Longest Increasing Path in a Matrix
Given an $m \times n$ integers matrix, find the length of the longest increasing path. You can move up, down, left, or right. Use DFS with memoization (DP).
Why Interviewers Ask This
Uber interviewers use this problem to assess a candidate's ability to combine graph traversal with dynamic programming optimization. They specifically evaluate whether you can identify overlapping subproblems in a grid-based DFS and implement memoization to prevent exponential time complexity, reflecting Uber's need for scalable algorithmic solutions in logistics.
How to Answer This Question
1. Clarify the constraints: Ask about matrix dimensions, negative numbers, and movement rules (up/down/left/right) to confirm the grid is treated as a directed acyclic graph based on values.
2. Define the state: Explain that the recursive function needs to return the longest path starting from the current cell (i, j).
3. Propose the brute force: Briefly mention a pure DFS approach would be O(4^mn), which is too slow, highlighting the need for optimization.
4. Introduce Memoization: Detail how to use a 2D array to cache results for each cell, ensuring each subproblem is solved only once.
5. Iterate and optimize: Describe looping through every cell to trigger the DFS if not already computed, tracking the global maximum. Conclude by stating the time complexity becomes O(m*n) due to memoization.
Key Points to Cover
- Identifying the problem as finding the longest path in a DAG derived from the matrix
- Explicitly stating the transition from O(4^mn) brute force to O(mn) with memoization
- Defining the base case where no larger neighbors exist returns 1
- Demonstrating knowledge of space complexity trade-offs using a separate DP table
- Iterating over all cells to ensure the starting point of the longest path is considered
Sample Answer
To solve the Longest Increasing Path in a Matrix efficiently, I would treat each cell as a node in a graph where edges exist only to neighbors with strictly greater values. A naive depth-first search would explore all paths, leading to exponential time complexity because we repeatedly recalculate paths for the same cells. To optimize this, I will apply Dynamic Programming with Memoization.
First, I'll initialize an m x n DP table with zeros to store the longest path length found so far for each coordinate. Then, I'll iterate through every cell in the matrix. For each unvisited cell, I'll invoke a helper DFS function. This function checks the four possible directions (up, down, left, right). If a neighbor has a higher value, the function recursively calculates the path length from that neighbor. Crucially, before performing any calculation, the function checks the DP table; if a value exists, it returns it immediately, avoiding redundant work.
The result for the current cell is one plus the maximum result returned from valid neighbors. We update the DP table and track the global maximum across all iterations. This approach ensures that every cell is processed exactly once, reducing the time complexity from exponential to O(m * n). Given Uber's focus on real-time data processing, this linear scaling relative to input size demonstrates the efficiency required for their large-scale routing algorithms.
Common Mistakes to Avoid
- Forgetting to check visited states via a memoization table, resulting in Time Limit Exceeded errors
- Assuming the path must start at the top-left corner instead of checking every cell as a potential start
- Failing to handle boundary conditions when accessing adjacent matrix elements
- Confusing 'increasing' with 'non-decreasing', allowing equal values to extend the path incorrectly
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