<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 {
//Toseperate two islands  make any one ->2
//apply bfs on all the nodes of any one island till you find the other
vector<int> rowDir = {-1, 0, +1, 0};
vector<int> colDir = {0, +1, 0, -1};
private:
    void bfsSperateIsland(int row, int col, vector<vector<int>>& grid, queue<vector<int>>&island1Queue) {
        queue<pair<int, int>>bfsQueue;
        bfsQueue.push({row, col});
        grid[row][col] = 2;
        island1Queue.push(vector<int>{row, col,0});
        while(!bfsQueue.empty()){
            pair<int, int>currNode = bfsQueue.front();
            bfsQueue.pop();
            for(int index = 0; index < 4; index++){
                int newRow = currNode.first + rowDir[index];
                int newCol = currNode.second + colDir[index];
                if(newRow >= 0 && newCol >= 0 && newRow < grid.size() && newCol < grid[0].size() && grid[newRow][newCol] == 1){
                    bfsQueue.push({newRow, newCol});
                    grid[newRow][newCol] = 2;
                    island1Queue.push(vector<int>{newRow, newCol,0});
                }
            }

        }
    }

    int bfs(vector<vector<int>>& grid, queue<vector<int>>bfsQueue){
        while(!bfsQueue.empty()){
            vector<int> currNode = bfsQueue.front();
            bfsQueue.pop();
            for(int index = 0; index < 4; index++){
                int newRow = currNode[0] + rowDir[index];
                int newCol = currNode[1] + colDir[index];
                if(newRow < 0 || newCol < 0 || newRow >= grid.size() || newCol >= grid[0].size() || grid[newRow][newCol] == 2){
                    continue;
                }
                if(grid[newRow][newCol] == 1){
                    return currNode[2];
                }
                bfsQueue.push(vector<int>{newRow, newCol,currNode[2]+1});
                grid[newRow][newCol] = 2;
            }
        }
        return 0;
    }
public:
    int shortestBridge(vector<vector<int>>& grid) {
        queue<vector<int>>island1Queue;
        bool flag = false;
        for(int row = 0; row < grid.size(); row++){
            for(int col = 0; col < grid[0].size(); col++){
                if(grid[row][col] == 1){
                    flag = true;
                    bfsSperateIsland(row, col, grid, island1Queue); 
                    break; 
                }
            }
            if (flag == true){
                break;
            }
            
        }

        return bfs(grid, island1Queue); 
    }
};
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