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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 g1 = {{0, 1}, {1, 0}};
+ vvd g2 = {{0, 0, 0}, {1, 1, 0}, {1, 1, 0}};
+ vvd 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
+}