
Problem Statement
You are given a 2D array grid
of size 2 x n
, representing a game board where each cell (r, c)
contains a number of points. Two robots play on this board starting from the top-left corner (0, 0)
and aim to reach the bottom-right corner (1, n - 1)
. Both robots can only move right or down.
Gameplay happens in two stages:
- The first robot moves from
(0, 0)
to(1, n - 1)
, collecting all points along its path. Each cell visited by this robot is set to0
afterward. - The second robot then moves on the updated grid and tries to collect as many points as possible from its own path, again moving only right or down.
The first robot plays optimally to minimize the second robot's score. The second robot plays optimally to maximize its score.
Return the number of points collected by the second robot after both have played optimally.
Examples
Example 1
Input:
grid = [[2, 5, 4], [1, 5, 1]]
Output:
4
Explanation:
The first robot clears a path minimizing the second robot's future score. After that path is zeroed, the second robot collects 4 points from the remaining grid.
Example 2
Input:
grid = [[3, 3, 1], [8, 5, 2]]
Output:
4
Explanation:
The first robot removes high-value cells optimally. The second robot picks the best possible remaining path, collecting 4 points.
Example 3
Input:
grid = [[1, 3, 1, 15], [1, 3, 3, 1]]
Output:
7
Explanation:
The second robot takes the highest-value remaining path, which gives 7 points.
Constraints
grid.length == 2
n == grid[0].length
1 <= n <= 5 * 10^4
1 <= grid[r][c] <= 10^5
Approach and Intuition
To solve this problem efficiently, observe that the only point where the first robot can switch from the top row to the bottom row is once at some column i
. Once it goes down, it must continue right along the bottom row. That means its path is:
- Move right along the top row to column
i
- Move down to the bottom row at column
i
- Continue moving right on the bottom row to the end
This gives us a crucial insight: there are only n
possible "transition points" (columns where the first robot might go down). For each of these n
possibilities:
Compute the sum of remaining points in the top row after column
i
(since those cells won’t be cleared by the first robot).Compute the sum of points collected before column
i
on the bottom row (since the first robot will clear them).The second robot can only collect from:
top[i + 1:]
(top row, to the right of where robot 1 turned down)bottom[:i]
(bottom row, to the left of where robot 1 turned down)
For each possible turning point i
, calculate the maximum of those two remaining segments, since the second robot can only take one path.
The minimum of these maximums (over all i
) is the best outcome the first robot can force. This value is the maximum number of points the second robot can collect under optimal play.
This is an optimization problem solved by computing prefix sums and iterating over all n
transition points. It runs in O(n) time and space.
Solutions
- C++
- Java
- Python
class Solution {
public:
long long optimalPathSum(vector<vector<int>>& board) {
long long topRowAccumulate = accumulate(begin(board[0]), end(board[0]), 0LL),
bottomRowAccumulate = 0;
long long smallestMax = LONG_LONG_MAX;
for (int index = 0; index < board[0].size(); ++index) {
topRowAccumulate -= board[0][index];
smallestMax = min(smallestMax, max(topRowAccumulate, bottomRowAccumulate));
bottomRowAccumulate += board[1][index];
}
return smallestMax;
}
};
The solution provided here addresses the Grid Game problem using C++. The code aims to determine the smallest maximum path sum on a two-row grid, ensuring that the approach is efficient for competitive programming scenarios.
In this code:
A class
Solution
is defined with one member function,optimalPathSum
, which accepts a 2D vectorboard
containing two rows representing the grid.Inside the
optimalPathSum
function:- Two
long long
variables,topRowAccumulate
andbottomRowAccumulate
, are initialized.topRowAccumulate
uses theaccumulate
function from the<numeric>
header to calculate the sum of the top row, whilebottomRowAccumulate
starts at zero. - Another
long long
variable,smallestMax
, is initialized to the maximum possible value for typelong long
.
- Two
The function then iterates through each column of the grid using a for-loop. For each column:
- The current value of the top row at the current index (
board[0][index]
) is subtracted fromtopRowAccumulate
. - The
smallestMax
value is updated by comparing it to the maximum oftopRowAccumulate
andbottomRowAccumulate
. - The value of the bottom row at the current index is added to
bottomRowAccumulate
.
