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    <p>For a given <code>N</code> we always have to start with an open brace. Now consider where it's corresponding closing brace is. It can be in the middle like in <code>()()</code> or at the end like <code>(())</code> for <code>N=2</code>.</p> <p>Now consider <code>N=3</code>:</p> <p>It can be at the end: <code>(()())</code> and <code>((()))</code>.</p> <p>Or in the middle: <code>()(())</code> and <code>()()()</code> where it's in position 2. Then it can also be in position 4: <code>(())()</code>.</p> <p>Now we can essentially combine the 2 cases by realising the case where the closing brace is at the end is the same as it being at the middle, but with all the possibilities for N=0 added to the end.</p> <p>Now to solve it you can work out all the possibilities for <code>n</code> between the begin and end brace and similarly you can work out all the possibilities for <code>m</code> after the end brace. (Note <code>m+n+1 = N</code>) Then you can just combine all possible combinations, append them to your list of possibilities and move on to the next possible location for the end brace.</p> <p>Just be warned an easy mistake to make with these types of problems, is to find all the possibilities for <code>i</code> and for <code>N-i</code> and just combine them, but this for <code>N=3</code> would double count <code>()()()</code> or at least print it twice.</p> <p>Here is some Python 2.x code that solves the problem:</p> <pre><code>memoise = {} memoise[0] = [""] memoise[1] = ["()"] def solve(N, doprint=True): if N in memoise: return memoise[N] memoise[N] = [] for i in xrange(1,N+1): between = solve(i-1, False) after = solve(N-i, False) for b in between: for a in after: memoise[N].append("("+b+")"+a) if doprint: for res in memoise[N]: print res return memoise[N] </code></pre>
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    1. PORecurrence Approach : How can we generate all possibilities on braces?
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    1. COThis is not home work and I am not school boy. This is Interview Question .
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    2. CO@siva My apologies then :)
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