How do you solve Out of Boundary Paths on LeetCode?

Out of Boundary Paths (LeetCode #576) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(m * n * maxMove) time complexity and O(m * n * maxMove) space complexity. Key insight: The number of paths grows exponentially with the number of moves.

What is the brute force solution for Out of Boundary Paths?

The brute force approach for Out of Boundary Paths is Brute Force, with O(4^maxMove) time complexity and O(maxMove) space complexity. The brute force approach involves exploring all possible paths the ball can take from its starting position. We recursively try to move the ball in all four directions until we either exceed the maximum number of moves or move out of the grid boundary. The optimal Optimal Solution approach improves this to O(m * n * maxMove).

What algorithm or data structure is used to solve Out of Boundary Paths?

Out of Boundary Paths uses the following concepts: Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Recursion.

What are the interview tips for Out of Boundary Paths?

Explain your thought process clearly before coding. Discuss edge cases and constraints upfront. Consider both time and space complexity in your solution.

What are common mistakes when solving Out of Boundary Paths?

Not considering out-of-bounds conditions properly. Failing to use memoization or DP to optimize recursive solutions.

#576

Out of Boundary Paths

Medium

Dynamic ProgrammingDynamic ProgrammingRecursion

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(4^maxMove)	O(m * n * maxMove)
Space	O(maxMove)	O(m * n * maxMove)

💡

Intuition

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

The optimal solution uses dynamic programming to store the number of ways to reach each cell in the grid for each number of moves left. This avoids recalculating paths for the same state multiple times, significantly improving efficiency.

⚙️

Algorithm

4 steps

1Step 1: Create a 3D DP array where dp[i][j][k] represents the number of ways to reach cell (i, j) with k moves left.
2Step 2: Initialize the base case where if the ball is out of bounds, it contributes 1 to the count.
3Step 3: Iterate through all possible moves from 0 to maxMove and update the DP table based on previous states.
4Step 4: Sum all the ways from the DP table for the starting position after maxMove moves.

solution.py12 lines

1def findPaths(m, n, maxMove, startRow, startColumn):
2    dp = [[[0] * (maxMove + 1) for _ in range(n)] for _ in range(m)]
3    mod = 10**9 + 7
4    for moves in range(1, maxMove + 1):
5        for r in range(m):
6            for c in range(n):
7                if r == 0: dp[r][c][moves] += 1
8                if r == m - 1: dp[r][c][moves] += 1
9                if c == 0: dp[r][c][moves] += 1
10                if c == n - 1: dp[r][c][moves] += 1
11                dp[r][c][moves] += (dp[r-1][c][moves-1] + dp[r+1][c][moves-1] + dp[r][c-1][moves-1] + dp[r][c+1][moves-1]) % mod
12    return dp[startRow][startColumn][maxMove] % mod

ℹ

Complexity note: The time complexity is O(m * n * maxMove) because we fill a 3D array where each dimension corresponds to the grid dimensions and the number of moves. The space complexity is also O(m * n * maxMove) due to the 3D DP array.

1The number of paths grows exponentially with the number of moves.
2Dynamic programming significantly reduces redundant calculations.

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