<p>You are given an <code>n x n</code> binary matrix <code>grid</code> where <code>1</code> represents land and <code>0</code> represents water.</p>

<p>An <strong>island</strong> is a 4-directionally connected group of <code>1</code>&#39;s not connected to any other <code>1</code>&#39;s. There are <strong>exactly two islands</strong> in <code>grid</code>.</p>

<p>You may change <code>0</code>&#39;s to <code>1</code>&#39;s to connect the two islands to form <strong>one island</strong>.</p>

<p>Return <em>the smallest number of </em><code>0</code><em>&#39;s you must flip to connect the two islands</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> grid = [[0,1],[1,0]]
<strong>Output:</strong> 1
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> grid = [[0,1,0],[0,0,0],[0,0,1]]
<strong>Output:</strong> 2
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> 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]]
<strong>Output:</strong> 1
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == grid.length == grid[i].length</code></li>
	<li><code>2 &lt;= n &lt;= 100</code></li>
	<li><code>grid[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
	<li>There are exactly two islands in <code>grid</code>.</li>
</ul>

class Solution {
public:
    void dfs(int row, int col, vector<vector<int>>& grid, queue<pair<int, pair<int, int>>>& q, int n, vector<vector<int>>& dist) {
        grid[row][col] = 2;
        q.push({0, {row, col}});
        dist[row][col] = 0;
        vector<int> dx = {-1, 0, +1, 0};
        vector<int> dy = {0, -1, 0, +1};

        for (int i = 0; i < 4; i++) {
            int nr = row + dx[i];
            int nc = col + dy[i];
            if (nr >= 0 && nc >= 0 && nr < n && nc < n && grid[nr][nc] != 2 && grid[nr][nc] == 1) {
                dfs(nr, nc, grid, q, n, dist);
            }
        }
    }

    int shortestBridge(vector<vector<int>>& grid) {
        int n = grid.size();
        vector<vector<int>> dist(n, vector<int>(n, INT_MAX));
        queue<pair<int, pair<int, int>>> q;
        bool found = false;
        
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 1) {
                    dfs(i, j, grid, q, n, dist);
                    found = true;
                    break;
                }
            }
            if (found) {
                break;
            }
        }

        while (!q.empty()) {
            int dis = q.front().first;
            int row = q.front().second.first;
            int col = q.front().second.second;
            q.pop();

            vector<int> dx = {-1, 0, +1, 0};
            vector<int> dy = {0, -1, 0, +1};

            for (int i = 0; i < 4; i++) {
                int nr = row + dx[i];
                int nc = col + dy[i];
                if (nr >= 0 && nc >= 0 && nr < n && nc < n) {
                    if (dist[nr][nc] > dis + 1) {
                        dist[nr][nc] = dis + 1;
                        q.push({dis + 1, {nr, nc}});
                    }
                }
            }
        }

        int ans = INT_MAX;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 1) {
                    ans = min(ans, dist[i][j]);
                }
            }
        }

        return ans-1;
    }
};

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