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POCombinations of positive integers uniq (order not important)
 

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  1. POCombinations of positive integers uniq (order not important)
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    <p>So, I was searching for a good solution for my problem.</p> <p>I need to generate(print) all the combination of a list of integers, for example: if the array contain integers from 0 to n-1, where n = 5:</p> <pre><code>int array[] = {0,1,2,3,4}; </code></pre> <p>the order of integers in the combination are NOT important, meaning {1,1,3}, {1,3,1} and {3,1,1} are actually the same combination because they all contain one 3 and two ones.</p> <p>so for the above array, all combination of length 3:</p> <pre><code>0,0,0 -&gt; the 1st combination 0,0,1 0,0,2 0,0,3 0,0,4 0,1,1 -&gt; this combination is 0,1,1, not 0,1,0 because we already have 0,0,1. 0,1,2 0,1,3 0,1,4 0,2,2 -&gt; this combination is 0,2,2, not 0,2,0 because we already have 0,0,2. 0,2,3 . . 0,4,4 1,1,1 -&gt; this combination is 1,1,1, not 1,0,0 because we already have 0,0,1. 1,1,2 1,1,3 1,1,4 1,2,2 -&gt; this combination is 1,2,2, not 1,2,0 because we already have 0,1,2. . . 4,4,4 -&gt; Last combination </code></pre> <p>For Now I Wrote Code for doing this, but my problem is: if the numbers in the array are not integer from 0 to n-1, lets say if the array was like this</p> <pre><code>int array[] = {1,3,6,7}; </code></pre> <p>my code doesn't work on this case, any algorithm or code for solving this problem,,</p> <p>Here is my code :</p> <pre><code>unsigned int next_combination(unsigned int *ar, int n, unsigned int k){ unsigned int finished = 0; unsigned int changed = 0; unsigned int i; for (i = k - 1; !finished &amp;&amp; !changed; i--) { if (ar[i] &lt; n - 1) { /* Increment this element */ ar[i]++; if (i &lt; k - 1) { /* Make the elements after it the same */ unsigned int j; for (j = i + 1; j &lt; k; j++) { ar[j] = ar[j - 1]; } } changed = 1; } finished = i == 0; } if (!changed) { /* Reset to first combination */ for (i = 0; i &lt; k; i++){ ar[i] = 0; } } return changed; } </code></pre> <p>And this is the main:</p> <pre><code>int main(){ unsigned int numbers[] = {0, 0, 0, 0, 0}; const unsigned int k = 3; unsigned int n = 5; do{ for(int i=0 ; i&lt;k ; ++i) cout &lt;&lt; numbers[i] &lt;&lt; " "; cout &lt;&lt; endl; }while (next_combination(numbers, n, k)); return 0; } </code></pre>
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    1. USRami Jarrar
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    1. COYou must rethink this from the beginning. What is `numbers`? Why does it have five elements when (as far as I can tell) the code uses only three? Where would you store `1,3,6,7`? If `next_combination` finds the kth combination, aren't you doing a lot of work over and over?
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      1. USBeta
    2. COthe result combination each time is stored in numbers,, sorry, its not a problem the five elements, my algorithm, is just increasing the numbers starting from zeros.
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      1. POCombinations of positive integers uniq (order not important)
      1. USRami Jarrar
 

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