How to fin the shortest pass in matrix C++ - algorithm

Question

Given a matrix of 0 and 1 (0 is free space, 1 is wall). Find the shortest path from one cell to another, passing only through 0 and also without touching 1.
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How can I do this using Lee's Algorithm?

Answer 1

class Solution {
public:
int shortestPathBinaryMatrix(vector<vector<int>>& grid) {
// edge case: start or end not accessible
if (grid[0][0] || grid.back().back()) return -1;
// support variables
int res = 2, len = 1, maxX = grid[0].size() - 1, maxY = grid.size() - 1;
queue<pair<int, int>> q;
// edge case: single cell matrix
if (!maxX && !maxY) return 1 - (grid[0][0] << 1);
// adding the starting point
q.push({0, 0});
// marking start as visited
grid[0][0] = -1;
while (len) {
while (len--) {
// reading and popping the coordinates on the front of the queue
auto [cx, cy] = q.front();
q.pop();
for (int x = max(0, cx - 1), lmtX = min(cx + 1, maxX); x <= lmtX; x++) {
for (int y = max(0, cy - 1), lmtY = min(cy + 1, maxY); y <= lmtY; y++) {
// check if we reached the target
if (x == maxX && y == maxY) return res;
// marking it as visited and adding it to the q if it was still a valid cell
if (!grid[y][x]) {
grid[y][x] = -1;
q.push({x, y});
}
}
}
}
// preparing for the next loop
res++;
len = q.size();
}
return -1;
}
};