Leetcode-3459

Length of Longest V-Shaped Diagonal Segment

Time complexity: $O(mn)$

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#include <iostream>
#include <vector>

class Solution {
public:
    int lenOfVDiagonal(std::vector<std::vector<int>>& grid) {
        int m = grid.size();
        int n = grid[0].size();
        
        std::memset(lens, 0, sizeof(lens));
        // lens.resize(m, std::vector<std::vector<std::pair<int, int>>>(n, std::vector<std::pair<int, int>>(4, std::pair<int, int>({0, 0}))));
        int max_len = 0;
        
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == 1) {
                    for (int k = 0; k < 4; ++k) {
                        max_len = std::max(max_len, DFS(grid, m, n, i, j, k, 1));
                    }
                }
            }
        }
        
        return max_len;
    }
    
private:
    int DFS(std::vector<std::vector<int>>& grid, int m, int n, int i, int j, int k, int t) {
        int ii = i + dirs[k][0];
        int jj = j + dirs[k][1];
        if (!lens[i][j][k].first) {
            lens[i][j][k].first = 1;
            if (ii >= 0 && ii < m && jj >= 0 && jj < n && (grid[i][j] * 2 - 3) * (grid[ii][jj] - 1) < 0) {
                lens[i][j][k].first += DFS(grid, m, n, ii, jj, k, 0);
            }
        }
        
        if (t) {
            if (!lens[i][j][k].second) {
                lens[i][j][k].second = 1;
                if (ii >= 0 && ii < m && jj >= 0 && jj < n && (grid[i][j] * 2 - 3) * (grid[ii][jj] - 1) < 0) {
                    lens[i][j][k].second += DFS(grid, m, n, ii, jj, k, 1);
                }
                lens[i][j][k].second = std::max(lens[i][j][k].second, DFS(grid, m, n, i, j, (k + 1) % 4, 0));
            }
            return lens[i][j][k].second;
        }
        else {
            return lens[i][j][k].first;
        }
    }
    
    std::pair<int, int> lens[500][500][4];
    const int dirs[4][2] = { {1, 1}, {1, -1}, {-1, -1}, {-1, 1} };
};

Leetcode-3459

Licensed under CC BY-NC-SA 4.0