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{"id":482,"date":"2021-07-13T11:31:06","date_gmt":"2021-07-13T11:31:06","guid":{"rendered":"https:\/\/www.wikitechy.com\/interview-questions\/?p=482"},"modified":"2021-09-13T07:21:08","modified_gmt":"2021-09-13T07:21:08","slug":"what-is-n-queen-problem","status":"publish","type":"post","link":"https:\/\/www.wikitechy.com\/interview-questions\/data-structure\/what-is-n-queen-problem\/","title":{"rendered":"What is N Queen Problem ?"},"content":{"rendered":"<div class=\"TextHeading\">\n<div class=\"hddn\">\n<h2 id=\"n-queen-problem\" class=\"color-pink\" style=\"text-align: justify;\">N Queen Problem<\/h2>\n<\/div>\n<\/div>\n<div class=\"Content\" style=\"text-align: justify;\">\n<div class=\"hddn\"><\/div>\n<\/div>\n<div class=\"Content\" style=\"text-align: justify;\">\n<div class=\"hddn\">\n<ul>\n<li>Let us discuss N Queen problem that can be solved using\u00a0<a href=\"https:\/\/www.wikitechy.com\/final-year-project\/dotnet\/server-hacking\/backtracking-algorithm\" target=\"_blank\" rel=\"noopener\">Backtracking<\/a>.<\/li>\n<li>The N Queen is the problem of placing N chess queens on an N\u00d7N chessboard so that no two queens attack each other. For example, following the solution for 4 Queen problem.<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div class=\"ImageContent\" style=\"text-align: justify;\">\n<div class=\"hddn\"><img decoding=\"async\" class=\"img-responsive center-block aligncenter\" src=\"https:\/\/cdn.wikitechy.com\/interview-questions\/data-structure\/what-is-n-queen-problem.jpg\" alt=\" N Queen Problem\" \/><\/div>\n<\/div>\n<div class=\"Content\" style=\"text-align: justify;\">\n<div class=\"hddn\">\n<ul>\n<li>Binary\u00a0<a href=\"https:\/\/www.wikitechy.com\/tutorials\/apache-pig\/apache-pig-tutorial\/matrix-multiplication-apache-pig\" target=\"_blank\" rel=\"noopener\">matrix<\/a>\u00a0which has 1&#8217;s for the blocks where queens are placed. For example, following is the output matrix for above 4 queen solution.<\/li>\n<\/ul>\n<div class=\"code-embed-wrapper\"> <div class=\"code-embed-infos\"> <\/div> <pre class=\"language-markdown code-embed-pre line-numbers\"  data-start=\"1\" data-line-offset=\"0\"><code class=\"language-markdown code-embed-code\">{ 0,  1,  0,  0 }<br\/>{ 0,  0,  0,  1 }<br\/>{ 1,  0,  0,  0 }<br\/>{ 0,  0,  1,  0 }<\/code><\/pre> <\/div>\n<div class=\"TextHeading\">\n<div class=\"hddn\">\n<h2 id=\"backtracking-algorithm\" class=\"color-green\">Backtracking Algorithm<\/h2>\n<\/div>\n<\/div>\n<div class=\"Content\">\n<div class=\"hddn\"><\/div>\n<\/div>\n<div class=\"Content\">\n<div class=\"hddn\">\n<ul>\n<li>The idea is to place queens one by one in different columns, starting from the leftmost column.<\/li>\n<li>When we place a queen in a column, we can checkout to clashes with already placed queens.<\/li>\n<li>In the current column, if we find a row for which there is no clash, we mark this row and column as part of the solution.<\/li>\n<li>If we do not find such a row due to clashes then we backtrack and return false.<\/li>\n<\/ul>\n<div class=\"code-embed-wrapper\"> <div class=\"code-embed-infos\"> <\/div> <pre class=\"language-markdown code-embed-pre line-numbers\"  data-start=\"1\" data-line-offset=\"0\"><code class=\"language-markdown code-embed-code\">1) Start from the leftmost column<br\/>2) If all queens are placed<br\/>    return true<br\/>3) Try all rows in the current column.  Do following for every tried row.<br\/>    a) If the queen can be placed safely in this row then mark this [row, <br\/>        column] as part of the solution and recursively check if placing queen here leads to a solution.<br\/>    b) If placing the queen in [row, column] leads to a solution then return <br\/>        true.<br\/>    c) If placing queen does not lead to a solution then umark this [row, <br\/>        column] (Backtrack) and go to step (a) to try other rows.<br\/>4) If all rows have been tried and nothing worked, return false to trigger <br\/>    backtracking.