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POPairWise matching millions of records
 

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    <p>I have an algorithmic problem at hand. To easily explain the problem, I will be using a simple analogy. I have an input file </p> <pre><code>Country,Exports Austrailia,Sheep US, Apple Austrialia,Beef </code></pre> <p>End Goal: I have to find the common products between the pairs of countries so</p> <pre><code>{"Austrailia,New Zealand"}:{"apple","sheep} {"Austrialia,US"}:{"apple"} {"New Zealand","US"}:{"apple","milk"} </code></pre> <p>Process :</p> <p>I read in the input and store it in a TreeMap > Where the List, the strings are interned due to many duplicates. Essentially, I am aggregating by country. where Key is country, Values are its Exports.</p> <pre><code>{"austrailia":{"apple","sheep","koalas"}} {"new zealand":{"apple","sheep","milk"}} {"US":{"apple","beef","milk"}} </code></pre> <p>I have about 1200 keys (countries) and total number of values(exports) is 80 million altogether. I sort all the values of each key:</p> <pre><code>{"austrailia":{"apple","sheep","koalas"}} -- &gt; {"austrailia":{"apple","koalas","sheep"}} </code></pre> <p>This is fast as there are only 1200 Lists to sort.</p> <pre><code>for(k1:keys) for(k2:keys) if(k1.compareTo(k2) &lt;0){ //Dont want to double compare List&lt;String&gt; intersectList = intersectList_func(k1's exports,k2's exports); countriespair.put({k1,k2},intersectList) } </code></pre> <p>This code block takes so long.I realise it O(n2) and around 1200*1200 comparisions.Thus,Running for almost 3 hours till now.. Is there any way, I can speed it up or optimise it. Algorithm wise is best option, or are there other technologies to consider.</p> <p><strong>Edit</strong>: <em>Since both List are sorted beforehand, the intersectList is O(n) where n is length of floor(listOne.length,listTwo.length) and NOT O(n2) as discussed below</em></p> <pre><code>private static List&lt;String&gt; intersectList(List&lt;String&gt; listOne,List&lt;String&gt; listTwo){ int i=0,j=0; List&lt;String&gt; listResult = new LinkedList&lt;String&gt;(); while(i!=listOne.size() &amp;&amp; j!=listTwo.size()){ int compareVal = listOne.get(i).compareTo(listTwo.get(j)); if(compareVal==0){ listResult.add(listOne.get(i)); i++;j++;} } else if(compareVal &lt; 0) i++; else if (compareVal &gt;0) j++; } return listResult; } </code></pre> <p><strong>Update 22 Nov</strong> My current implementation is still running for almost 18 hours. :|</p> <p><strong>Update 25 Nov</strong> I had run the new implementation as suggested by Vikram and a few others. It's been running this Friday. My question, is that how does grouping by exports rather than country save computational complexity. I find that the complexity is the same. As Groo mentioned, I find that the complexity for the second part is O(E*C^2) where is E is exports and C is country.</p>
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    1. COUse a SQL DB and query would be a possible solution.
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    2. CO@prog_guy Give your input file to test my code on
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