{"id":37982,"date":"2026-01-20T14:31:01","date_gmt":"2026-01-20T11:31:01","guid":{"rendered":"https:\/\/fatihsoysal.com\/blog\/baslangic-dostu-rehber-minimum-bitwise-dizi-olusturma-i-problem-3314\/"},"modified":"2026-01-20T14:31:01","modified_gmt":"2026-01-20T11:31:01","slug":"baslangic-dostu-rehber-minimum-bitwise-dizi-olusturma-i-problem-3314","status":"publish","type":"post","link":"https:\/\/fatihsoysal.com\/blog\/baslangic-dostu-rehber-minimum-bitwise-dizi-olusturma-i-problem-3314\/","title":{"rendered":"Ba\u015flang\u0131\u00e7 Dostu Rehber: Minimum Bitwise Dizi Olu\u015fturma I (Problem 3314)"},"content":{"rendered":"<h2>Ba\u015flang\u0131\u00e7 Dostu Rehber: Minimum Bitwise Dizi Olu\u015fturma I (Problem 3314)<\/h2>\n<p>\n    Yaz\u0131l\u0131m geli\u015ftirme d\u00fcnyas\u0131nda, \u00f6zellikle performans\u0131n kritik oldu\u011fu veya d\u00fc\u015f\u00fck seviyeli manip\u00fclasyonlar\u0131n gerekti\u011fi alanlarda, bitwise (bit d\u00fczeyinde) i\u015flemler hayati bir rol oynar. Bu i\u015flemler, say\u0131lar\u0131n ikili g\u00f6sterimleri \u00fczerinde do\u011frudan \u00e7al\u0131\u015farak, veri s\u0131k\u0131\u015ft\u0131rma, \u015fifreleme, grafik programlama ve algoritmik problemler gibi \u00e7e\u015fitli alanlarda g\u00fc\u00e7l\u00fc ve verimli \u00e7\u00f6z\u00fcmler sunar. Bu makalede, bitwise i\u015flemlerin temel prensiplerini ke\u015ffedecek ve &#8220;Minimum Bitwise Dizi Olu\u015fturma I&#8221; (Problem 3314) ad\u0131n\u0131 verdi\u011fimiz, ba\u015flang\u0131\u00e7 seviyesi dostu ancak d\u00fc\u015f\u00fcnd\u00fcr\u00fcc\u00fc bir problemi ele alaca\u011f\u0131z. Amac\u0131m\u0131z, verilen bir tam say\u0131dan yola \u00e7\u0131karak, bitwise OR ko\u015fulunu sa\u011flayan ve elemanlar\u0131n\u0131n toplam\u0131 minimum olan bir dizi olu\u015fturman\u0131n mant\u0131\u011f\u0131n\u0131, algoritmas\u0131n\u0131 ve C++, Python, JavaScript dillerindeki uygulamalar\u0131n\u0131 ad\u0131m ad\u0131m \u00f6\u011frenmektir. Haz\u0131rsan\u0131z, bitlerin b\u00fcy\u00fcl\u00fc d\u00fcnyas\u0131na dalal\u0131m!\n<\/p>\n<h2>Bitwise \u0130\u015flemlerine H\u0131zl\u0131 Bir Bak\u0131\u015f<\/h2>\n<p>\n    Bitwise i\u015flemler, say\u0131lar\u0131n bellekteki ikili (binary) g\u00f6sterimleri \u00fczerinde do\u011frudan \u00e7al\u0131\u015f\u0131r. Her say\u0131 bir dizi bit olarak temsil edilir (0 veya 1). Bu i\u015flemler, programc\u0131lara veriler \u00fczerinde \u00e7ok hassas kontrol sa\u011flar.\n<\/p>\n<h3>AND (&#038;) Operat\u00f6r\u00fc<\/h3>\n<p>\n    Bitwise AND operat\u00f6r\u00fc, iki say\u0131n\u0131n kar\u015f\u0131l\u0131kl\u0131 bitlerini kar\u015f\u0131la\u015ft\u0131r\u0131r. Yaln\u0131zca her iki bit de 1 ise sonu\u00e7 biti 1 olur, aksi takdirde 0 olur. Genellikle belirli bir bitin set (1) olup olmad\u0131\u011f\u0131n\u0131 kontrol etmek veya bir say\u0131n\u0131n belirli bitlerini maskelemek i\u00e7in kullan\u0131l\u0131r.