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-rw-r--r--far_from_land.c56
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+// 1162. As Far from Land as Possible
+#include <stdbool.h>
+#include <stdio.h>
+#include <stdlib.h>
+
+/*
+ * given 'n * n grid' containing only values 0 and 1, where 0 represents water
+ * and 1 represents land, find a water cell such that its distance to the
+ * nearest land cell is maximised, and returh the distance. if no land or water
+ * exists, return -1. the distance used is the manhattan distance: the distance
+ * between two cells '(x0, y0)' and '(x1, y1)' is '|x0 - x1| + |y0 - y1|'
+ */
+
+int maxDistance(int **grid, int gridSize, int *gridColSize) {
+ int queue[10000][2] = {0}, dir[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
+ int head = 0, tail = 0, step = -1;
+ bool has_land = false, has_water = false;
+ for (int i = 0; i < gridSize; i++)
+ for (int j = 0; j < *gridColSize; j++)
+ if (grid[i][j] == 1) {
+ has_land = true;
+ grid[i][j] = 2;
+ queue[tail][0] = i;
+ queue[tail][1] = j;
+ tail += 1;
+ } else
+ has_water = true;
+ while (head < tail) {
+ int size = tail - head;
+ step += 1;
+ for (int i = 0; i < size; i++) {
+ int x = queue[head][0], y = queue[head][1];
+ head += 1;
+ for (int j = 0; j < 4; j++) {
+ int new_x = x + dir[j][0], new_y = y + dir[j][1];
+ if (new_x >= 0 && new_y >= 0 && new_x < gridSize &&
+ new_y < *gridColSize && grid[new_x][new_y] == 0) {
+ grid[new_x][new_y] = 2;
+ queue[tail][0] = new_x;
+ queue[tail][1] = new_y;
+ tail += 1;
+ }
+ }
+ }
+ }
+ if (!has_land || !has_water)
+ return -1;
+ return step;
+}
+
+int main() {
+ int grid1[3][3] = {{1, 0, 1}, {0, 0, 0}, {1, 0, 1}}, gs1 = 3, gcs1[] = {3};
+ int grid2[3][3] = {{1, 0, 0}, {0, 0, 0}, {0, 0, 0}}, gs2 = 3, gcs2[] = {3};
+ printf("%d\n", maxDistance(grid1, gs1, gcs1)); // expect: 2
+ printf("%d\n", maxDistance(grid2, gs2, gcs2)); // expect: 2
+}