<p>You are given an <code>n x n</code> integer matrix <code>grid</code> where each value <code>grid[i][j]</code> represents the elevation at that point <code>(i, j)</code>.</p>

<p>The rain starts to fall. At time <code>t</code>, the depth of the water everywhere is <code>t</code>. You can swim from a square to another 4-directionally adjacent square if and only if the elevation of both squares individually are at most <code>t</code>. You can swim infinite distances in zero time. Of course, you must stay within the boundaries of the grid during your swim.</p>

<p>Return <em>the least time until you can reach the bottom right square </em><code>(n - 1, n - 1)</code><em> if you start at the top left square </em><code>(0, 0)</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/06/29/swim1-grid.jpg" style="width: 164px; height: 165px;" />
<pre>
<strong>Input:</strong> grid = [[0,2],[1,3]]
<strong>Output:</strong> 3
Explanation:
At time 0, you are in grid location (0, 0).
You cannot go anywhere else because 4-directionally adjacent neighbors have a higher elevation than t = 0.
You cannot reach point (1, 1) until time 3.
When the depth of water is 3, we can swim anywhere inside the grid.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/06/29/swim2-grid-1.jpg" style="width: 404px; height: 405px;" />
<pre>
<strong>Input:</strong> grid = [[0,1,2,3,4],[24,23,22,21,5],[12,13,14,15,16],[11,17,18,19,20],[10,9,8,7,6]]
<strong>Output:</strong> 16
<strong>Explanation:</strong> The final route is shown.
We need to wait until time 16 so that (0, 0) and (4, 4) are connected.
</pre>

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

<ul>
	<li><code>n == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= n &lt;= 50</code></li>
	<li><code>0 &lt;= grid[i][j] &lt;&nbsp;n<sup>2</sup></code></li>
	<li>Each value <code>grid[i][j]</code> is <strong>unique</strong>.</li>
</ul>

class Solution {

private:
    int dijkstra(vector<vector<int>>& grid){
        int n = grid.size();

        vector<vector<int>> dist(n,vector<int>(n,1e9)) ;
        priority_queue<pair<int,pair<int,int>> , vector<pair<int,pair<int,int>>> , greater<pair<int,pair<int,int>>>> pq ;
        dist[0][0] = 0 ;
        pq.push({grid[0][0],{0,0}}) ;

        int dx[] = {0,-1,0,1} ;
        int dy[] = {-1,0,1,0} ;

        while(!pq.empty()){
            auto a = pq.top() ;

            pq.pop() ;

            int t = a.first ;
            int x = a.second.first ;
            int y = a.second.second ;

            if(x == n-1 && y == n-1)
                return t ;

            for(int i=0 ; i< 4 ; i++){
                int nx = x + dx[i] ;
                int ny = y + dy[i] ;

                if(nx>=0 && ny>=0 && nx<n && ny <n){
                    int maxLevel = max(grid[nx][ny] , t) ;
                    if(maxLevel < dist[nx][ny] ) {
                        dist[nx][ny] = maxLevel ;
                        pq.push({maxLevel,{nx,ny}}) ;
                    }
                }
            }
        }

         return -1 ;
    }

    int findPar(vector<int>& parent , int node){
        if(parent[node] == node)
            return node ;
        return parent[node] = findPar(parent , parent[node]) ;
    }

    void unionBySize(vector<int>& parent , vector<int>& size , int nodeA , int nodeB){
        
        int ult_A = findPar(parent,nodeA);
        int ult_B = findPar(parent,nodeB);

        if(size[ult_A] < size[ult_B]){
            parent[ult_A] = ult_B ;
            size[ult_B] += size[ult_A] ;
        }
        else{
            parent[ult_B] = ult_A ;
            size[ult_A] += size[ult_B] ;
        }
    }

    int disjointSet(vector<vector<int>>& grid){
        int n = grid.size();
        int totalNodes = n*n ;
        vector<int> parent(totalNodes) , size(totalNodes , 1)  ;
        vector<pair<int,int>> hash(totalNodes) ;

        for(int i=0 ; i<n ; i++){
            for(int j=0 ; j<n ; j++){
                hash[grid[i][j]] = {i,j} ;
                parent[i*n+j] = i*n+j ;
            }
        }
        int dx[] = {0,-1,0,1};
        int dy[] = {-1,0,1,0};
        for(int i=0 ; i<totalNodes; i++){
            int x = hash[i].first ;
            int y = hash[i].second ;
            int nodeA = x * n + y ;
            for(int k=0 ; k<4 ; k++){
                int nx = x + dx[k] ;
                int ny = y + dy[k] ;
                
                if(nx >=0 && ny>=0 && nx<n && ny<n){
                    int t = grid[nx][ny] ;
                    int nodeB = nx * n + ny ;
                    if(t < i && findPar(parent ,nodeA) != findPar(parent , nodeB))    
                        unionBySize(parent,size,nodeA,nodeB) ;
                }
            }

            if(findPar(parent,0) == findPar(parent , totalNodes -1))
                return i;
        } 
        return -1;

    }

public:
    int swimInWater(vector<vector<int>>& grid) {
        return disjointSet(grid);
    }
};
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