Today I have done leetcode's June LeetCoding Challenge with cpp.

June LeetCoding Challenge 1

Description

Shortest Path in Binary Matrix

Given an n x n binary matrix grid, return the length of the shortest clear path in the matrix. If there is no clear path, return -1.

A clear path in a binary matrix is a path from the top-left cell (i.e., (0, 0)) to the bottom-right cell (i.e., (n - 1, n - 1)) such that:

• All the visited cells of the path are 0.
• All the adjacent cells of the path are 8-directionally connected (i.e., they are different and they share an edge or a corner).

The length of a clear path is the number of visited cells of this path.

Example 1:

Input: grid = [[0,1],[1,0]]
Output: 2

Example 2:

Input: grid = [[0,0,0],[1,1,0],[1,1,0]]
Output: 4

Example 3:

Input: grid = [[1,0,0],[1,1,0],[1,1,0]]
Output: -1

Constraints:

• n == grid.length
• n == grid[i].length
• 1 <= n <= 100
• grid[i][j] is 0 or 1

Solution

class Solution {
  int move[8][2] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1}};
public:
  int shortestPathBinaryMatrix(vector<vector<int>>& grid) {
    if(grid.front().front() || grid.back().back()) return -1;
    int n = grid.size();
    vector<vector<int>> dis(n, vector<int>(n));
    dis[0][0] = 1;
    queue<pair<int, int>> q;
    q.push(make_pair(0, 0));
    while(!q.empty()) {
      auto [row, col] = q.front();
      q.pop();
      for(int i = 0; i < 8; ++i) {
        int newRow = row + move[i][0];
        int newCol = col + move[i][1];
        if(newRow < 0 || newRow >= n || newCol < 0 || newCol >= n || grid[newRow][newCol]) continue;
        if(dis[newRow][newCol]) continue;
        dis[newRow][newCol] = dis[row][col] + 1;
        q.push(make_pair(newRow, newCol));
      }
    }
    return dis.back().back() ? dis.back().back() : -1;
  }
};

// Accepted
// 89/89 cases passed (69 ms)
// Your runtime beats 79.27 % of cpp submissions
// Your memory usage beats 61.94 % of cpp submissions (20.9 MB)