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// 1091. Shortest Path in Binary Matrix
#include "leetcode.h"
/*
* 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 (ie. 0,0) to the bottom-right
* cell (ie. n - 1, n - 1) such thatyt
* - all the visited cells are of the path 0
* - all the adjacent cells of the path are 8-directionally connected (ie. they
* are different and they share an edge of a corner) the length of a clear path
* is the number of visited cells of this path
*/
class Solution {
public:
int shortestPathBinaryMatrix(vector<vector<int>> &grid) {
if (grid[0][0] || grid.back().back())
return -1;
int ans = 2, n = 1, nr = grid[0].size() - 1, nc = grid.size() - 1;
queue<pair<int, int>> qp;
if (!nr && !nc)
return 1 - (grid[0][0] << 1);
qp.push({0, 0});
grid[0][0] = -1;
while (n) {
while (n--) {
auto [cx, cy] = qp.front();
qp.pop();
for (int i = max(0, cx - 1), lmtx = min(cx + 1, nr); i <= lmtx; i++)
for (int j = max(0, cy - 1), lmty = min(cy + 1, nc); j <= lmty; j++) {
if (i == nr && j == nc)
return ans;
if (!grid[j][i]) {
grid[j][i] = -1;
qp.push({i, j});
}
}
}
ans++;
n = qp.size();
}
return -1;
}
};
int main() {
Solution obj;
vvd(int) g1 = {{0, 1}, {1, 0}};
vvd(int) g2 = {{0, 0, 0}, {1, 1, 0}, {1, 1, 0}};
vvd(int) g3 = {{1, 0, 0}, {1, 1, 0}, {1, 1, 0}};
printf("%d\n", obj.shortestPathBinaryMatrix(g1)); // expect: 2
printf("%d\n", obj.shortestPathBinaryMatrix(g2)); // expect: 4
printf("%d\n", obj.shortestPathBinaryMatrix(g3)); // expect: -1
}
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