Determine Whether Matrix Can Be Obtained By Rotation
<p>Given two <code>n x n</code> binary matrices <code>mat</code> and <code>target</code>, return <code>true</code><em> if it is possible to make </em><code>mat</code><em> equal to </em><code>target</code><em> by <strong>rotating</strong> </em><code>mat</code><em> in <strong>90-degree increments</strong>, or </em><code>false</code><em> otherwise.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1800-1899/1886.Determine%20Whether%20Matrix%20Can%20Be%20Obtained%20By%20Rotation/images/grid3.png" style="width: 301px; height: 121px;" />
<pre>
<strong>Input:</strong> mat = [[0,1],[1,0]], target = [[1,0],[0,1]]
<strong>Output:</strong> true
<strong>Explanation: </strong>We can rotate mat 90 degrees clockwise to make mat equal target.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1800-1899/1886.Determine%20Whether%20Matrix%20Can%20Be%20Obtained%20By%20Rotation/images/grid4.png" style="width: 301px; height: 121px;" />
<pre>
<strong>Input:</strong> mat = [[0,1],[1,1]], target = [[1,0],[0,1]]
<strong>Output:</strong> false
<strong>Explanation:</strong> It is impossible to make mat equal to target by rotating mat.
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1800-1899/1886.Determine%20Whether%20Matrix%20Can%20Be%20Obtained%20By%20Rotation/images/grid4.png" style="width: 661px; height: 184px;" />
<pre>
<strong>Input:</strong> mat = [[0,0,0],[0,1,0],[1,1,1]], target = [[1,1,1],[0,1,0],[0,0,0]]
<strong>Output:</strong> true
<strong>Explanation: </strong>We can rotate mat 90 degrees clockwise two times to make mat equal target.
</pre>

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

<ul>
	<li><code>n == mat.length == target.length</code></li>
	<li><code>n == mat[i].length == target[i].length</code></li>
	<li><code>1 &lt;= n &lt;= 10</code></li>
	<li><code>mat[i][j]</code> and <code>target[i][j]</code> are either <code>0</code> or <code>1</code>.</li>
</ul>
Array; Matrix
class Solution {
public:
    bool findRotation(vector<vector<int>>& mat, vector<vector<int>>& target) {
        int n = mat.size();
        for (int k = 0; k < 4; ++k) {
            vector<vector<int>> g(n, vector<int>(n));
            for (int i = 0; i < n; ++i)
                for (int j = 0; j < n; ++j)
                    g[i][j] = mat[j][n - i - 1];
            if (g == target) return true;
            mat = g;
        }
        return false;
    }
};