- The current value of the top row at the current index (
After the loop completes,
smallestMax
, which represents the minimum of the maximum possible accumulated values during the traversal, is returned.The solution efficiently determines the optimal path sum by accumulating values as it progresses through the grid, minimizing the maximum possible accumulated sum that a player can be forced to collect.
This approach ensures a time-efficient traversal of the grid, updating accumulated sums and minimal maximum values in a single pass, leveraging dynamic programming principles for optimal performance.
class Solution {
public long optimalPath(int[][] matrix) {
// Calculate total sum of first row elements
long topRowTotal = 0;
for (int value : matrix[0]) {
topRowTotal += value;
}
long bottomRowTotal = 0, leastMaxSum = Long.MAX_VALUE;
for (int i = 0; i < matrix[0].length; i++) {
topRowTotal -= matrix[0][i];
leastMaxSum = Math.min(
leastMaxSum,
Math.max(topRowTotal, bottomRowTotal)
);
bottomRowTotal += matrix[1][i];
}
return leastMaxSum;
}
}
The Java solution to the "Grid Game" problem involves calculating the optimal path a player can take in a grid represented by a 2-row matrix, aiming for a strategy that minimizes the maximum possible score an opponent can achieve in the upper row after the player takes their turn in the lower row.
The method optimalPath(int[][] matrix)
computes this by:
- Initializing sums of the elements for both the top and bottom rows of the matrix.
- Iteratively updating these sums as if the current column’s top row value is removed (since the player has moved past it), and adding the current column’s bottom row value to the bottom sum (since the player now stands there).
- At each iteration, the method computes the maximum total of remaining values in the two rows (indicating a potential score for an opponent moving only forward).
- Keeping track of the minimal value among these maximum totals, which effectively becomes the least possible score that an opponent could assure if they played optimally given the player's path.
The approach efficiently determines an optimal playing strategy by leveraging simple arithmetic operations and comparisons, ensuring both high performance and clarity. By iterating through the columns just once and using no extra space beyond a few variables, the solution also exhibits an optimal time complexity of O(n) and space complexity of O(1), where (n) is the number of columns in the grid.
- Use arrays to handle matrix operations for a clear and efficient computation.
- Implement simple loop constructs and minimization checks to evaluate the game conditions dynamically.
- Maintain low space usage by avoiding additional data structure beyond necessary variables.
class Solution:
def gridGame(self, grid: List[List[int]]) -> int:
upper_row_sum = sum(grid[0])
lower_row_sum = 0
min_max_value = float("inf")
for idx in range(len(grid[0])):
upper_row_sum -= grid[0][idx]
min_max_value = min(min_max_value, max(upper_row_sum, lower_row_sum))
lower_row_sum += grid[1][idx]
return min_max_value
The given Python solution addresses a "Grid Game" problem, where two dimensional matrix of integers is used as input. The main objective is to determine the minimum possible value by moving through the grid based on specific rules. The solution employs an efficient algorithm to achieve this by leveraging a combination of sum arithmetic and loop-based calculations. Here's an explanation of how the given code achieves this:
- Initiate two accumulative variables (
upper_row_sum
andlower_row_sum
) to keep track of the sum of elements for top and bottom rows, respectively. - Start by calculating the sum of all elements in the upper row.
- As you iterate over each column through a loop:
- Subtract the current column's element in the top row from
upper_row_sum
, effectively updating the sum to represent only the sum of elements to the right of current column. - Compute the maximum of the current
upper_row_sum
andlower_row_sum
and then consider it for the minimum max value (min_max_value
) encountered so far in the loop, which is updated using the Pythonmin
function. - Add the current column's element in the bottom row to
lower_row_sum
, updating it to include all elements from the start to the current column.
- Subtract the current column's element in the top row from
- At the end of iterations,
min_max_value
holds the answer: the minimized maximum value possible from sum of rows based on described rules.
This approach ensures you analyze all relevant scenarios where the robot moves from left to right to induce minimum potential maximum row sum. The method efficiently calculates this with O(n) complexity, where n is the number of columns in the grid, making it suitable for large datasets.
No comments yet.