934. Shortest Bridge

934. Shortest Bridge (Medium)

In a given 2D binary array grid, 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 1:

Input: grid = [[0,1],[1,0]]
Output: 1

Example 2:

Input: grid = [[0,1,0],[0,0,0],[0,0,1]]
Output: 2

Example 3:

Input: grid = [[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

 

Constraints:

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

Companies:
Microsoft, Google, Uber, Bloomberg, Snapchat

Related Topics:
Depth-first Search, Breadth-first Search

Solution 1.

// OJ: https://leetcode.com/problems/shortest-bridge/
// Author: github.com/lzl124631x
// Time: O(MN)
// Space: O(MN)
class Solution {
    int M, N, dirs[4][2] = {{0,1},{0,-1},{1,0},{-1,0}};
    queue<pair<int, int>> qa, qb;
    void dfs(vector<vector<int>> &A, int x, int y, int color) {
        A[x][y] = color;
        if (color == 2) qa.emplace(x, y);
        else qb.emplace(x, y);
        for (auto &[dx, dy] : dirs) {
            int a = x + dx, b = y + dy;
            if (a < 0 || b < 0 || a >= M || b >= N || A[a][b] != 1) continue;
            dfs(A, a, b, color);
        }
    }
    void bfs(vector<vector<int>> &A, queue<pair<int, int>> &q, vector<vector<int>> &dist) {
        int step = 1;
        while (q.size()) {
            int cnt = q.size();
            while (cnt--) {
                auto [x, y] = q.front();
                q.pop();
                for (auto &[dx, dy] : dirs) {
                    int a = x + dx, b = y + dy;
                    if (a < 0 || b < 0 || a >= M || b >= N || A[a][b] != 0 || dist[a][b] != INT_MAX) continue;
                    dist[a][b] = step;
                    q.emplace(a, b);
                }
            }
            ++step;
        }
    }
public:
    int shortestBridge(vector<vector<int>>& A) {
        M = A.size(), N = A[0].size();
        int color = 2, ans = INT_MAX;
        vector<vector<int>> da(M, vector<int>(N, INT_MAX)), db(M, vector<int>(N, INT_MAX));
        for (int i = 0; i < M; ++i) {
            for (int j = 0; j < N; ++j) {
                if (A[i][j] == 1) dfs(A, i, j, color++);
            }
        }
        bfs(A, qa, da);
        bfs(A, qb, db);
        for (int i = 0; i < M; ++i) {
            for (int j = 0; j < N; ++j) {
                if (A[i][j] == 0 && da[i][j] != INT_MAX && db[i][j] != INT_MAX) ans = min(ans, da[i][j] + db[i][j] - 1);
            }
        }
        return ans;
    }
};