\n<\/p>\n<pre><code class=\"language-cpp\">\n\/\/ \u00d6rnek: 5 (0101) & 3 (0011) = 1 (0001)\nint a = 5; \/\/ 0101\nint b = 3; \/\/ 0011\nint result = a & b; \/\/ result = 1 (0001)\n<\/pre>\n<p><\/code><\/p>\n<h3>OR (|) Operat\u00f6r\u00fc<\/h3>\n<p>\n    Bitwise OR operat\u00f6r\u00fc, iki say\u0131n\u0131n kar\u015f\u0131l\u0131kl\u0131 bitlerini kar\u015f\u0131la\u015ft\u0131r\u0131r. Kar\u015f\u0131l\u0131kl\u0131 bitlerden en az biri 1 ise sonu\u00e7 biti 1 olur, aksi takdirde 0 olur. Genellikle bir say\u0131n\u0131n belirli bitlerini set etmek (1 yapmak) i\u00e7in kullan\u0131l\u0131r.\n<\/p>\n<pre><code class=\"language-cpp\">\n\/\/ \u00d6rnek: 5 (0101) | 3 (0011) = 7 (0111)\nint a = 5; \/\/ 0101\nint b = 3; \/\/ 0011\nint result = a | b; \/\/ result = 7 (0111)\n<\/pre>\n<p><\/code><\/p>\n<h3>XOR (^) Operat\u00f6r\u00fc<\/h3>\n<p>\n    Bitwise XOR (Exclusive OR) operat\u00f6r\u00fc, iki say\u0131n\u0131n kar\u015f\u0131l\u0131kl\u0131 bitlerini kar\u015f\u0131la\u015ft\u0131r\u0131r. Bitler farkl\u0131 ise (biri 0, di\u011feri 1) sonu\u00e7 biti 1 olur, aksi takdirde (her ikisi de 0 veya her ikisi de 1) 0 olur. Genellikle bitleri de\u011fi\u015ftirmek veya bir say\u0131n\u0131n belirli bitlerini tersine \u00e7evirmek i\u00e7in kullan\u0131l\u0131r.\n<\/p>\n<pre><code class=\"language-cpp\">\n\/\/ \u00d6rnek: 5 (0101) ^ 3 (0011) = 6 (0110)\nint a = 5; \/\/ 0101\nint b = 3; \/\/ 0011\nint result = a ^ b; \/\/ result = 6 (0110)\n<\/pre>\n<p><\/code><\/p>\n<h3>NOT (~) ve Kayd\u0131rma (<<, >>) Operat\u00f6rleri<\/h3>\n<p>\n    Bitwise NOT (~) operat\u00f6r\u00fc, bir say\u0131n\u0131n t\u00fcm bitlerini tersine \u00e7evirir (0'lar\u0131 1, 1'leri 0 yapar). Bit kayd\u0131rma operat\u00f6rleri ise bir say\u0131n\u0131n bitlerini sola (<code><<<\/code>) veya sa\u011fa (<code>>><\/code>) belirli bir miktar kayd\u0131r\u0131r. Sola kayd\u0131rma, say\u0131y\u0131 2'nin kuvvetleriyle \u00e7arpmakla e\u015fde\u011ferdir (<code>x << y<\/code> = <code>x * 2^y<\/code>). Sa\u011fa kayd\u0131rma ise say\u0131y\u0131 2'nin kuvvetleriyle b\u00f6lmekle e\u015fde\u011ferdir (<code>x >> y<\/code> = <code>x \/ 2^y<\/code>). Bu operat\u00f6rler, belirli bitleri olu\u015fturmak veya izole etmek i\u00e7in \u00e7ok kullan\u0131\u015fl\u0131d\u0131r.\n<\/p>\n<h2>Problem 3314: Minimum Bitwise Dizi Nedir?<\/h2>\n<p>\n    \u015eimdi, bu makalenin odak noktas\u0131 olan \"Minimum Bitwise Dizi Olu\u015fturma I\" problemine yak\u0131ndan bakal\u0131m.\n<\/p>\n<h3>Problem Tan\u0131m\u0131 ve K\u0131s\u0131tlamalar<\/h3>\n<p>\n    Problem 3314'\u00fcn temel tan\u0131m\u0131 \u015fu \u015fekildedir:\n<\/p>\n<blockquote><p>\n    Verilen bir <code>K<\/code> tam say\u0131s\u0131 i\u00e7in, <code>N<\/code> uzunlu\u011funda bir <code>A<\/code> dizisi olu\u015fturun \u00f6yle ki, <code>A[0] | A[1] | ... | A[N-1] = K<\/code> ko\u015fulu sa\u011flans\u0131n ve <code>A<\/code> dizisindeki elemanlar\u0131n toplam\u0131 (<code>sum(A)<\/code>) minimum olsun. <b>Ayr\u0131ca, <code>N<\/code> de\u011feri, <code>K<\/code> say\u0131s\u0131ndaki set (1) bitlerinin say\u0131s\u0131na e\u015fit olmal\u0131d\u0131r.