[LeetCode] Problem 934 - Shortest Bridge

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In a given 2D binary array A, there are two islands. (An island is a 4-directionally connected group of 1s not connected to any other 1s.)

Now, we may change 0s to 1s so as to connect the two islands together to form 1 island.

Return the smallest number of 0s that must be flipped. (It is guaranteed that the answer is at least 1.)

Example

No.1

Input: [[0,1],[1,0]]

Output: 1

No.2

Input: [[0,1,0],[0,0,0],[0,0,1]]

Output: 2

No.3

Input: [[1,1,1,1,1],[1,0,0,0,1],[1,0,1,0,1],[1,0,0,0,1],[1,1,1,1,1]]

Output: 1

Note

  1. 1 <= A.length = A[0].length <= 100
  2. A[i][j] == 0 or A[i][j] == 1

Code

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public int shortestBridge(int[][] A) {
    int result = 0;
    int m = A.length;
    int n = A[0].length;
    int[][] directions = new int[][] {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
    Queue<int[]> queue = new LinkedList<>();

    findIsland:
    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            if (A[i][j] == 1) {
                dfs(queue, A, i, j, m, n);
                break findIsland;
            }
        }
    }

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

        for (int i = 0; i < size; i++) {
            int[] pos = queue.poll();

            for (int[] direction : directions) {
                int x = pos[0] + direction[0];
                int y = pos[1] + direction[1];

                if (x < 0 || y < 0 || x >= m || y >= n || A[x][y] == 2)
                    continue;

                if (A[x][y] == 1)
                    return result;

                A[x][y] = 2;
                queue.offer(new int[] {x, y});
            }
        }

        result++;
    }

    return -1;
}

private void dfs(Queue<int[]> queue, int[][] A, int x, int y, int m, int n) {
    if (x < 0 || y < 0 || x >= m || y >= n || A[x][y] != 1)
        return;

    queue.offer(new int[] {x, y});
    A[x][y] = 2;
    dfs(queue, A, x + 1, y, m, n);
    dfs(queue, A, x - 1, y, m, n);
    dfs(queue, A, x, y + 1, m, n);
    dfs(queue, A, x, y - 1, m, n);
}