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<p>A string is a <em>valid parentheses string</em> (denoted VPS) if and only if it consists of <code>"("</code> and <code>")"</code> characters only, and:</p>
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<ul>
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<li>It is the empty string, or</li>
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<li>It can be written as <code>AB</code> (<code>A</code> concatenated with <code>B</code>), where <code>A</code> and <code>B</code> are VPS's, or</li>
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<li>It can be written as <code>(A)</code>, where <code>A</code> is a VPS.</li>
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</ul>
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<p>We can similarly define the <em>nesting depth</em> <code>depth(S)</code> of any VPS <code>S</code> as follows:</p>
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<ul>
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<li><code>depth("") = 0</code></li>
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<li><code>depth(A + B) = max(depth(A), depth(B))</code>, where <code>A</code> and <code>B</code> are VPS's</li>
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<li><code>depth("(" + A + ")") = 1 + depth(A)</code>, where <code>A</code> is a VPS.</li>
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</ul>
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<p>For example, <code>""</code>, <code>"()()"</code>, and <code>"()(()())"</code> are VPS's (with nesting depths 0, 1, and 2), and <code>")("</code> and <code>"(()"</code> are not VPS's.</p>
<li><code>'U'</code> moves our position up one row, if the position exists on the board;</li>
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<li><code>'D'</code> moves our position down one row, if the position exists on the board;</li>
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<li><code>'L'</code> moves our position left one column, if the position exists on the board;</li>
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<li><code>'R'</code> moves our position right one column, if the position exists on the board;</li>
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<li><code>'!'</code> adds the character <code>board[r][c]</code> at our current position <code>(r, c)</code> to the answer.</li>
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</ul>
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<p>(Here, the only positions that exist on the board are positions with letters on them.)</p>
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<p>Return a sequence of moves that makes our answer equal to <code>target</code> in the minimum number of moves. You may return any path that does so.</p>
<p>Given a 2D <code>grid</code> of <code>0</code>s and <code>1</code>s, return the number of elements in the largest <strong>square</strong> subgrid that has all <code>1</code>s on its <strong>border</strong>, or <code>0</code> if such a subgrid doesn't exist in the <code>grid</code>.</p>
<p>Given 2 integers <code>n</code> and <code>start</code>. Your task is return <strong>any</strong> permutation <code>p</code> of <code>(0,1,2.....,2^n -1) </code>such that :</p>
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<ul>
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<li><code>p[0] = start</code></li>
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<li><code>p[i]</code> and <code>p[i+1]</code> differ by only one bit in their binary representation.</li>
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<li><code>p[0]</code> and <code>p[2^n -1]</code> must also differ by only one bit in their binary representation.</li>
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