<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>

// Approach 1: Depth-First-Search + Breadth-First-Search
class Solution {
public:
    int dx[4] = {1, -1, 0, 0};
    int dy[4] = {0, 0, 1, -1};

    void dfs(vector<vector<int>>& grid, int x, int y, int n, vector<vector<int>>& bfsq) {
        grid[x][y] = 2;
        bfsq.push_back({x, y});
        for(int k = 0; k < 4; k++) {
            int i = dx[k] + x;
            int j = dy[k] + y;
            if(i >= 0 && i < n && j >= 0 && j < n && grid[i][j] == 1) {
                dfs(grid, i, j, n, bfsq);
            }
        }
    }
    int shortestBridge(vector<vector<int>>& grid) {
        int n = grid.size();
        int firstX = -1;
        int firstY = -1;
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < n; j++) {
                if(grid[i][j] == 1) {
                    firstX = i;
                    firstY = j;
                    break;
                }
            }
        }
        // Add all land cells of island grid to bfsQueue.
        vector<vector<int>> bfsq;
        dfs(grid, firstX, firstY, n, bfsq);

        int distance = 0;

        while(!bfsq.empty()) {
            vector<vector<int>> newBfs;
            for(auto pair : bfsq) {
                int x = pair[0];
                int y = pair[1];
                for(int k = 0; k < 4; k++) {
                    int currx = dx[k] + x;
                    int curry = dy[k] + y;
                    if(currx >= 0 && currx < n && curry >= 0 && curry < n) {
                        if(grid[currx][curry] == 1) {
                            return distance;
                        }
                        else if(grid[currx][curry] == 0) {
                            newBfs.push_back({currx, curry});
                            grid[currx][curry] = -1;
                        }
                    }
                }
            }
            bfsq = newBfs;
            distance++;
        }
        return distance;
    }
};
// Complexity Analysis
// Let n×n be the size of the input matrix grid.
// Time complexity: O(n^2)
// The general time complexity of Depth-First-Search is O(V+E), where V stands for the number of vertices. 
// The maximum number of cells in the first island is n^2, so iterating over its cells will take O(n^2) time.
// E is a constant here since we are only allowed to traverse in up to 4 directions.
// The general time complexity of Breadth-First-Search is O(V+E), where V stands for the number of vertices.
// The maximum number of water cells we need to check before reaching the second island is n^2, 
// which will take O(n^2) time.

// Space complexity: O(n^2)
// The general space complexity of Depth-First-Search is O(V), where V stands for the number of vertices. 
// The maximum number of cells in the first island is n^2, thus the space used by the recursive stack 
// during DFS is O(n^2)
// The general space complexity of Breadth-First-Search is O(V), where V stands for the number of vertices.
// The maximum number of water cells we need to check using BFS before reaching the second island is n^2, 
// thus the space used by the queue is O(n^2).
// To sum up, the overall space complexity is O(n^2)
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