How do you solve Grid Game on LeetCode?

Grid Game (LeetCode #2017) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding the impact of the first robot's path on the second robot's collection is crucial.

What is the brute force solution for Grid Game?

The brute force approach for Grid Game is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves exploring all possible paths for the first robot and calculating the points collected by the second robot after the first robot has traversed its path. This is straightforward but inefficient as it checks every possible combination. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Grid Game?

Grid Game uses the following concepts: Array, Matrix, Prefix Sum. The recommended patterns to study are: Prefix Sum, Dynamic Programming.

What are the interview tips for Grid Game?

Clearly explain your thought process and the reasoning behind each step. Consider edge cases, such as minimum grid sizes, during your explanation. Practice explaining your code as you write it to simulate the interview environment.

What are common mistakes when solving Grid Game?

Not considering all possible paths for the first robot. Failing to optimize the calculation of points for the second robot.

#2017

Grid Game

Medium

ArrayMatrixPrefix SumPrefix SumDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution leverages prefix sums to efficiently calculate the maximum points the second robot can collect after the first robot has made its move. By precomputing the prefix sums, we can quickly determine the points collected by the second robot without simulating every path.

⚙️

Algorithm

4 steps

1Step 1: Compute prefix sums for both rows to facilitate quick calculations of points collected.
2Step 2: For each possible column index where the first robot can move down, calculate the total points collected by the second robot using the prefix sums.
3Step 3: The total points for the second robot is the sum of points in the first row up to the index and the second row from that index to the end.
4Step 4: Keep track of the maximum points collected by the second robot across all possible paths of the first robot.

solution.py14 lines

1def gridGame(grid):
2    n = len(grid[0])
3    prefix_row1 = [0] * n
4    prefix_row2 = [0] * n
5    prefix_row1[0] = grid[0][0]
6    prefix_row2[0] = grid[1][0]
7    for i in range(1, n):
8        prefix_row1[i] = prefix_row1[i-1] + grid[0][i]
9        prefix_row2[i] = prefix_row2[i-1] + grid[1][i]
10    max_points = 0
11    for i in range(n):
12        points = prefix_row1[i] + (prefix_row2[n-1] - prefix_row2[i])
13        max_points = max(max_points, points)
14    return max_points

ℹ

Complexity note: The time complexity is O(n) because we compute prefix sums in a single pass and then evaluate the points in another pass, leading to a linear complexity overall.

1Understanding the impact of the first robot's path on the second robot's collection is crucial.
2Using prefix sums allows for efficient calculation of points without redundant computations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.