1568. Minimum Number of Days to Disconnect Island

You are given an m x n binary grid grid where 1 represents land and 0 represents water. An island is a maximal 4-directionally (horizontal or vertical) connected group of 1’s.

The grid is said to be connected if we have exactly one island, otherwise is said disconnected.

In one day, we are allowed to change any single land cell (1) into a water cell (0).

Return the minimum number of days to disconnect the grid.

class Solution {
    bool bfs(vector<vector<int>> g, int sy, int sx, int islands) {
        if(g[sy][sx] == 0) return true;
        int n = g.size(), m = g[0].size();
        int dy[4]{-1,0,1,0}, dx[4]{0,1,0,-1};
        queue<pair<int,int>> q;
        q.push({sy,sx});
        g[sy][sx] = 0;
        islands--;
        while(!q.empty()) {
            auto [y, x] = q.front(); q.pop();
            for(int i = 0; i < 4; i++) {
                int ny = y + dy[i], nx = x + dx[i];
                if(0 <= ny and ny < n and 0 <= nx and nx < m and g[ny][nx]) {
                    g[ny][nx] = 0;
                    q.push({ny,nx});
                    islands--;
                }
            }
        }
        
        return islands;
    }
public:
    int minDays(vector<vector<int>>& grid) {
        int n = grid.size(), m = grid[0].size();
        vector<pair<int,int>> islands;
        int res = 2;
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < m; j++) {
                if(grid[i][j]) {
                    islands.push_back({i,j});
                }
            }
        }
        if(islands.size() == 1) return 1;
        if(islands.empty() or bfs(grid,islands[0].first, islands[0].second, islands.size())) return 0;
        int sy = islands.back().first, sx = islands.back().second;
        for(auto [y, x] : islands) {
            grid[y][x] = 0;
            if(bfs(grid, sy, sx, islands.size() - 1)) return 1;
            sy = y, sx = x;
            grid[y][x] = 1;
        }
        
        return 2;
    }
};