# Java solution using BFS

• ``````public int shortestDistance(int[][] grid) {
int[][] dist = new int[grid.length][grid[0].length];
int walk = 1, minDist = -1;
for (int i = 0; i < grid.length; i++) {
for (int j = 0; j < grid[i].length; j++) {
// if the current cell is a building time to find the adjacent
// free space
if (grid[i][j] == 1) {
// travesing to all 0 from first building, then to all -1 from second building and so on.
minDist = bfs(dist, grid, --walk, i, j);
}
}
}

return minDist;
}

// Iterative BFS
private int bfs(int[][] dist, int[][] grid, int walk, int i, int j) {
queue.offer(new int[] { i, j });

int depth = 0, minDist = -1;
int[][] delta = { { 0, 1 }, { 0, -1 }, { 1, 0 }, { -1, 0 } };

while (!queue.isEmpty()) {
depth++;
int lenght = queue.size();

// this loop is required so that same level depth cell can be visited
for (int k = 0; k < lenght; k++) {
int[] pos = queue.poll();
for (int l = 0; l < 4; l++) {
int y = pos[0] + delta[l][0];
int x = pos[1] + delta[l][1];

// if the adjacent cell is out of bound or unreachable
if (y < 0 || y >= grid.length || x < 0 || x >= grid[y].length || grid[y][x] != walk) {
continue;
}

grid[y][x] = walk - 1;

// calculates distence from each building to other building.
dist[y][x] += depth;

queue.offer(new int[] { y, x });

// This will always get the min val assigned.
if (minDist < 0 || minDist > dist[y][x]) {
minDist = dist[y][x];
}
}
}
}

return minDist;
}
``````

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