LeetCode 3459: Length of Longest V-Shaped Diagonal Segment Solution

Master LeetCode problem 3459 (Length of Longest V-Shaped Diagonal Segment), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

3459. Length of Longest V-Shaped Diagonal Segment

Hard

Array Dynamic Programming Memoization Matrix

Problem Explanation

Explanation:

To solve this problem, we can iterate through all possible starting points in the grid and explore in all 4 diagonal directions to find the longest V-shaped diagonal segment starting from that point. We can use DFS (Depth First Search) to explore the valid segments.

Start by iterating through each cell in the grid.
For each cell with value 1, perform a DFS in all 4 diagonal directions considering the clockwise 90-degree turn.
During the DFS, keep track of the length of the V-shaped diagonal segment.
Update the maximum length found so far.

Time Complexity: The time complexity of this approach is O(n^2) where n is the size of the grid.

Space Complexity: The space complexity is O(n^2) for the recursive DFS calls.

Solution Code

class Solution {
    public int longestVShape(int[][] grid) {
        int n = grid.length;
        int m = grid[0].length;
        int maxLen = 0;
        
        int[][] directions = {{-1, -1}, {-1, 1}, {1, -1}, {1, 1}};
        
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (grid[i][j] == 1) {
                    for (int[] dir : directions) {
                        maxLen = Math.max(maxLen, dfs(grid, i, j, dir[0], dir[1]));
                    }
                }
            }
        }
        
        return maxLen;
    }
    
    private int dfs(int[][] grid, int x, int y, int dx, int dy) {
        int len = 1;
        int n = grid.length;
        int m = grid[0].length;
        
        while (x + dx >= 0 && x + dx < n && y + dy >= 0 && y + dy < m && grid[x + dx][y + dy] == 2) {
            len++;
            x += dx;
            y += dy;
            int temp = dx;
            dx = dy;
            dy = temp;
        }
        
        return len;
    }
}

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