{"id":28422,"date":"2022-12-30T21:09:28","date_gmt":"2022-12-30T15:39:28","guid":{"rendered":"https:\/\/jassweb.com\/solved\/solved-counting-efficiently-pairs-of-strings-which-together-contain-all-the-vowels\/"},"modified":"2022-12-30T21:09:28","modified_gmt":"2022-12-30T15:39:28","slug":"solved-counting-efficiently-pairs-of-strings-which-together-contain-all-the-vowels","status":"publish","type":"post","link":"https:\/\/jassweb.com\/solved\/solved-counting-efficiently-pairs-of-strings-which-together-contain-all-the-vowels\/","title":{"rendered":"[Solved] Counting efficiently pairs of strings, which together contain all the vowels"},"content":{"rendered":"<p> [ad_1]<br \/>\n<\/p>\n<div id=\"answer-54971700\" class=\"answer js-answer accepted-answer js-accepted-answer\" data-answerid=\"54971700\" data-parentid=\"54970876\" data-score=\"0\" data-position-on-page=\"1\" data-highest-scored=\"1\" data-question-has-accepted-highest-score=\"1\" itemprop=\"acceptedAnswer\" itemscope itemtype=\"https:\/\/schema.org\/Answer\">\n<div class=\"post-layout\">\n<div class=\"votecell post-layout--left\"><\/div>\n<div class=\"answercell post-layout--right\">\n<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>Suppose you have a very large number of strings.  Your algorithm will compare all the strings between them, and thats a terrible number of iterations.  <\/p>\n<h1>Radical improvement:<\/h1>\n<h2>New approach<\/h2>\n<p>Build a map that associates a string of ordered unique vowels (e.g. &#8220;ae&#8221;) to a list of all the string found that contains exactly these unique vowels, whatever the number of repetition and the order. For example:  <\/p>\n<pre><code>ao -&gt; aaooaoaooa, aoa, aaooaaooooooo  (3 words)\nuie -&gt; uiieieiieieuuu, uuuuiiiieeee   (2 words)\naeio -&gt; aeioooeeiiaiei                (1 word)\n<\/code><\/pre>\n<p>Of course, that&#8217;s a lot of strings, so in your code, you would use your bitmap rather than the string of ordered unique vowels.  Also note that you don&#8217;t want to produce the list of combined string, but only their count.  So you don&#8217;t need the list of all string occurenes, but just to maintain the count of strings matching the bitmap: <\/p>\n<pre><code>16385 -&gt; 3 \n1048848 -&gt; 2\n16657 -&gt; 1  \n<\/code><\/pre>\n<p>Then look at the winning combinations between the mapping existing indexes, like you did it. For a large list of strings, you would have a much smaller list of mapping indexes, so that would be a significant improvement.  <\/p>\n<p>For each winning combination take the size of the first list of strings time the size of the second list of strings to increase your count.  <\/p>\n<pre><code>16385 1048848 is complete -&gt; 3 x 2 = 6 combinations\n1048848 16657 is complete -&gt; 2 x 1 = 2 combinations \n                                    ---\n                                     8 potential combinations  \n<\/code><\/pre>\n<h2>What&#8217;s the improvement ?<\/h2>\n<p>These combinations were found by analysing 3&#215;2 bitmaps, rather than looking at 6&#215;5 bitmaps corresponding to the unique strings.  A gain by a significant order of magnitude if you have larger number of strings.<\/p>\n<p>To be more general, as you have 5 wovels and there must be at least one, you can have only a maximum of <code>2&lt;&lt;5-1<\/code> so 31 different bitmaps and therefore a maximum of C(31,2) that&#8217;s 465 <strike><code>31*30<\/code> that&#8217;s 930<\/strike> combinations to check whether you have 100 strings or 10 million strings as input.  Would I be correct to state that it&#8217;s roughly  O(n) ?   <\/p>\n<p>Possible implementation:  <\/p>\n<pre><code>map&lt;int, int&gt; mymap; \nint n;\ncin&gt;&gt;n;\n\nfor(int i=0;i&lt;n;i++) {\n    string s;\n    cin&gt;&gt;s;\n    int bm=0;\n    for(int j=0;j&lt;s.length();j++)\n        bm |= (1&lt;&lt;(s[j]-'a'));\n    mymap[bm]++;\n}\nint complete = (1&lt;&lt;('a'-'a')) | (1&lt;&lt;('e'-'a')) | (1&lt;&lt;('i'-'a')) | (1&lt;&lt;('o'-'a')) | (1&lt;&lt;('u'-'a'));\nint count = 0;\nint comparisons = 0; \nfor (auto i=mymap.begin(); i!=mymap.end(); i++)  {\n    auto j=i; \n    for(++j;j!=mymap.end();j++) {\n        comparisons++; \n        if((i-&gt;first | j-&gt;first)==complete) {\n            count += i-&gt;second * j-&gt;second; \n            cout &lt;&lt; i-&gt;first &lt;&lt;\" \"&lt;&lt;j-&gt;first&lt;&lt;\" :\"&lt;&lt;i-&gt;second&lt;&lt;\" \"&lt;&lt;j-&gt;second&lt;&lt;endl;\n        }\n    }\n}\nauto special = mymap.find(complete);  \/\/ special case: all strings having all letters\nif (special!=mymap.end()) {           \/\/ can be combined with themselves as well\n    count += special-&gt;second * (special-&gt;second -1) \/ 2; \n}\ncout&lt;&lt;\"Result: \"&lt;&lt;count&lt;&lt;\" (found in \"&lt;&lt;comparisons&lt;&lt;\" comparisons)\\n\";\n<\/code><\/pre>\n<p><a rel=\"nofollow noopener\" target=\"_blank\" href=\"https:\/\/ideone.com\/1anawY\">Online demo with 4 examples<\/a> (only strings with all letters, your initial example, my example above, and an example with couple of more strings)<\/p>\n<\/p><\/div>\n<div class=\"mt24\"><\/div>\n<\/div>\n<p>            <span class=\"d-none\" itemprop=\"commentCount\">6<\/span> <\/p><\/div>\n<\/div>\n<p>[ad_2]<\/p>\n<p>solved Counting efficiently pairs of strings, which together contain all the vowels <\/p>\n","protected":false},"excerpt":{"rendered":"<p>[ad_1] Suppose you have a very large number of strings. Your algorithm will compare all the strings between them, and thats a terrible number of iterations. Radical improvement: New approach Build a map that associates a string of ordered unique vowels (e.g. &#8220;ae&#8221;) to a list of all the string found that contains exactly these &#8230; <a title=\"[Solved] Counting efficiently pairs of strings, which together contain all the vowels\" class=\"read-more\" href=\"https:\/\/jassweb.com\/solved\/solved-counting-efficiently-pairs-of-strings-which-together-contain-all-the-vowels\/\" aria-label=\"More on [Solved] Counting efficiently pairs of strings, which together contain all the vowels\">Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[320],"tags":[362],"class_list":["post-28422","post","type-post","status-publish","format-standard","hentry","category-solved","tag-string"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.4 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>[Solved] Counting efficiently pairs of strings, which together contain all the vowels - JassWeb<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/jassweb.com\/solved\/solved-counting-efficiently-pairs-of-strings-which-together-contain-all-the-vowels\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"[Solved] Counting efficiently pairs of strings, which together contain all the vowels - JassWeb\" \/>\n<meta property=\"og:description\" content=\"[ad_1] Suppose you have a very large number of strings. Your algorithm will compare all the strings between them, and thats a terrible number of iterations. Radical improvement: New approach Build a map that associates a string of ordered unique vowels (e.g. &#8220;ae&#8221;) to a list of all the string found that contains exactly these ... 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