<\/b>\n<\/p><\/blockquote>\n<p>\n    Bu problemde, <code>N<\/code> de\u011feri bize do\u011frudan verilmez; <code>K<\/code> say\u0131s\u0131n\u0131n ikili g\u00f6sterimindeki 1'lerin say\u0131s\u0131 (popcount) ile belirlenir. \u00d6rne\u011fin, <code>K = 13<\/code> (ikili olarak <code>1101<\/code>) ise, 3 adet set bit oldu\u011fundan <code>N = 3<\/code> olmal\u0131d\u0131r.\n<\/p>\n<h3>Neden Minimum Toplam?<\/h3>\n<p>\n    Bitwise OR operat\u00f6r\u00fcn\u00fcn temel \u00f6zelli\u011fi, e\u011fer bir bit <code>K<\/code>'da set (1) ise, bu bitin dizideki en az bir eleman <code>A[i]<\/code>'de de set olmas\u0131 gerekti\u011fidir. Elemanlar\u0131n toplam\u0131n\u0131 minimize etmek i\u00e7in, <code>K<\/code>'n\u0131n her set bitinin m\u00fcmk\u00fcn oldu\u011funca az elemana da\u011f\u0131t\u0131lmas\u0131 gerekir. \u0130deal senaryo, <code>K<\/code>'n\u0131n her set bitinin dizideki <b>yaln\u0131zca bir<\/b> elemana ait olmas\u0131d\u0131r. Bu durumda, dizideki elemanlar\u0131n toplam\u0131, <code>K<\/code>'n\u0131n kendisine e\u015fit olacakt\u0131r. \u00d6rne\u011fin, <code>K = 13 (1101_2)<\/code> i\u00e7in, <code>8 (1000_2)<\/code>, <code>4 (0100_2)<\/code> ve <code>1 (0001_2)<\/code> elemanlar\u0131n\u0131 i\u00e7eren bir dizi olu\u015fturursak, bu elemanlar\u0131n OR'u <code>13<\/code>'\u00fc verir ve toplamlar\u0131 da <code>13<\/code>'t\u00fcr. Bu, elde edebilece\u011fimiz en k\u00fc\u00e7\u00fck toplamd\u0131r.\n<\/p>\n<h3>\u00d6rnek Senaryo<\/h3>\n<p>\n    <code>K = 13<\/code> (ikili olarak <code>1101<\/code>)\n<\/p>\n<ul>\n<li><code>K<\/code>'daki set bitler: 2<sup>0<\/sup> (1), 2<sup>2<\/sup> (4), 2<sup>3<\/sup> (8).<\/li>\n<li>Set bit say\u0131s\u0131 (<code>N<\/code>) = 3.<\/li>\n<li>Olu\u015fturulacak dizi <code>A<\/code>: <code>[1, 4, 8]<\/code> (veya farkl\u0131 s\u0131ralama).<\/li>\n<li>Dizinin OR'u: <code>1 | 4 | 8 = 13<\/code>.<\/li>\n<li>Dizinin toplam\u0131: <code>1 + 4 + 8 = 13<\/code>.<\/li>\n<\/ul>\n<p>\n    Bu \u00f6rnek, problem tan\u0131m\u0131na ve minimum toplam hedefine m\u00fckemmel bir \u015fekilde uyar.\n<\/p>\n<h2>\u00c7\u00f6z\u00fcme Giden Yol: Bitleri<\/p>\n","protected":false},"excerpt":{"rendered":"Yaz\u0131l\u0131m geli\u015ftirme d\u00fcnyas\u0131nda, \u00f6zellikle performans\u0131n kritik oldu\u011fu veya d\u00fc\u015f\u00fck seviyeli manip\u00fclasyonlar\u0131n gerekti\u011fi alanlarda, bitwise (bit d\u00fc&#8230;","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"csco_page_header_type":"","csco_page_load_nextpost":"","csco_page_subscribe_form":"","csco_page_contact_form":"","footnotes":""},"categories":[1],"tags":[],"class_list":{"0":"post-37982","1":"post","2":"type-post","3":"status-publish","4":"format-standard","6":"category-genel","7":"cs-entry","8":"cs-video-wrap"},"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v20.5 (Yoast SEO v25.3.1) - 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