<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>

#include <iostream>
#include <vector>
#include <queue>
#include <algorithm> // For max function

using namespace std;

class Solution {
public:
    int maximumSafenessFactor(vector<vector<int>>& grid) {
        // Calculate safeness factors for each cell
        vector<vector<int>> safeness = calculateSafenessFactors(grid);
        // Find the maximum safeness value from the safeness matrix
        int maxSafeness = findMaxSafeness(safeness);
        // Perform binary search to find the maximum safeness factor
        return findMaxSafenessFactor(grid, safeness, maxSafeness);
    }

private:
    vector<vector<int>> calculateSafenessFactors(vector<vector<int>>& grid) {
        int n = grid.size();
        vector<vector<int>> safeness(n, vector<int>(n, INT_MAX)); 
        queue<pair<int, int>> q;

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 1) {
                    q.push({i, j});
                    safeness[i][j] = 0;
                }
            }
        }

        vector<int> dirX = {-1, 1, 0, 0};
        vector<int> dirY = {0, 0, -1, 1};

        // Perform BFS to calculate safeness factors
        while (!q.empty()) {
            auto cell = q.front(); q.pop();
            int x = cell.first;
            int y = cell.second;

            for (int i = 0; i < 4; i++) {
                int newX = x + dirX[i];
                int newY = y + dirY[i];

                if (newX >= 0 && newX < n && newY >= 0 && newY < n && safeness[newX][newY] > safeness[x][y] + 1) {
                    safeness[newX][newY] = safeness[x][y] + 1;
                    q.push({newX, newY});
                }
            }
        }

        return safeness;
    }

    int findMaxSafeness(vector<vector<int>>& safeness) {
        int maxSafeness = 0;
        for (auto& row : safeness) {
            for (auto val : row) {
                maxSafeness = max(maxSafeness, val);
            }
        }
        return maxSafeness;
    }

    int findMaxSafenessFactor(vector<vector<int>>& grid, vector<vector<int>>& safeness, int maxSafeness) {
        int left = 0, right = maxSafeness, result = 0;

        // Binary search to find the maximum safeness factor
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (canAchieveSafeness(grid, safeness, mid)) {
                result = mid; 
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }

        return result;
    }

    // Method to check if a path with the given safeness factor exists using BFS
    bool canAchieveSafeness(vector<vector<int>>& grid, vector<vector<int>>& safeness, int threshold) {
        int n = grid.size();
        if (safeness[0][0] < threshold) return false; 

        queue<pair<int, int>> q;
        vector<vector<bool>> visited(n, vector<bool>(n, false));
        q.push({0, 0});
        visited[0][0] = true;

        vector<int> dirX = {-1, 1, 0, 0};
        vector<int> dirY = {0, 0, -1, 1};

        while (!q.empty()) {
            auto cell = q.front(); q.pop();
            int x = cell.first;
            int y = cell.second;

            if (x == n - 1 && y == n - 1) return true; // If we reach the bottom-right corner

            for (int i = 0; i < 4; i++) {
                int newX = x + dirX[i];
                int newY = y + dirY[i];

                if (newX >= 0 && newX < n && newY >= 0 && newY < n && !visited[newX][newY] && safeness[newX][newY] >= threshold) {
                    q.push({newX, newY});
                    visited[newX][newY] = true;
                }
            }
        }

        return false;
    }
};
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