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POHow many valid parenthesis combinations?
 

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  1. POHow many valid parenthesis combinations?
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    <p>We have:</p> <ul> <li><p><code>n1</code> number of <code>{}</code> brackets , </p></li> <li><p><code>n2</code> number of <code>()</code> brackets , </p></li> <li><p><code>n3</code> number of <code>[]</code> brackets , </p></li> </ul> <p>How many different valid combination of these brackets we can have? </p> <p>What I thought: I wrote a brute force code in java (which comes in the following) and counted all possible combinations, I know it's the worst solution possible, </p> <p>(the code is for general case in which we can have different types of brackets)</p> <p>Any mathematical approach ? </p> <p>Note 1: valid combination is defined as usual, e.g. <code>{{()}}</code> : valid , <code>{(}){}</code> : invalid</p> <p>Note 2: let's assume that we have 2 pairs of <code>{}</code> , 1 pair of <code>()</code> and 1 pair of <code>[]</code>, the number of valid combinations would be 168 and the number of all possible (valid &amp; invalid) combinations would be 840</p> <pre><code>static void paranthesis_combination(char[] open , char[] close , int[] arr){ int l = 0; for (int i = 0 ; i &lt; arr.length ; i++) l += arr[i]; l *= 2; paranthesis_combination_sub(open , close , arr , new int[arr.length] , new int[arr.length], new StringBuilder(), l); System.out.println(paran_count + " : " + valid_paran_count); return; } static void paranthesis_combination_sub(char[] open , char[] close, int[] arr , int[] open_so_far , int[] close_so_far, StringBuilder strbld , int l){ if (strbld.length() == l &amp;&amp; valid_paran(open , close , strbld)){ System.out.println(new String(strbld)); valid_paran_count++; return; } for (int i = 0 ; i &lt; open.length ; i++){ if (open_so_far[i] &lt; arr[i]){ strbld.append(open[i]); open_so_far[i]++; paranthesis_combination_sub(open , close, arr , open_so_far , close_so_far, strbld , l); open_so_far[i]--; strbld.deleteCharAt(strbld.length() -1 ); } } for (int i = 0 ; i &lt; open.length ; i++){ if (close_so_far[i] &lt; open_so_far[i]){ strbld.append(close[i]); close_so_far[i]++; paranthesis_combination_sub(open , close, arr , open_so_far , close_so_far, strbld , l); close_so_far[i]--; strbld.deleteCharAt(strbld.length() -1 ); } } return; } </code></pre>
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    1. USArian Hosseinzadeh
    1. USArian Hosseinzadeh
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    1. COHow do you define a valid combination? Depending on the parsing rules, there could be many different answers to this question.
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      1. POHow many valid parenthesis combinations?
      1. USMatt Bryant
    2. COQuestion is now edited
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      1. POHow many valid parenthesis combinations?
      1. USArian Hosseinzadeh
    3. COLook up Catalan numbers. Is ({}) valid?
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      1. POHow many valid parenthesis combinations?
      1. USJoni
 

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