Shortest Path in Binary Matrix
Shortest Path in Binary Matrix is a medium graphs problem solved with the bfs for shortest path pattern.
The best approach, optimal (bfs), runs in O(n²) time and O(n²) space.
Below are 2 approaches in Java, from dfs exploring all paths up.
Problem
In an n × n binary grid, find the length of the shortest clear path from the top-left to the bottom-right cell, moving in any of 8 directions through 0 cells only. The length counts cells visited. Return -1 if there is no path.
Examples
Example 1
- Input
grid = [[0,0,0],[1,1,0],[1,1,0]]- Output
4
Example 2
- Input
grid = [[1,0],[0,0]]- Output
-1- Why
- The start cell is blocked.
Constraints
1 <= n <= 100; cells are 0 (open) or 1 (blocked).- 8-directional moves; path length counts cells.
Animated walkthrough
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Solutions
Try it yourself first. Then compare: each approach lists its idea, the steps, its time and space, and the Java code.
| Approach | Time | Space |
|---|---|---|
| DFS exploring all paths | O(8^(n²)) | O(n²) |
| Optimal (BFS) | O(n²) | O(n²) |
1DFS exploring all paths
O(8^(n²))O(n²)Try every simple path with backtracking and keep the shortest. Correct, but exponential.
- DFS from (0, 0), marking cells on the current path, updating the best when reaching the end.
class Solution {
private int best;
public int shortestPathBinaryMatrix(int[][] grid) {
int n = grid.length;
if (grid[0][0] == 1 || grid[n - 1][n - 1] == 1) return -1;
best = Integer.MAX_VALUE;
dfs(grid, 0, 0, 1);
return best == Integer.MAX_VALUE ? -1 : best;
}
private void dfs(int[][] g, int r, int c, int len) {
int n = g.length;
if (len >= best) return;
if (r == n - 1 && c == n - 1) { best = len; return; }
g[r][c] = 1;
for (int dr = -1; dr <= 1; dr++)
for (int dc = -1; dc <= 1; dc++) {
int nr = r + dr, nc = c + dc;
if (nr >= 0 && nc >= 0 && nr < n && nc < n && g[nr][nc] == 0) dfs(g, nr, nc, len + 1);
}
g[r][c] = 0;
}
}2Optimal (BFS)
O(n²)O(n²)BFS from the top-left over 8 directions, storing each cell's distance. The first time the bottom-right is dequeued, its distance is the shortest path length.
- If start or end is blocked, return -1.
- Queue (0, 0) with distance 1; mark visited by setting the cell to 1.
- Expand 8 neighbours; return the distance on reaching the end.
class Solution {
public int shortestPathBinaryMatrix(int[][] grid) {
int n = grid.length;
if (grid[0][0] == 1 || grid[n - 1][n - 1] == 1) return -1;
Queue<int[]> q = new ArrayDeque<>();
q.add(new int[] { 0, 0, 1 });
grid[0][0] = 1;
while (!q.isEmpty()) {
int[] c = q.poll();
if (c[0] == n - 1 && c[1] == n - 1) return c[2];
for (int dr = -1; dr <= 1; dr++)
for (int dc = -1; dc <= 1; dc++) {
int r = c[0] + dr, k = c[1] + dc;
if (r >= 0 && k >= 0 && r < n && k < n && grid[r][k] == 0) {
grid[r][k] = 1;
q.add(new int[] { r, k, c[2] + 1 });
}
}
}
return -1;
}
}Edge cases to test
- Start or end blocked
- n = 1
Hints
Hint 1
Every move costs the same, so BFS reaches the end first along a shortest path.
FAQ
What is the best time complexity for Shortest Path in Binary Matrix?
Optimal (BFS) runs in O(n²) time and O(n²) extra space.
Which pattern does Shortest Path in Binary Matrix use?
It is a graphs problem that uses the bfs for shortest path pattern.
Is there a brute force solution for Shortest Path in Binary Matrix?
Yes. DFS exploring all paths takes O(8^(n²)) time and O(n²) space. Try every simple path with backtracking and keep the shortest.
Which edge cases should I test for Shortest Path in Binary Matrix?
Start or end blocked; n = 1.