Grid Game
Problem Description​
You are given a 0-indexed 2D array
grid
of size 2 x n
, where grid[r][c]
represents the number of points at position (r, c)
on the matrix. Two robots initially start at (0, 0)
and want to reach (1, n-1)
. Each robot may only move to the right ((r, c) to (r, c + 1))
or down ((r, c)
to (r + 1, c))
.
At the start of the game, the first robot moves from (0, 0) to (1, n-1)
, collecting all the points from the cells on its path. For all cells (r, c)
traversed on the path, grid[r][c]
is set to 0
. Then, the second robot moves from (0, 0) to (1, n-1)
, collecting the points on its path. Note that their paths may intersect with one another.
The first robot wants to minimize the number of points collected by the second robot. In contrast, the second robot wants to maximize the number of points it collects. If both robots play optimally, return the number of points collected by the second robot.
Examples​
Example 1:
Input: `grid = [[2,5,4],[1,5,1]]`
Output: `4`
Explanation:
- The optimal path taken by the first robot is shown in red, and the optimal path taken by the second robot is shown in blue.
- The cells visited by the first robot are set to 0.
- The second robot will collect 0 + 0 + 4 + 0 = 4 points.
Example 2:
Input: `grid = [[3,3,1],[8,5,2]]`
Output: `4`
Explanation:
- The optimal path taken by the first robot is shown in red, and the optimal path taken by the second robot is shown in blue.
- The cells visited by the first robot are set to 0.
- The second robot will collect 0 + 3 + 1 + 0 = 4 points.
Example 3:
Input: `grid = [[1,3,1,15],[1,3,3,1]]`
Output: `7`
Explanation:
- The optimal path taken by the first robot is shown in red, and the optimal path taken by the second robot is shown in blue.
- The cells visited by the first robot are set to 0.
- The second robot will collect 0 + 1 + 3 + 3 + 0 = 7 points.
Approach​
In solving the problem, we employ two dynamic programming arrays, dp1
and dp2
. dp1[j]
tracks the maximum points collected by the first robot as it moves across the grid from left to right, starting at (0, 0)
. Simultaneously, dp2[j]
records the maximum points collected by the second robot moving from right to left, starting at (0, n-1)
. The approach ensures optimal play: after dp1
is computed, dp2
is filled backwards while calculating the maximum points max_points
that the second robot can collect after the first has minimized its score.
C++​
class Solution {
public:
long long gridGame(vector<vector<int>>& grid) {
int n = grid[0].size();
// Initialize dp arrays
vector<long long> dp1(n), dp2(n);
// Initialize dp1
dp1[0] = grid[0][0];
for (int j = 1; j < n; ++j) {
dp1[j] = dp1[j - 1] + grid[0][j];
}
// Initialize dp2
dp2[n - 1] = grid[1][n - 1];
for (int j = n - 2; j >= 0; --j) {
dp2[j] = dp2[j + 1] + grid[1][j];
}
// Find maximum points for the second robot
long long max_points = LLONG_MAX;
for (int j = 0; j < n; ++j) {
max_points = min(max_points, max(dp1[j], (j > 0 ? dp2[j - 1] : LLONG_MAX)));
}
return max_points;
}
};