{"id":15457,"date":"2022-10-11T16:14:47","date_gmt":"2022-10-11T10:44:47","guid":{"rendered":"https:\/\/jassweb.com\/solved\/solved-time-complexity-of-binary-search\/"},"modified":"2022-10-11T16:14:47","modified_gmt":"2022-10-11T10:44:47","slug":"solved-time-complexity-of-binary-search","status":"publish","type":"post","link":"https:\/\/jassweb.com\/solved\/solved-time-complexity-of-binary-search\/","title":{"rendered":"[Solved] Time Complexity of Binary Search?"},"content":{"rendered":"

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With binary search you typically search in a sorted random access data structure like an array, by discarding half of the array with each comparison. Hence, in k<\/code> steps you effectively cover 2^k<\/code> entries. This yields a complexity of at most log2(n)<\/code> of n<\/code> elements.<\/p>\n

With landau symbols, the base of the logarithm disappears because it is a constant: O(log2(n)) = O(log(n) \/ log(2)) = O(log(n))<\/code>.<\/p>\n

Now, if you, for some reason, can not only discard half of the values, but two thirds, by always knowing in which third the needle will end up in, this means you cover 3^k<\/code> many entries in k<\/code> steps.<\/p>\n

Hence, you get log3(n)<\/code>. But this again reduces to the same time complexity as log(3)<\/code> is a constant: O(log3(n)) = O(log(n)\/log(3)) = O(log(n))<\/code>.<\/p>\n<\/p><\/div>\n

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solved Time Complexity of Binary Search? <\/p>\n","protected":false},"excerpt":{"rendered":"

[ad_1] With binary search you typically search in a sorted random access data structure like an array, by discarding half of the array with each comparison. Hence, in k steps you effectively cover 2^k entries. This yields a complexity of at most log2(n) of n elements. With landau symbols, the base of the logarithm disappears … 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":[457,1382],"class_list":["post-15457","post","type-post","status-publish","format-standard","hentry","category-solved","tag-algorithm","tag-binary-search"],"yoast_head":"\n[Solved] Time Complexity of Binary Search? - 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-time-complexity-of-binary-search\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"[Solved] Time Complexity of Binary Search? - JassWeb\" \/>\n<meta property=\"og:description\" content=\"[ad_1] With binary search you typically search in a sorted random access data structure like an array, by discarding half of the array with each comparison. 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