Minimum Moves to Move a Box to Their Target Location
<p>A storekeeper is a game in which the player pushes boxes around in a warehouse trying to get them to target locations.</p>

<p>The game is represented by an <code>m x n</code> grid of characters <code>grid</code> where each element is a wall, floor, or box.</p>

<p>Your task is to move the box <code>&#39;B&#39;</code> to the target position <code>&#39;T&#39;</code> under the following rules:</p>

<ul>
	<li>The character <code>&#39;S&#39;</code> represents the player. The player can move up, down, left, right in <code>grid</code> if it is a floor (empty cell).</li>
	<li>The character <code>&#39;.&#39;</code> represents the floor which means a free cell to walk.</li>
	<li>The character<font face="monospace">&nbsp;</font><code>&#39;#&#39;</code><font face="monospace">&nbsp;</font>represents the wall which means an obstacle (impossible to walk there).</li>
	<li>There is only one box <code>&#39;B&#39;</code> and one target cell <code>&#39;T&#39;</code> in the <code>grid</code>.</li>
	<li>The box can be moved to an adjacent free cell by standing next to the box and then moving in the direction of the box. This is a <strong>push</strong>.</li>
	<li>The player cannot walk through the box.</li>
</ul>

<p>Return <em>the minimum number of <strong>pushes</strong> to move the box to the target</em>. If there is no way to reach the target, return <code>-1</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1200-1299/1263.Minimum%20Moves%20to%20Move%20a%20Box%20to%20Their%20Target%20Location/images/sample_1_1620.png" style="width: 500px; height: 335px;" />
<pre>
<strong>Input:</strong> grid = [[&quot;#&quot;,&quot;#&quot;,&quot;#&quot;,&quot;#&quot;,&quot;#&quot;,&quot;#&quot;],
               [&quot;#&quot;,&quot;T&quot;,&quot;#&quot;,&quot;#&quot;,&quot;#&quot;,&quot;#&quot;],
               [&quot;#&quot;,&quot;.&quot;,&quot;.&quot;,&quot;B&quot;,&quot;.&quot;,&quot;#&quot;],
               [&quot;#&quot;,&quot;.&quot;,&quot;#&quot;,&quot;#&quot;,&quot;.&quot;,&quot;#&quot;],
               [&quot;#&quot;,&quot;.&quot;,&quot;.&quot;,&quot;.&quot;,&quot;S&quot;,&quot;#&quot;],
               [&quot;#&quot;,&quot;#&quot;,&quot;#&quot;,&quot;#&quot;,&quot;#&quot;,&quot;#&quot;]]
<strong>Output:</strong> 3
<strong>Explanation:</strong> We return only the number of times the box is pushed.</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> grid = [[&quot;#&quot;,&quot;#&quot;,&quot;#&quot;,&quot;#&quot;,&quot;#&quot;,&quot;#&quot;],
               [&quot;#&quot;,&quot;T&quot;,&quot;#&quot;,&quot;#&quot;,&quot;#&quot;,&quot;#&quot;],
               [&quot;#&quot;,&quot;.&quot;,&quot;.&quot;,&quot;B&quot;,&quot;.&quot;,&quot;#&quot;],
               [&quot;#&quot;,&quot;#&quot;,&quot;#&quot;,&quot;#&quot;,&quot;.&quot;,&quot;#&quot;],
               [&quot;#&quot;,&quot;.&quot;,&quot;.&quot;,&quot;.&quot;,&quot;S&quot;,&quot;#&quot;],
               [&quot;#&quot;,&quot;#&quot;,&quot;#&quot;,&quot;#&quot;,&quot;#&quot;,&quot;#&quot;]]
<strong>Output:</strong> -1
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> grid = [[&quot;#&quot;,&quot;#&quot;,&quot;#&quot;,&quot;#&quot;,&quot;#&quot;,&quot;#&quot;],
               [&quot;#&quot;,&quot;T&quot;,&quot;.&quot;,&quot;.&quot;,&quot;#&quot;,&quot;#&quot;],
               [&quot;#&quot;,&quot;.&quot;,&quot;#&quot;,&quot;B&quot;,&quot;.&quot;,&quot;#&quot;],
               [&quot;#&quot;,&quot;.&quot;,&quot;.&quot;,&quot;.&quot;,&quot;.&quot;,&quot;#&quot;],
               [&quot;#&quot;,&quot;.&quot;,&quot;.&quot;,&quot;.&quot;,&quot;S&quot;,&quot;#&quot;],
               [&quot;#&quot;,&quot;#&quot;,&quot;#&quot;,&quot;#&quot;,&quot;#&quot;,&quot;#&quot;]]
<strong>Output:</strong> 5
<strong>Explanation:</strong> push the box down, left, left, up and up.
</pre>

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

<ul>
	<li><code>m == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 20</code></li>
	<li><code>grid</code> contains only characters <code>&#39;.&#39;</code>, <code>&#39;#&#39;</code>, <code>&#39;S&#39;</code>, <code>&#39;T&#39;</code>, or <code>&#39;B&#39;</code>.</li>
	<li>There is only one character <code>&#39;S&#39;</code>, <code>&#39;B&#39;</code>, and <code>&#39;T&#39;</code> in the <code>grid</code>.</li>
</ul>
Breadth-First Search; Array; Matrix; Heap (Priority Queue)
class Solution {
    private int m;
    private int n;
    private char[][] grid;

    public int minPushBox(char[][] grid) {
        m = grid.length;
        n = grid[0].length;
        this.grid = grid;
        int si = 0, sj = 0, bi = 0, bj = 0;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == 'S') {
                    si = i;
                    sj = j;
                } else if (grid[i][j] == 'B') {
                    bi = i;
                    bj = j;
                }
            }
        }
        int[] dirs = {-1, 0, 1, 0, -1};
        Deque<int[]> q = new ArrayDeque<>();
        boolean[][] vis = new boolean[m * n][m * n];
        q.offer(new int[] {f(si, sj), f(bi, bj), 0});
        vis[f(si, sj)][f(bi, bj)] = true;
        while (!q.isEmpty()) {
            var p = q.poll();
            int d = p[2];
            bi = p[1] / n;
            bj = p[1] % n;
            if (grid[bi][bj] == 'T') {
                return d;
            }
            si = p[0] / n;
            sj = p[0] % n;
            for (int k = 0; k < 4; ++k) {
                int sx = si + dirs[k], sy = sj + dirs[k + 1];
                if (!check(sx, sy)) {
                    continue;
                }
                if (sx == bi && sy == bj) {
                    int bx = bi + dirs[k], by = bj + dirs[k + 1];
                    if (!check(bx, by) || vis[f(sx, sy)][f(bx, by)]) {
                        continue;
                    }
                    vis[f(sx, sy)][f(bx, by)] = true;
                    q.offer(new int[] {f(sx, sy), f(bx, by), d + 1});
                } else if (!vis[f(sx, sy)][f(bi, bj)]) {
                    vis[f(sx, sy)][f(bi, bj)] = true;
                    q.offerFirst(new int[] {f(sx, sy), f(bi, bj), d});
                }
            }
        }
        return -1;
    }

    private int f(int i, int j) {
        return i * n + j;
    }

    private boolean check(int i, int j) {
        return i >= 0 && i < m && j >= 0 && j < n && grid[i][j] != '#';
    }
}