What is the most efficient way to count all paths in an mXn grid?
This problem explores combinatorial mathematics and pathfinding algorithms within a constrained grid environment. It tests logical reasoning and mathematical derivation skills.
Why Interviewers Ask This
Candidates are asked this to evaluate their grasp of combinatorics and basic recursion. Microsoft interviewers look for the ability to derive the mathematical formula nCr rather than just writing a brute-force recursive code. It demonstrates whether a candidate can optimize a solution from exponential time to constant time using mathematical insights.
How to Answer This Question
Begin by describing the movement constraints, typically moving only right or down. Propose a simple recursive solution first to show understanding of the traversal. Then, explain the observation that any valid path consists of exactly (m-1) downs and (n-1) rights. Derive the combination formula C(m+n-2, m-1). Discuss the implementation of factorials or iterative calculation to avoid overflow issues.
Key Points to Cover
- Total steps equal sum of rows and columns minus two
- Path counting reduces to a combination problem
- Mathematical formula avoids exponential recursion
- Iterative calculation prevents stack overflow
Sample Answer
In an mXn grid, moving only right or down, every path requires exactly m-1 downward moves and n-1 rightward moves. The total number of steps is always m+n-2. Therefore, the problem reduces to choosing which of these steps are 'down' moves. This is a direct application of combinations, specifically C(m+n-2, m-1). Instead of using recursion which is slow, I would calculate this using an iterative loop to compute the factorial ratio efficiently, ensuring the solution runs in linear time relative to the grid size.
Common Mistakes to Avoid
- Using recursion without memoization leading to TLE
- Incorrectly calculating the number of required moves
- Not handling edge cases where m or n is 1
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