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Message: <blockquote data-quote="Tom Riddle" data-source="post: 8900899" data-attributes="member: 47610"><p><span style="font-size: 15px"><span style="color: Blue">Sort the array first (you can do it in worst case O(n log n) using QuickSort ). Then read the array from the beginning till the end ONCE ( O<img src="data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7" class="smilie smilie--sprite smilie--sprite23" alt="(n)" title="Thumbs down    (n)" loading="lazy" data-shortname="(n)" /> ), and if you come across a duplicate remove it.</span></span></p><p><span style="font-size: 15px"><span style="color: Blue"></span></span></p><p><span style="font-size: 15px"><span style="color: Blue">So the total complexity is O(n log n), which is much better than O(n^2) which would result when using the obvious algorithm of taking each word and comparing it with all the other words in the array.</span></span></p><p><span style="font-size: 15px"><span style="color: Blue"></span></span></p><p><span style="font-size: 15px"><span style="color: Blue">University of Moratuwa.</span></span></p></blockquote><p></p>

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