Length of Longest V-Shaped Diagonal Segment

1. Initialize maxLength = 0.
2. Define a helper function getExpectedValue(index) that returns the expected grid value for a given position in the sequence (1 for index 0, 2 for odd indices > 0, 0 for even indices > 0).
3. Create four 2D DP arrays of size n x m: L_tl, L_tr, L_bl, L_br to store lengths of straight segments ending at (r, c) from the four diagonal directions.
4. Pre-computation Pass:
   • Calculate L_tl and L_tr by iterating r from 0 to n-1 and c from 0 to m-1.
   • Calculate L_bl and L_br by iterating r from n-1 to 0 and c from m-1 to 0 (or any other valid traversal order).
   • For each cell (r, c), update the DP value based on the previous diagonal cell's DP value and grid[r][c]. If grid[r][c] == 1, a new segment of length 1 starts.
   • Update maxLength = max(maxLength, dp_value) after each calculation.
5. V-Shape Pass:
   • Iterate through each cell (r, c) from (0, 0) to (n-1, m-1).
   • For each of the four incoming directions (e.g., from Top-Left): a. Get the length of the first leg, len1, from the corresponding DP table (e.g., L_tl[r][c]). b. If len1 > 0: i. Determine the new direction (dr, dc) after a 90-degree clockwise turn. ii. Initialize len2 = 0, currentIndex = len1, currR = r, currC = c. iii. Loop: Move to (currR + dr, currC + dc). If it's in bounds and the value matches getExpectedValue(currentIndex), increment len2 and currentIndex, and update (currR, currC). Otherwise, break. iv. Update maxLength = max(maxLength, len1 + len2).
6. Return maxLength.

The core idea is to break down the problem into two main parts: finding all valid straight segments and then combining them to form V-shapes.

1. Pre-computation of Straight Segments: We use four 2D DP arrays, one for each of the four diagonal directions (Top-Left to Bottom-Right, Top-Right to Bottom-Left, etc.). Let's denote them L_tl, L_tr, L_bl, L_br. L_tl[r][c] will store the length of a valid straight diagonal segment ending at cell (r, c) and arriving from the top-left.

   The calculation is as follows: To compute L_tl[r][c], we look at the previous cell in that diagonal, (r-1, c-1). If a valid segment of length k ends there (i.e., L_tl[r-1][c-1] = k > 0), and the value at grid[r][c] matches the expected value at index k of the sequence (1, 2, 0, 2, 0, ...), then we can extend the segment. The new length at (r, c) becomes k + 1. If grid[r][c] is 1, it can start a new segment of length 1. Otherwise, no valid segment ends at (r, c) from this direction.

   We fill these four DP tables by iterating through the grid in appropriate orders. While filling, we also keep track of the maximum length found so far, which covers all straight-line segments (V-shapes with zero turns).

2. Finding V-Shapes: After the DP tables are filled, we iterate through every cell (r, c) in the grid, treating it as a potential pivot point of a V-shape. For each of the four possible incoming diagonal directions, we get the length of the first leg from our pre-computed DP tables (e.g., len1 = L_tl[r][c]).

   If a valid first leg exists (len1 > 0), we then simulate the second leg. A 90-degree clockwise turn from the incoming direction determines the direction of the second leg. We then iteratively traverse from (r, c) in this new direction, checking if the grid values match the required sequence, which now continues from index len1. The length of this second leg (len2) is calculated. The total length of the V-shape is len1 + len2. We update our global maximum length with this value.

This process is repeated for all cells and all four possible turns at each cell.

1. Initialize maxLength = 0, grid dimensions n, m.
2. Define direction vectors dr, dc for the four diagonal directions.
3. Create DP tables:
   • L[n][m][4]: For first legs (sequence 1, 2, 0, ...).
   • S[n][m][4]: For second legs starting with 2 (2, 0, 2, ...).
   • T[n][m][4]: For second legs starting with 0 (0, 2, 0, ...).
4. Compute L tables:
   • Iterate through the grid (e.g., top-to-bottom, left-to-right for directions from top) and fill L tables. Update maxLength with each L value.
5. Compute S and T tables:
   • Iterate through the grid in reverse orders (e.g., bottom-to-top, right-to-left for directions towards bottom-right) and fill S and T tables based on their recursive definitions.
6. Combine for V-shapes:
   • Iterate r from 0 to n-1 and c from 0 to m-1.
   • For each incoming direction d_in from 0 to 3: a. len1 = L[r][c][d_in]. b. If len1 > 0: i. d_out = (d_in + 1) % 4 (clockwise turn). ii. nextR = r + dr[d_out], nextC = c + dc[d_out]. iii. If (nextR, nextC) is in bounds: - expected = (len1 % 2 == 1) ? 2 : 0. - If grid[nextR][nextC] == expected: - len2 = (expected == 2) ? S[nextR][nextC][d_out] : T[nextR][nextC][d_out]. - maxLength = max(maxLength, len1 + len2).
7. Return maxLength.

This approach extends the previous one by eliminating the iterative search for the second leg of the V-shape. It does so by pre-calculating the lengths of all possible second legs as well.

We define three sets of DP tables:

1. L[r][c][dir]: Stores the length of a valid segment starting with 1 (1, 2, 0, ... sequence) that ends at (r, c), arriving from direction dir. This is identical to the DP tables in the previous approach and represents the first leg of a V-shape.

2. S[r][c][dir]: Stores the length of a valid segment that starts at (r, c) with the value 2 and continues with the sequence 0, 2, 0, ... along direction dir.

3. T[r][c][dir]: Stores the length of a valid segment that starts at (r, c) with the value 0 and continues with the sequence 2, 0, 2, ... along direction dir.

The S and T tables are mutually dependent. For example, to calculate S for a direction (e.g., towards bottom-right), S_br[r][c] will be 1 + T_br[r+1][c+1] if grid[r][c] is 2, and 0 otherwise. Similarly, T_br[r][c] depends on S_br at the next cell. These tables can be filled by iterating through the grid in the reverse order of the direction of traversal.

The Algorithm:

1. Pre-computation:

   • First, compute the four L tables, one for each incoming direction, just like in the previous approach. While doing so, update a maxLength variable to account for all straight-line segments.
   • Next, compute the S and T tables. For each of the four diagonal directions, we need one S and one T table (total of 8 tables). These are computed by iterating in the reverse direction (e.g., for segments going towards the bottom-right, we iterate from the bottom-right of the grid backwards).

2. Combination:

   • Iterate through every cell (r, c) of the grid, considering it as the pivot of a V-shape.
   • For each of the four possible incoming directions d_in: a. Get the length of the first leg: len1 = L[r][c][d_in]. b. If len1 > 0, it means a valid first leg ends at (r, c). c. The second leg starts at the cell adjacent to (r, c) in the turned direction d_out. The expected value at this next cell depends on len1. If len1 is odd, the next value should be 2; if even, it should be 0. d. We check if the value at the next cell matches the expectation. If it does, we can find the length of the rest of the second leg (len2) in O(1) time by looking it up in the pre-computed S or T tables for direction d_out at that next cell's coordinates. e. The total length is len1 + len2. We update maxLength with this combined length.

After checking all cells as potential pivots, maxLength will hold the length of the longest V-shaped segment.