> Here is a complete editorial breaking down the mathematical intuition, the algorithmic approach, tradeoffs, and the C++ implementation. > Intuition: The "Infinite Loop" Illusion > At first glance, the rule that allows "revisiting a cell" suggests the path could loop infinitely, which would normally require complex cycle-detection or graph algorithms. However, a closer mathematical inspection of the transition rule reveals this is impossible. > Let a valid path be a sequence of values > > $v_1, v_2, v_3, \dots, v_k$ > . The problem states that for a new cell > $v_i$ to be valid (where > $i \ge 3$ ), it must satisfy: > > $v_i < v_{i-1}$ > OR > > $v_i < v_{i-2}$ > This is logically equivalent to stating that > > $v_i$ > must be strictly less than the maximum of the two previously visited values: > > $$v_i < \max(v_{i-1}, v_{i-2})$$ > Because the next value is always strictly smaller than the maximum of the previous adjacent pair, the "maximum adjacent value" in our path is strictly decreasing. You cannot traverse a cycle because returning to the same sequence of cells would require the values to eventually increase or remain perfectly flat, violating the strict inequality. > Therefore, the state space forms a > > Directed Acyclic Graph (DAG) > . When working with a DAG, we can safely use Depth First Search (DFS) with Memoization without fear of infinite recursion. > The Approach: 3D DP / Memoized DFS > Since the validity of the next move depends on the current cell and the > > immediately preceding > cell, our DP state must capture both. 1. > > State Definition: > Instead of storing the exact coordinates of the previous cell (which is clunky), we only need to know the > > direction > we moved to arrive at the current cell. Our state is > (r, c, prev_dir) , where > prev_dir is an integer > $0 \dots 3$ representing Up, Right, Down, and Left. 2. > > Transitions: > From > > (r, c, prev_dir) > , we calculate the value of the previous cell. Then, we look at all 4 valid neighbors. If a neighbor's value is strictly less than the current cell's value or the previous cell's value, we recursively call our DFS. 3. > > Base Case / Start: > The first move has a special strict rule: > > $v_2 < v_1$ > . We iterate over every cell in the grid, check its 4 neighbors for a valid first move, and initiate our DFS from there. > Tradeoffs and Alternatives > DFS + Memoization vs. BFS > > > DFS (Chosen): > > Explores a path to its end, caching the longest path from that state backward. It perfectly fits the DAG structure and avoids redundant state processing natively through the memo table. - > BFS: Standard BFS calculates shortest paths. To find the > longest path in a DAG using BFS, you would need Kahn's Algorithm (Topological Sort). Constructing an explicit adjacency list for all > $R \times C \times 4$ states just to run Kahn's algorithm would introduce massive memory and time overhead. > State Representation - We track > prev_dir ( > $\mathcal{O}(1)$ space per cell) rather than > prev_r and > prev_c . Storing previous coordinates implicitly creates an > $R \times C \times 4$ state space, but doing so explicitly as a 4D array > memo[r][c][pr][pc] would demand > $\mathcal{O}((R \times C)^2)$ memory, which would result in a Memory Limit Exceeded (MLE) error on large grids. > C++ Implementation > > C++ ``` > # > > include > > > # > > include > > > using > namespace > std ; > > class > > Solution > { > // Directions: 0=Up, 1=Right, 2=Down, 3=Left > int dr[ > 4 ] = { > -1 , > 0 , > 1 , > 0 }; > int dc[ > 4 ] = { > 0 , > 1 , > 0 , > -1 }; > int R, C; > // memo[r][c][prev_dir] stores the max additional cells we can visit > vector < > vector < > vector < > int >>> memo; > int > dfs ( > int r, > int c, > int prev_dir, > const > vector < > vector < > int >>& grid) { > if (memo[r][c][prev_dir] != > -1 ) { > return memo[r][c][prev_dir]; } > // Reconstruct the previous cell's coordinates based on where we came from > int pr = r + dr[prev_dir]; > int pc = c + dc[prev_dir]; > int val_curr = grid[r][c]; > int val_prev = grid[pr][pc]; > int max_extra = > 0 ; > // Try all 4 directions for the next move > for ( > int nd = > 0 ; nd < > 4 ; ++nd) { > int nr = r + dr[nd]; > int nc = c + dc[nd]; > if (nr >= > 0 && nr < R && nc >= > 0 && nc < C) { > int val_next = grid[nr][nc]; > // The core transition rule > if (val_next < val_curr || val_next < val_prev) { > // If we move in direction 'nd', the next cell sees us > // coming from the opposite direction. > int next_prev_dir = (nd + > 2 ) % > 4 ; max_extra = max(max_extra, > 1 + dfs(nr, nc, next_prev_dir, grid)); } } } > return memo[r][c][prev_dir] = max_extra; } > public : > int > maxPathLength ( > vector < > vector < > int >>& grid) { > if (grid.empty() || grid[ > 0 ].empty()) > return > 0 ; R = grid.size(); C = grid[ > 0 ].size(); > // Initialize 3D memo table with -1 memo.assign(R, > vector < > vector < > int >>(C, > vector < > int >( > 4 , > -1 ))); > int max_len = > 1 ; > // Minimum path length (just the starting cell) > // Try starting the path from every possible cell > for ( > int r = > 0 ; r < R; ++r) { > for ( > int c = > 0 ; c < C; ++c) { > // Try all 4 possible first moves > for ( > int nd = > 0 ; nd < > 4 ; ++nd) { > int nr = r + dr[nd]; > int nc = c + dc[nd]; > if (nr >= > 0 && nr < R && nc >= > 0 && nc < C) { > // The explicit rule for the very first move: b < a > if (grid[nr][nc] < grid[r][c]) { > int next_prev_dir = (nd + > 2 ) % > 4 ; > // 2 cells visited so far: the start cell + this first move max_len = max(max_len, > 2 + dfs(nr, nc, next_prev_dir, grid)); } } } } } > return max_len; } }; ``` > Complexity Analysis - > > Time Complexity: > > $\mathcal{O}(R \times C)$ There are > $R \times C$ cells in the grid, and each cell has exactly > $4$ possible entry directions. This gives us a maximum of > $4 \times R \times C$ states. Resolving each state requires checking > $4$ neighbors, an > $\mathcal{O}(1)$ operation. Thus, the total time bound is directly proportional to the number of cells. - > > Space Complexity: > > $\mathcal{O}(R \times C)$ The 3D > memo array requires exactly > $R \times C \times 4$ integers. The recursion stack for the DFS will go at most as deep as the longest possible path, which is bounded by the total number of states, strictly keeping space complexity linear with respect to the grid size.

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