<\/code><\/pre> <\/div>\n<h2 id=\"sample-code-in-c\" class=\"color-blue\">Sample Code in C<\/h2>\n<div class=\"code-embed-wrapper\"> <div class=\"code-embed-infos\"> <\/div> <pre class=\"language-c code-embed-pre line-numbers\"  data-start=\"1\" data-line-offset=\"0\"><code class=\"language-c code-embed-code\">\/* C\/C++ program to solve N Queen Problem using <br\/>backtracking *\/<br\/>#define N 4 <br\/>#include&lt;stdio.h&gt; <br\/>#include&lt;stdbool.h&gt; <br\/><br\/>\/* A utility function to print solution *\/<br\/>void printSolution(int board[N][N]) <br\/>{ <br\/>\tfor (int i = 0; i &lt; N; i++) <br\/>\t{ <br\/>\t\tfor (int j = 0; j &lt; N; j++) <br\/>\t\t\tprintf(&quot; %d &quot;, board[i][j]); <br\/>\t\tprintf(&quot;\\n&quot;); <br\/>\t} <br\/>} <br\/><br\/>\/* A utility function to check if a queen can be placed on board[row][col]. <br\/>Note that this function is called when &quot;col&quot; queens are already placed in columns<br\/> from 0 to col -1. So we need to check only left side for attacking queens *\/<br\/>bool isSafe(int board[N][N], int row, int col) <br\/>{ <br\/>\tint i, j; <br\/><br\/>\t\/* Check this row on left side *\/<br\/>\tfor (i = 0; i &lt; col; i++) <br\/>\t\tif (board[row][i]) <br\/>\t\t\treturn false; <br\/><br\/>\t\/* Check upper diagonal on left side *\/<br\/>\tfor (i=row, j=col; i&gt;=0 &amp;&amp; j&gt;=0; i--, j--) <br\/>\t\tif (board[i][j]) <br\/>\t\t\treturn false; <br\/><br\/>\t\/* Check lower diagonal on left side *\/<br\/>\tfor (i=row, j=col; j&gt;=0 &amp;&amp; i&lt;N; i++, j--) <br\/>\t\tif (board[i][j]) <br\/>\t\t\treturn false; <br\/><br\/>\treturn true; <br\/>} <br\/><br\/>\/* A recursive utility function to solve N <br\/>Queen problem *\/<br\/>bool solveNQUtil(int board[N][N], int col) <br\/>{ <br\/>\t\/* base case: If all queens are placed <br\/>\tthen return true *\/<br\/>\tif (col &gt;= N) <br\/>\t\treturn true; <br\/><br\/>\t\/* Consider this column and try placing <br\/>\tthis queen in all rows one by one *\/<br\/>\tfor (int i = 0; i &lt; N; i++) <br\/>\t{ <br\/>\t\t\/* Check if the queen can be placed on <br\/>\t\tboard[i][col] *\/<br\/>\t\tif ( isSafe(board, i, col) ) <br\/>\t\t{ <br\/>\t\t\t\/* Place this queen in board[i][col] *\/<br\/>\t\t\tboard[i][col] = 1; <br\/><br\/>\t\t\t\/* recur to place rest of the queens *\/<br\/>\t\t\tif ( solveNQUtil(board, col + 1) ) <br\/>\t\t\t\treturn true; <br\/><br\/>\t\t\t\/* If placing queen in board[i][col] <br\/>\t\t\tdoesn&#039;t lead to a solution, then <br\/>\t\t\tremove queen from board[i][col] *\/<br\/>\t\t\tboard[i][col] = 0; \/\/ BACKTRACK <br\/>\t\t} <br\/>\t} <br\/><br\/>\t\/* If the queen cannot be placed in any row in <br\/>\t\tthis colum col then return false *\/<br\/>\treturn false; <br\/>} <br\/><br\/>\/* This function solves the N Queen problem using Backtracking. It mainly uses <br\/>solveNQUtil() to solve the problem. It returns false if queens cannot be placed, <br\/>otherwise, return true and prints placement of queens in the form of 1s.<br\/><br\/> Please note that there may be more than one solutions, this function prints one of the <br\/>feasible solutions.*\/<br\/>bool solveNQ() <br\/>{ <br\/>\tint board[N][N] = { {0, 0, 0, 0}, <br\/>\t\t{0, 0, 0, 0}, <br\/>\t\t{0, 0, 0, 0}, <br\/>\t\t{0, 0, 0, 0} <br\/>\t}; <br\/><br\/>\tif ( solveNQUtil(board, 0) == false ) <br\/>\t{ <br\/>\tprintf(&quot;Solution does not exist&quot;); <br\/>\treturn false; <br\/>\t} <br\/><br\/>\tprintSolution(board); <br\/>\treturn true; <br\/>} <br\/><br\/>\/\/ driver program to test above function <br\/>int main() <br\/>{ <br\/>\tsolveNQ(); <br\/>\treturn 0; <br\/>}<\/code><\/pre> <\/div>\n<h2 id=\"output\" class=\"color-blue\">Output<\/h2>\n<div class=\"code-embed-wrapper\"> <div class=\"code-embed-infos\"> <\/div> <pre class=\"language-c code-embed-pre line-numbers\"  data-start=\"1\" data-line-offset=\"0\"><code class=\"language-c code-embed-code\">0  0  1  0 <br\/> 1  0  0  0 <br\/> 0  0  0  1 <br\/> 0  1  0  0 <\/code><\/pre> <\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Answer : Let us discuss N Queen problem that can be solved using 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