{"id":12898,"date":"2022-10-02T09:00:04","date_gmt":"2022-10-02T03:30:04","guid":{"rendered":"https:\/\/jassweb.com\/solved\/solved-convert-nested-for-loop-to-single-for-loop-in-java-closed\/"},"modified":"2022-10-02T09:00:04","modified_gmt":"2022-10-02T03:30:04","slug":"solved-convert-nested-for-loop-to-single-for-loop-in-java-closed","status":"publish","type":"post","link":"https:\/\/jassweb.com\/solved\/solved-convert-nested-for-loop-to-single-for-loop-in-java-closed\/","title":{"rendered":"[Solved] Convert nested ‘for’ loop to single ‘for’ loop in java [closed]"},"content":{"rendered":"

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I don’t think you can avoid two loops, but you can make it sequential<\/em>, instead of nested.<\/p>\n

long[] delta = new long[ar];\nfor(i=0;i<k;i++)\n{\n    l=sc.nextLong();\n    r=sc.nextLong();\n    v=sc.nextLong();\n\n    if (l-1 < r) {\n        m=(int)(l-1);\n        delta[m] = v;\n\n        m=(int)r;\n        if (m < delta.length) {\n          delta[m] = -v;\n        }\n    }\n}\n\nlong cumulative = 0;\nfor (int a = 0; a < ar.length; ++a) {\n  cumulative += delta[a];\n  ar[a] += cumulative;\n}\n<\/code><\/pre>\n
Rather than keeping on incrementing the values in delta<\/code> for each of the l,r<\/code> ranges, this just stores a “delta” array: this stores a +v<\/code> at the index where you’d start incrementing the array by v<\/code>; and -v<\/code> at the index where you’d stop.  So, recording that the range between l<\/code> and r<\/code> should be incremented by v<\/code> is now an O(1) operation, rather than an O(r-l) operation.<\/p>\n
So, some portion of this array looks like:<\/p>\n
     2    \n 0 0   0 0 0 0 0   0 0 0\n                -2\n<\/code><\/pre>\n
(I’m just vertically shifting the 2s to make it clearer)<\/p>\n
If you calculate the cumulative sum of these elements:<\/p>\n
     2 2 2 2 2 2  \n 0 0             0 0 0 0\n<\/code><\/pre>\n
In other words, you can show a range where you are going to increment by 2 by storing just the start and end positions of this range.<\/p>\n
This is what the cumulative<\/code> variable stores: it’s just the sum of all the elements in the delta<\/code> array to the left of and including the current position. And that’s the amount to increment the corresponding element of ar<\/code> by.<\/p>\n<\/p><\/div>\n
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solved Convert nested ‘for’ loop to single ‘for’ loop in java [closed] <\/p>\n","protected":false},"excerpt":{"rendered":"
[ad_1] I don’t think you can avoid two loops, but you can make it sequential, instead of nested. long[] delta = new long[ar]; for(i=0;i<k;i++) { l=sc.nextLong(); r=sc.nextLong(); v=sc.nextLong(); if (l-1 < r) { m=(int)(l-1); delta[m] = v; m=(int)r; if (m < delta.length) { delta[m] = -v; } } } long cumulative = 0; for (int … 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":[323,391,431],"class_list":["post-12898","post","type-post","status-publish","format-standard","hentry","category-solved","tag-java","tag-loops","tag-nested"],"yoast_head":"\n[Solved] Convert nested 'for' loop to single 'for' loop in java [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-convert-nested-for-loop-to-single-for-loop-in-java-closed\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"[Solved] Convert nested 'for' loop to single 'for' loop in java [closed] - JassWeb\" \/>\n<meta property=\"og:description\" content=\"[ad_1] I don’t think you can avoid two loops, but you can make it sequential, instead of nested. long[] delta = new long[ar]; for(i=0;i<k;i++) { l=sc.nextLong(); r=sc.nextLong(); v=sc.nextLong(); if (l-1 < r) { m=(int)(l-1); delta[m] = v; m=(int)r; if (m < delta.length) { delta[m] = -v; } } } long cumulative = 0; for (int ... 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