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Length of Longest V-Shaped Diagonal Segment

Difficulty: Hard

Problem Description

You are given a 2D integer matrix grid of size n x m, where each element is either 0, 1, or 2. A V-shaped diagonal segment is defined as:

The segment starts with 1.
The subsequent elements follow the infinite sequence: 2, 0, 2, 0, ....
The segment starts along a diagonal direction (top-left to bottom-right, bottom-right to top-left, top-right to bottom-left, or bottom-left to top-right).
Continues the sequence in the same diagonal direction.
Makes at most one clockwise 90-degree turn to another diagonal direction while maintaining the sequence.

Return the length of the longest V-shaped diagonal segment. If no valid segment exists, return 0.

Key Insights

A valid V-shaped diagonal segment must start with 1.
The subsequent elements must alternate between 2 and 0.
The traversal can change direction only once, allowing for a maximum of two diagonal segments.
The possible diagonal movements can be represented as changes in row and column indices.

Space and Time Complexity

Time Complexity: O(n * m)
Space Complexity: O(1)

Solution

To find the longest V-shaped diagonal segment, we can use a brute-force approach combined with dynamic programming principles:

Iterate through each cell in the grid.
For each 1 found, attempt to extend the segment in all four diagonal directions.
Track the length of each segment while checking for valid transitions (from 2 to 0 and vice versa).
Allow a single turn to check for the potential of forming a longer segment.
Keep track of the maximum length encountered.

We will use a nested loop to go through the matrix and use direction vectors for diagonal movements.

Code Solutions

def longestVShapeSegment(grid):
    n, m = len(grid), len(grid[0])
    max_length = 0

    # Directions: (dx, dy) for top-left, top-right, bottom-left, bottom-right
    directions = [(-1, -1), (-1, 1), (1, -1), (1, 1)]
    
    for i in range(n):
        for j in range(m):
            if grid[i][j] == 1:
                # Check each direction
                for dx, dy in directions:
                    length = 1
                    x, y = i, j
                    turn_used = False
                    
                    while True:
                        x += dx
                        y += dy
                        if 0 <= x < n and 0 <= y < m:
                            if grid[x][y] == 2:
                                length += 1
                            elif grid[x][y] == 0:
                                break
                            else:
                                if not turn_used:
                                    # Try to turn in the opposite direction
                                    for turn_dx, turn_dy in directions:
                                        if (turn_dx, turn_dy) != (dx, dy):
                                            turn_x, turn_y = x, y
                                            while True:
                                                turn_x += turn_dx
                                                turn_y += turn_dy
                                                if 0 <= turn_x < n and 0 <= turn_y < m:
                                                    if grid[turn_x][turn_y] == 2:
                                                        length += 1
                                                    elif grid[turn_x][turn_y] == 0:
                                                        break
                                                    else:
                                                        turn_used = True
                                                else:
                                                    break
                                break
                        else:
                            break
                    max_length = max(max_length, length)

    return max_length

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