How do you solve Maximum Number of Moves in a Grid on LeetCode?

Maximum Number of Moves in a Grid (LeetCode #2684) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(m * n) time complexity and O(m * n) space complexity. Key insight: Dynamic programming can significantly reduce the time complexity by avoiding redundant calculations.

What is the brute force solution for Maximum Number of Moves in a Grid?

The brute force approach for Maximum Number of Moves in a Grid is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we will explore all possible paths starting from each cell in the first column. For each cell, we will recursively check all valid moves and count the maximum moves we can make. The optimal Optimal Solution approach improves this to O(m * n).

What algorithm or data structure is used to solve Maximum Number of Moves in a Grid?

Maximum Number of Moves in a Grid uses the following concepts: Array, Dynamic Programming, Matrix. The recommended patterns to study are: Dynamic Programming, Matrix Traversal.

What are the interview tips for Maximum Number of Moves in a Grid?

Always clarify the problem constraints and edge cases before diving into the solution. Discuss your thought process and approach before coding to ensure clarity. Test your solution with various test cases, including edge cases, to validate correctness.

What are common mistakes when solving Maximum Number of Moves in a Grid?

Not considering edge cases where no moves can be made. Failing to properly initialize the DP table or forgetting to check bounds during transitions.

#2684

Maximum Number of Moves in a Grid

Medium

ArrayDynamic ProgrammingMatrixDynamic ProgrammingMatrix Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(m * n)Space O(m * n)

In the optimal approach, we use dynamic programming to store the maximum moves possible from each cell. This avoids redundant calculations and allows us to build solutions incrementally.

⚙️

Algorithm

3 steps

1Step 1: Create a DP table where dp[i][j] represents the maximum moves from cell (i, j).
2Step 2: Initialize the first column of the DP table to 0 since no moves can be made from the first column.
3Step 3: Iterate through the grid from left to right, updating the DP table based on valid moves from the previous column.

solution.py9 lines

1def maxMoves(grid):
2    m, n = len(grid), len(grid[0])
3    dp = [[0] * n for _ in range(m)]
4    for j in range(1, n):
5        for i in range(m):
6            for r in [i-1, i, i+1]:
7                if 0 <= r < m and grid[r][j] > grid[i][j-1]:
8                    dp[i][j] = max(dp[i][j], dp[r][j-1] + 1)
9    return max(dp[i][-1] for i in range(m))

ℹ

Complexity note: The time complexity is O(m * n) because we iterate through each cell in the grid and check up to 3 possible previous cells for each move. The space complexity is also O(m * n) due to the DP table.

1Dynamic programming can significantly reduce the time complexity by avoiding redundant calculations.
2Understanding the movement constraints helps in designing the DP state transitions.

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