Shortest Bridge - Google Top Interview Questions


Problem Statement :


Given a two-dimensional list of integers matrix containing 0s and 1s, 0 represents water and 1 represents land.

An island is a group of connecting 1s in 4 directions that are either surrounded by 0s or by the edges.
Find the shortest bridge that connects two islands.

It is guaranteed that there are two and only two islands.

Constraints

n, m ≤ 250 where n and m are the number of rows and columns in matrix

Example 1

Input

matrix = [

    [0, 1],

    [1, 0]

]

Output

1

Explanation

Either of the two water elements can be used as the bridge.



Example 2

Input

matrix = [

    [1, 0, 0],

    [0, 0, 0],

    [0, 0, 1]

]

Output

3

Explanation

There's six shortest bridges such as [(0, 1), (0, 2), (1, 2)] or [(1, 0), (2, 0), (2, 1)]



Solution :



title-img




                        Solution in C++ :

void dfs(vector<vector<int>> &matrix, int x, int y, queue<pair<int, int>> &q) {
    // Invalid position or cell-state
    if (x < 0 || y < 0 || x >= matrix.size() || y >= matrix[0].size() || matrix[x][y] == 2 ||
        matrix[x][y] < 0) {
        return;
    }

    // If the cell is a water cell adjacent to the island
    // mark it and place it into the bfs queue and return
    if (matrix[x][y] == 0) {
        matrix[x][y] = -1;
        q.emplace(make_pair(x, y));
        return;
    }

    // Mark island cell with different color
    if (matrix[x][y] == 1) {
        matrix[x][y] = 2;
    }

    // run dfs to mark neighbors
    vector<vector<int>> dirs{{-1, 0}, {0, -1}, {1, 0}, {0, 1}};

    for (auto &dir : dirs) {
        int x1 = x + dir[0];
        int y1 = y + dir[1];

        dfs(matrix, x1, y1, q);
    }
}

int solve(vector<vector<int>> &matrix) {
    int m = matrix.size(), n = m == 0 ? 0 : matrix[0].size();

    // Find the first land cell (can be any of the two islands)
    // and run dfs to mark the island with a different color
    bool landFound = false;
    queue<pair<int, int>> q;
    for (int x = 0; x < m && !landFound; ++x) {
        for (int y = 0; y < n && !landFound; ++y) {
            if (matrix[x][y] == 1) {
                landFound = true;
                dfs(matrix, x, y, q);
            }
        }
    }

    // Means that there's no island at all
    // so return error
    if (q.empty()) {
        return -1;
    }

    // Run bfs from the edges of the marked island
    vector<vector<int>> dirs{{-1, 0}, {0, -1}, {1, 0}, {0, 1}};

    while (!q.empty()) {
        pair<int, int> curr = q.front();
        q.pop();

        for (auto &dir : dirs) {
            int x = curr.first + dir[0];
            int y = curr.second + dir[1];

            if (x >= 0 && y >= 0 && x < m && y < n) {
                // If we have found the second island
                // return the distance
                // distance is -ve of the current cell
                // since we've marked water cells with
                // increasing negative values
                if (matrix[x][y] == 1) {
                    return 0 - matrix[curr.first][curr.second];
                }
                // If it is a water cell, update its distance
                // and then add it to the queue
                else if (matrix[x][y] == 0) {
                    matrix[x][y] = matrix[curr.first][curr.second] - 1;
                    q.emplace(make_pair(x, y));
                }
            }
        }
    }

    // We didn't find the second island so return error
    return -1;
}
                    


                        Solution in Java :

import java.util.*;

class Solution {
    private static final int[] DIRS = {-1, 0, 1, 0, -1};
    private static final int NEWCOLOR = 2;

    public int solve(int[][] grid) {
        final int R = grid.length, C = grid[0].length;

        Queue<int[]> q = new ArrayDeque<>();
        findIsland(grid, R, C, q);

        for (int step = 0; q.isEmpty() == false; step++) {
            for (int b = q.size(); b != 0; b--) {
                int[] now = q.poll();
                for (int i = 0; i != 4; i++) {
                    final int nr = now[0] + DIRS[i], nc = now[1] + DIRS[i + 1];
                    if (nr == -1 || nr == R || nc == -1 || nc == C || grid[nr][nc] == NEWCOLOR)
                        continue;
                    if (grid[nr][nc] == 1)
                        return step;
                    grid[nr][nc] = NEWCOLOR;
                    q.offer(new int[] {nr, nc});
                }
            }
        }

        return -1;
    }

    private void findIsland(int[][] grid, final int R, final int C, Queue<int[]> q) {
        for (int r = 0; r != R; r++)
            for (int c = 0; c != C; c++)
                if (grid[r][c] == 1) {
                    drown(grid, R, C, r, c, NEWCOLOR, q);
                    return;
                }
    }

    private void drown(int[][] grid, int R, int C, int r, int c, int color, Queue<int[]> q) {
        if (r == R || c == C || r == -1 || c == -1 || grid[r][c] != 1)
            return;
        grid[r][c] = color;
        q.offer(new int[] {r, c});
        for (int i = 0; i != 4; i++) drown(grid, R, C, r + DIRS[i], c + DIRS[i + 1], color, q);
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, matrix):
        cnt = 0

        rows = len(matrix)
        columns = len(matrix[0])
        visited = {}

        for i in range(2):
            flag = True

            while flag:
                column = cnt % columns
                row = cnt // columns

                if (row, column) not in visited:
                    visited[(row, column)] = 0

                    if matrix[row][column] == 1:
                        flag = False
                        matches = {(row, column): 0}
                        self.find_touching(matrix, visited, matches, row, column, rows, columns)

                cnt += 1

            if i == 0:
                island_one = self.find_outline(matches)
                print(island_one)
            else:
                island_two = self.find_outline(matches)
                print(island_two)

        best = -1

        for pos1 in island_one:
            for pos2 in island_two:
                needed = abs(pos1[0] - pos2[0]) + abs(pos1[1] - pos2[1]) - 1
                if best == -1 or needed < best:
                    best = needed

        return best

    def find_touching(self, matrix, visited, matches, row, column, rows, columns):
        moves = [(-1, 0), (0, -1), (0, 1), (1, 0)]

        for move in moves:
            new_row = row + move[0]
            new_column = column + move[1]

            if (new_row, new_column) not in visited:
                if new_row >= 0 and new_row < rows:
                    if new_column >= 0 and new_column < columns:
                        visited[(new_row, new_column)] = 0
                        if matrix[new_row][new_column] == 1:
                            matches[(new_row, new_column)] = 0
                            self.find_touching(
                                matrix, visited, matches, new_row, new_column, rows, columns
                            )

        return matrix

    def find_outline(self, matches):
        keys = list(matches.keys())
        for item in keys:
            flag = True
            for direction in [(-1, 0), (1, 0), (0, -1), (0, 1)]:
                if (item[0] + direction[0], item[1] + direction[1]) not in matches:
                    flag = False
                    break
            if flag:
                matches[item] = -1

        return [x for x in matches if matches[x] == 0]
                    


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