{"id":24087,"date":"2022-11-30T08:11:14","date_gmt":"2022-11-30T02:41:14","guid":{"rendered":"https:\/\/jassweb.com\/solved\/solved-python-reverse-list-with-k-steps-in-place-operator-vs-standard-operator-closed\/"},"modified":"2022-11-30T08:11:14","modified_gmt":"2022-11-30T02:41:14","slug":"solved-python-reverse-list-with-k-steps-in-place-operator-vs-standard-operator-closed","status":"publish","type":"post","link":"https:\/\/jassweb.com\/solved\/solved-python-reverse-list-with-k-steps-in-place-operator-vs-standard-operator-closed\/","title":{"rendered":"[Solved] Python reverse list with k steps, in-place operator vs standard operator [closed]"},"content":{"rendered":"

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The easiest way is to check how nums<\/code> looks like after every operation (with print(nums)<\/code> in scratch or calling it (nums<\/code>) in terminal). Next time, try it yourself.<\/p>\n

In A<\/strong>:
\nWith the first line (nums[:k] = nums[len(nums)-k:]<\/code>): nums = [5, 6, 7, 4, 5, 6, 7]<\/code> – you’ve substituted nums[0:3]<\/code> (1, 2, 3<\/code>) with nums[4:7]<\/code> (5, 6, 7<\/code>).<\/p>\n

Then, with the second line (nums[k:] = nums[:len(nums)-k]<\/code>): nums = [5, 6, 7, 5, 6, 7, 4]<\/code> – you’ve substituted nums[3:7]<\/code> (4, 5, 6, 7<\/code>) with nums[0:4]<\/code> (5, 6, 7, 4<\/code>).<\/p>\n

In B<\/strong>:
\nYou’ve done this operations “in place”, without changing the nums<\/code> between them. You’ve “grabbed” 1, 2, 3, 4<\/code> and 5, 6, 7<\/code>, made list [5,6,7,1,2,3,4]<\/code> and assigned it to nums<\/code> overwriting its value ([1,2,3,4,5,6,7]<\/code>).<\/p>\n<\/p><\/div>\n

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3<\/span> <\/p><\/div>\n<\/div>\n

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solved Python reverse list with k steps, in-place operator vs standard operator [closed] <\/p>\n","protected":false},"excerpt":{"rendered":"

[ad_1] The easiest way is to check how nums looks like after every operation (with print(nums) in scratch or calling it (nums) in terminal). Next time, try it yourself. In A: With the first line (nums[:k] = nums[len(nums)-k:]): nums = [5, 6, 7, 4, 5, 6, 7] – you’ve substituted nums[0:3] (1, 2, 3) with … 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":[540,349],"class_list":["post-24087","post","type-post","status-publish","format-standard","hentry","category-solved","tag-list","tag-python"],"yoast_head":"\n[Solved] Python reverse list with k steps, in-place operator vs standard operator [closed] - 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-python-reverse-list-with-k-steps-in-place-operator-vs-standard-operator-closed\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"[Solved] Python reverse list with k steps, in-place operator vs standard operator [closed] - JassWeb\" \/>\n<meta property=\"og:description\" content=\"[ad_1] The easiest way is to check how nums looks like after every operation (with print(nums) in scratch or calling it (nums) in terminal). Next time, try it yourself. In A: With the first line (nums[:k] = nums[len(nums)-k:]): nums = [5, 6, 7, 4, 5, 6, 7] – you’ve substituted nums[0:3] (1, 2, 3) with ... 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