<p>You are given a <strong>0-indexed</strong> 2D matrix <code>grid</code> of size <code>n x n</code>, where <code>(r, c)</code> represents:</p>

<ul>
	<li>A cell containing a thief if <code>grid[r][c] = 1</code></li>
	<li>An empty cell if <code>grid[r][c] = 0</code></li>
</ul>

<p>You are initially positioned at cell <code>(0, 0)</code>. In one move, you can move to any adjacent cell in the grid, including cells containing thieves.</p>

<p>The <strong>safeness factor</strong> of a path on the grid is defined as the <strong>minimum</strong> manhattan distance from any cell in the path to any thief in the grid.</p>

<p>Return <em>the <strong>maximum safeness factor</strong> of all paths leading to cell </em><code>(n - 1, n - 1)</code><em>.</em></p>

<p>An <strong>adjacent</strong> cell of cell <code>(r, c)</code>, is one of the cells <code>(r, c + 1)</code>, <code>(r, c - 1)</code>, <code>(r + 1, c)</code> and <code>(r - 1, c)</code> if it exists.</p>

<p>The <strong>Manhattan distance</strong> between two cells <code>(a, b)</code> and <code>(x, y)</code> is equal to <code>|a - x| + |b - y|</code>, where <code>|val|</code> denotes the absolute value of val.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2023/07/02/example1.png" style="width: 362px; height: 242px;" />
<pre>
<strong>Input:</strong> grid = [[1,0,0],[0,0,0],[0,0,1]]
<strong>Output:</strong> 0
<strong>Explanation:</strong> All paths from (0, 0) to (n - 1, n - 1) go through the thieves in cells (0, 0) and (n - 1, n - 1).
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2023/07/02/example2.png" style="width: 362px; height: 242px;" />
<pre>
<strong>Input:</strong> grid = [[0,0,1],[0,0,0],[0,0,0]]
<strong>Output:</strong> 2
<strong>Explanation:</strong> The path depicted in the picture above has a safeness factor of 2 since:
- The closest cell of the path to the thief at cell (0, 2) is cell (0, 0). The distance between them is | 0 - 0 | + | 0 - 2 | = 2.
It can be shown that there are no other paths with a higher safeness factor.
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2023/07/02/example3.png" style="width: 362px; height: 242px;" />
<pre>
<strong>Input:</strong> grid = [[0,0,0,1],[0,0,0,0],[0,0,0,0],[1,0,0,0]]
<strong>Output:</strong> 2
<strong>Explanation:</strong> The path depicted in the picture above has a safeness factor of 2 since:
- The closest cell of the path to the thief at cell (0, 3) is cell (1, 2). The distance between them is | 0 - 1 | + | 3 - 2 | = 2.
- The closest cell of the path to the thief at cell (3, 0) is cell (3, 2). The distance between them is | 3 - 3 | + | 0 - 2 | = 2.
It can be shown that there are no other paths with a higher safeness factor.
</pre>

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

<ul>
	<li><code>1 &lt;= grid.length == n &lt;= 400</code></li>
	<li><code>grid[i].length == n</code></li>
	<li><code>grid[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
	<li>There is at least one thief in the <code>grid</code>.</li>
</ul>

class Solution {
    void multisourceBFS(vector<vector<int>>& arr, vector<int> &dx, vector<int> &dy)
    {
        int m = arr.size(), n = arr[0].size();
        queue<pair<int,pair<int,int>>> q;       //{dist, {x,y}}
        vector<vector<int>> vis(m, vector<int>(n,0));
        
        for(int i=0;i<m;i++)
            for(int j=0;j<n;j++)
            {
                if(arr[i][j] == 1)      //If thief, insert into queue
                {
                    q.push({0, {i,j}});
                    vis[i][j] = 1;
                    arr[i][j] = 0;      //Dist from itself = 0
                }
            }
        
        while(!q.empty())
        {
            auto cur = q.front();
            q.pop();
            
            int curDist = cur.first;
            int x = cur.second.first;
            int y = cur.second.second;
            
            arr[x][y] = curDist;
            
            for(int i=0;i<4;i++)
            {
                int nx = x + dx[i];
                int ny = y + dy[i];
                
                if(nx>=0 and ny>=0 and nx<m and ny<n and !vis[nx][ny])
                {
                    vis[nx][ny] = 1;
                    q.push({curDist+1, {nx, ny}});
                }
            }
        }
    }
public:
    /*
    Combination of:
    01 Matrix
    Path with Minimum Effort
    Walls and Gates
    778. Swim in Rising Water
    */
    
    /*
    Thought Process:
    Dijkstra Modified to find the longest path, so here we will be using a visited array
    to visit each node only once.
    We maximise the min in each path using a MAX HEAP
    */
    
    //NEETCODE
    int maximumSafenessFactor(vector<vector<int>>& arr) {
        int m = arr.size(), n = arr[0].size();
        vector<int> dx{-1,0,1,0};
        vector<int> dy{0,1,0,-1};
        
        //Multisource BFS 
        multisourceBFS(arr, dx, dy);
            
        //Modified Dijsktra
        priority_queue<pair<int,pair<int,int>>> pq;//Max heap -> {minManhattanDist in that path, {x,y}}
        vector<vector<int>> vis(m, vector<int>(n,0));
        int minManhattanDist = 0;
        
        pq.push({arr[0][0], {0,0}});
        vis[0][0] = 1;
        
        while(pq.size())
        {
            auto cur = pq.top();
            pq.pop();
            
            int curDist = cur.first;
            int x = cur.second.first;
            int y = cur.second.second;
            
            if(x == m-1 and y==n-1)
                return curDist;
            
            for(int i=0;i<4;i++)
            {
                int nx = x+dx[i];
                int ny = y+dy[i];
                
                if(nx>=0 and ny>=0 and nx<m and ny<n and !vis[nx][ny])
                {
                    int newDist = arr[nx][ny];
                    minManhattanDist = min(curDist, newDist);
                    vis[nx][ny] = 1;
                    pq.push({minManhattanDist, {nx, ny}});
                }
            }
            
        }
        
        return -1;
        
    }
};
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