How do you solve Count Routes to Climb a Rectangular Grid on LeetCode?

Count Routes to Climb a Rectangular Grid (LeetCode #3797) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * m * d) time complexity and O(n * m) space complexity. Key insight: Dynamic programming helps avoid redundant calculations.

What is the time complexity of Count Routes to Climb a Rectangular Grid?

The optimal solution for Count Routes to Climb a Rectangular Grid runs in O(n * m * d) time and uses O(n * m) space. The complexity is driven by the need to fill the DP table, iterating through rows, columns, and possible moves within distance d.

What is the brute force solution for Count Routes to Climb a Rectangular Grid?

The brute force approach for Count Routes to Climb a Rectangular Grid is Brute Force, with O(n²) time complexity and O(1) space complexity. We can explore all possible paths from the bottom to the top row, checking each cell's availability and distance constraints. This method is straightforward but inefficient due to the exponential number of paths. The optimal Optimal Solution approach improves this to O(n * m * d).

What algorithm or data structure is used to solve Count Routes to Climb a Rectangular Grid?

Count Routes to Climb a Rectangular Grid uses the following concepts: Array, Dynamic Programming, Matrix, Prefix Sum. The recommended patterns to study are: Dynamic Programming, Graph Traversal.

What are the interview tips for Count Routes to Climb a Rectangular Grid?

Explain your thought process clearly. Discuss edge cases and constraints upfront. Practice similar dynamic programming problems.

What are common mistakes when solving Count Routes to Climb a Rectangular Grid?

Not considering blocked cells properly. Ignoring distance constraints during path calculations.

#3797

Count Routes to Climb a Rectangular Grid

Hard

ArrayDynamic ProgrammingMatrixPrefix SumDynamic ProgrammingGraph Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

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

We can use dynamic programming to efficiently count paths by storing results of subproblems. This avoids redundant calculations and allows us to build up the solution from the bottom row to the top.

⚙️

Algorithm

3 steps

1Step 1: Initialize a DP table where dp[r][c][0] counts paths from below and dp[r][c][1] counts paths staying on the same row.
2Step 2: Iterate from the bottom row to the top, updating the DP table based on valid moves within distance d.
3Step 3: Sum paths from the top row to get the total count.

solution.py14 lines

1def countRoutes(grid, d):
2    n, m = len(grid), len(grid[0])
3    dp = [[[0, 0] for _ in range(m)] for _ in range(n)]
4    for c in range(m):
5        if grid[n-1][c] == '.': dp[n-1][c][0] = 1
6    for r in range(n-2, -1, -1):
7        for c in range(m):
8            if grid[r][c] == '.':
9                for nc in range(max(0, c-d), min(m, c+d+1)):
10                    if grid[r+1][nc] == '.':
11                        dp[r][c][0] += dp[r+1][nc][0]
12                        dp[r][c][1] += dp[r+1][nc][0]
13                dp[r][c][0] += dp[r][c][1]
14    return sum(dp[0][c][0] for c in range(m))

ℹ

Complexity note: The complexity is driven by the need to fill the DP table, iterating through rows, columns, and possible moves within distance d.

1Dynamic programming helps avoid redundant calculations.
2Understanding movement constraints is crucial for path counting.

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