Twisted Mirror Path Count - Solution & Explanation

Problem Overview: You are given a matrix where each cell may contain a mirror that twists the direction of movement. Starting from the top-left cell, you need to count how many valid ways reach the bottom-right cell while respecting how mirrors redirect the path.

Approach 1: Brute Force DFS Traversal (Exponential Time, O(2^(m+n)) time, O(m+n) space)

The most direct way is to simulate every possible path from the start cell. At each position, you follow the allowed movement determined by the mirror orientation in that cell. A recursive DFS explores both possible directions whenever the mirror allows branching. This approach is useful for understanding how the mirror redirection affects movement across the matrix. However, many subproblems repeat because different paths often reach the same cell with identical conditions.

Approach 2: DFS with Memoization (Dynamic Programming) (O(m*n) time, O(m*n) space)

Instead of recomputing the number of ways from the same cell repeatedly, store results in a memo table. When DFS reaches a cell (i, j), compute the number of valid paths from that state once and cache it. If the same state appears again, return the stored value immediately. This converts the exponential recursion into a linear number of states equal to the number of cells. The idea mirrors classic grid path counting but adjusts transitions depending on the mirror orientation stored in the grid. Memoization works well here because the number of unique states in the dynamic programming formulation is bounded by m * n.

Approach 3: Bottom-Up DP on the Grid (O(m*n) time, O(m*n) space)

A tabulation approach builds the solution iteratively. Create a DP table where dp[i][j] represents the number of ways to reach cell (i, j). Initialize the start cell with 1. Then iterate through the grid row by row. For each cell, propagate its path count to the next cell(s) determined by the mirror orientation. For example, a mirror may redirect the path to the right neighbor or the cell below. This avoids recursion and gives predictable performance. Since the grid structure is fixed, each state contributes to at most two transitions, making the total work proportional to the number of cells.

Approach 4: Space Optimized DP (O(m*n) time, O(n) space)

If transitions only depend on the current row and the next row, the DP table can be compressed into a rolling array. Instead of storing the entire m x n table, maintain only the current row and update values while iterating through the grid. This technique is common in array-based grid DP problems where dependencies are limited to adjacent cells.

Recommended for interviews: The bottom-up dynamic programming approach is usually what interviewers expect. A brute force DFS demonstrates that you understand the path rules and mirror behavior, but the DP optimization shows you recognize overlapping subproblems and can reduce the complexity to O(m*n).