LeetCode 2017: Grid Game Solution
Master LeetCode problem 2017 (Grid Game), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
2017. Grid Game
Problem Explanation
Explanation
To solve this problem, we can use dynamic programming to keep track of the maximum points that can be collected by the second robot at each cell. We iterate through the grid from left to right and top to bottom, calculating the maximum points the second robot can collect at each cell based on the choices made by the first robot. The optimal strategy for the first robot is to minimize the points left for the second robot to collect.
- Initialize a 2D array
dpof size 2 x n to store the maximum points the second robot can collect at each cell. - Iterate through the grid from left to right and top to bottom:
- At each cell (r, c), update
dp[r][c]to be the maximum of:- The points in the current cell plus the maximum points the second robot can collect from the cell above (if not in the first row).
- The points in the current cell plus the maximum points the second robot can collect from the cell on the left (if not in the first column).
- At each cell (r, c), update
- The maximum points the second robot can collect will be in
dp[1][n-1].
Time complexity: O(n), where n is the number of columns in the grid. Space complexity: O(n) for storing the dynamic programming array.
Solution Code
class Solution {
public long gridGame(int[][] grid) {
int n = grid[0].length;
long[][] dp = new long[2][n];
dp[1][n-1] = grid[1][n-1];
for (int r = 0; r < 2; r++) {
for (int c = n - 2; c >= 0; c--) {
dp[r][c] = grid[r][c] + Math.min(dp[r][c + 1], (r == 0) ? Long.MAX_VALUE : dp[r - 1][c]);
}
}
return dp[1][0];
}
}Try It Yourself
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