<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 {
private:
    bool isValid(int nRow,int nCol,int n){
        return nRow>= 0 && nRow<n && nCol>=0 && nCol<n;
    }
    void dfs(int row,int col, vector<vector<int>>& grid,vector<pair<int,int>>& island,
            vector<int>& delRow,vector<int>& delCol){
        int n=grid.size();
        grid[row][col]=0;
        island.push_back({row,col});
        for(int i=0;i<4;i++){
            int nRow=row+delRow[i];
            int nCol=col+delCol[i];
            if(isValid(nRow,nCol,n) && grid[nRow][nCol]==1){
                dfs(nRow,nCol,grid,island,delRow,delCol);
            }
        }
    }
    
    int getMinFlips(vector<pair<int,int>> &island1,vector<pair<int,int>> &island2){
        int minDist=INT_MAX;
        for(int i=0;i<island1.size();i++){
            for(int j=0;j<island2.size();j++){
                int x1=island1[i].first;
                int x2=island2[j].first;
                int y1=island1[i].second;
                int y2=island2[j].second;
                
                int dist=abs(x1-x2)+abs(y1-y2)-1;
                minDist=min(minDist,dist);
            }
        }
        return minDist;
    }
public:   
    int shortestBridge(vector<vector<int>>& grid) {

        int n=grid.size();
        vector<pair<int,int>> island1;
        vector<pair<int,int>> island2;

        int islandCnt=0;
        for(int row=0;row<n;row++){
            for(int col=0;col<n;col++){
                vector<int> delRow{-1,0,1,0};
                vector<int> delCol{0,1,0,-1};
                if(grid[row][col]==1){
                    islandCnt++;
                    if(islandCnt==1){
                        dfs(row,col,grid,island1,delRow,delCol);
                    }
                    else if(islandCnt==2){
                        dfs(row,col,grid,island2,delRow,delCol);
                    }
                }
            }
        }
        int minDist=getMinFlips(island1,island2);

        return minDist;
    }
};
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