934. Shortest Bridge
Explanation
To solve this problem, we can follow these steps:
- Use depth-first search (DFS) to label the two islands separately with different numbers.
- Use breadth-first search (BFS) to expand from one island towards the other, keeping track of the number of flips needed to connect the two islands.
- Stop the BFS when we reach the other island, and return the number of flips needed.
Time complexity: O(n^2)
Space complexity: O(n^2)
class Solution {
public int shortestBridge(int[][] grid) {
int n = grid.length;
boolean[][] visited = new boolean[n][n];
Queue<int[]> queue = new LinkedList<>();
boolean found = false;
// DFS to label the two islands with different numbers
for (int i = 0; i < n; i++) {
if (found) break;
for (int j = 0; j < n; j++) {
if (grid[i][j] == 1) {
dfs(grid, i, j, visited, queue);
found = true;
break;
}
}
}
int steps = 0;
int[][] directions = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
// BFS to expand from one island towards the other
while (!queue.isEmpty()) {
int size = queue.size();
for (int i = 0; i < size; i++) {
int[] curr = queue.poll();
for (int[] dir : directions) {
int x = curr[0] + dir[0];
int y = curr[1] + dir[1];
if (x >= 0 && x < n && y >= 0 && y < n && !visited[x][y]) {
if (grid[x][y] == 1) {
return steps;
}
visited[x][y] = true;
queue.offer(new int[]{x, y});
}
}
}
steps++;
}
return -1;
}
private void dfs(int[][] grid, int i, int j, boolean[][] visited, Queue<int[]> queue) {
if (i < 0 || i >= grid.length || j < 0 || j >= grid[0].length || visited[i][j] || grid[i][j] == 0) {
return;
}
visited[i][j] = true;
queue.offer(new int[]{i, j});
dfs(grid, i + 1, j, visited, queue);
dfs(grid, i - 1, j, visited, queue);
dfs(grid, i, j + 1, visited, queue);
dfs(grid, i, j - 1, visited, queue);
}
}Code Editor (Testing phase)
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