I knew this was DFS with memoization pretty fast, the tricky part is realizing you don't need visited tracking because the strictly increasing constraint already prevents cycles.
Use depth-first search with memoization to compute the longest increasing path starting from each cell. For each cell, recursively explore all four directions where the neighbor's value is strictly greater, caching the result to avoid redundant computations. The answer is the maximum path length found across all cells.
Pro tip: Emphasize that memoization reduces the time complexity from exponential to O(m*n), and mention that this is essentially finding the longest path in a directed acyclic graph (DAG) formed by the increasing condition. Also, note that you can avoid recursion depth issues by using iterative topological sort if needed.
Confirm that the path must be strictly increasing, moves are only up/down/left/right, and the grid dimensions. Ask about edge cases like empty grid or single cell.
For a cell (i,j), the longest increasing path starting there is 1 + max(longest path from valid neighbors). Valid neighbors are those within bounds and with value > grid[i][j].
Use a memo table (2D array) initialized to 0. For each cell, if memo[i][j] is not computed, recursively compute it by exploring four directions and taking the maximum. Store and return the result.
Loop through every cell, call the DFS function, and keep track of the global maximum path length. Return that maximum as the answer.
Time complexity is O(m*n) because each cell is visited once. Space complexity is O(m*n) for memoization and recursion stack. Mention that iterative topological sort is an alternative.
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