grid[row][col1] + grid[row][col2];
dp[row][col1][col2] = cherries + prevCherries;
res = Math.max(res, dp[row][col1][col2]);
}
}
}
return res;
}
}
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