\u0130ki \u0130\u015faret\u00e7i Tekni\u011fi, algoritmik verimlilik arayan yaz\u0131l\u0131mc\u0131lar i\u00e7in vazge\u00e7ilmez bir ara\u00e7t\u0131r. Naif O(N\u00b2) \u00e7\u00f6z\u00fcmlerden zarif O(N) optimizasyonlara ge\u00e7i\u015f yaparak kodunuzu h\u0131zland\u0131r\u0131r. Bu teknikle, veri yap\u0131lar\u0131n\u0131 daha etkin i\u015fleyebilir, zaman ve bellekten tasarruf edebilirsiniz.<\/p>\n
Yaz\u0131l\u0131m geli\u015ftirme s\u00fcre\u00e7lerinde, \u00f6zellikle b\u00fcy\u00fck veri setleriyle \u00e7al\u0131\u015f\u0131rken, algoritmalar\u0131n performans\u0131 kritik bir rol oynar. Kimi zaman basit g\u00f6r\u00fcnen bir problem, do\u011fru yakla\u015f\u0131m se\u00e7ilmedi\u011finde sistem kaynaklar\u0131n\u0131 a\u015f\u0131r\u0131 t\u00fcketerek uygulaman\u0131n yava\u015flamas\u0131na hatta kilitlenmesine neden olabilir. \u0130\u015fte tam da bu noktada, \u201c\u0130ki \u0130\u015faret\u00e7i Tekni\u011fi\u201d (Two-Pointer Technique) adeta bir sihirli de\u011fnek gibi devreye girer. Bu teknik, belirli t\u00fcrdeki problemlerin \u00e7\u00f6z\u00fcm\u00fcn\u00fc sadece daha h\u0131zl\u0131 de\u011fil, ayn\u0131 zamanda daha az kaynak t\u00fcketerek ger\u00e7ekle\u015ftirmemizi sa\u011flayan zarif bir y\u00f6ntem sunar. Gelin, bu g\u00fc\u00e7l\u00fc algoritma tasar\u0131m kal\u0131b\u0131n\u0131 s\u0131f\u0131rdan ba\u015flayarak detayl\u0131ca inceleyelim ve kod \u00f6rnekleriyle nas\u0131l hayat buldu\u011funu g\u00f6relim.<\/p>\n
\u0130ki \u0130\u015faret\u00e7i Tekni\u011fi, temel olarak bir veri yap\u0131s\u0131 (genellikle dizi veya ba\u011fl\u0131 liste) \u00fczerinde ayn\u0131 anda hareket eden iki farkl\u0131 i\u015faret\u00e7i (indeks veya referans) kullanarak problemleri \u00e7\u00f6zmeye odaklanan bir algoritma desenidir. Bu i\u015faret\u00e7iler, belirli ko\u015fullara g\u00f6re ilerleyerek veya geri \u00e7ekilerek veri yap\u0131s\u0131n\u0131n farkl\u0131 b\u00f6lgelerini ayn\u0131 anda inceleme f\u0131rsat\u0131 sunar. Bu e\u015f zamanl\u0131 inceleme, naif yakla\u015f\u0131mlar\u0131n birden fazla i\u00e7 i\u00e7e d\u00f6ng\u00fc kullanmas\u0131n\u0131 gerektiren durumlar\u0131 ortadan kald\u0131rarak, zaman karma\u015f\u0131kl\u0131\u011f\u0131n\u0131 genellikle O(N\u00b2) gibi y\u00fcksek de\u011ferlerden O(N) gibi \u00e7ok daha verimli de\u011ferlere indirmemizi sa\u011flar.<\/p>\n
Peki, bu teknik neden bu kadar \u00f6nemlidir? En ba\u015fta, h\u0131z ve verimlilik modern yaz\u0131l\u0131m\u0131n temel ta\u015flar\u0131d\u0131r. Kullan\u0131c\u0131lar, web sitelerinden mobil uygulamalara kadar her alanda an\u0131nda yan\u0131t beklerler. O(N\u00b2) bir algoritma, girdi boyutu N artt\u0131k\u00e7a performans\u0131n\u0131n katlanarak k\u00f6t\u00fcle\u015fti\u011fi anlam\u0131na gelir. \u00d6rne\u011fin, 1.000 elemanl\u0131 bir dizide 1.000.000 i\u015flem yaparken, 100.000 elemanl\u0131 bir dizide 10.000.000.000 i\u015flem yapman\u0131z gerekebilir. Bu t\u00fcr bir b\u00fcy\u00fcme, ger\u00e7ek d\u00fcnya senaryolar\u0131nda kabul edilemez gecikmelere yol a\u00e7ar. \u0130ki \u0130\u015faret\u00e7i Tekni\u011fi ise, her eleman\u0131 en fazla sabit say\u0131da (genellikle bir veya iki kez) ziyaret ederek bu problemi k\u00f6kten \u00e7\u00f6zer. Bu sayede, girdi boyutu b\u00fcy\u00fcd\u00fck\u00e7e performans kayb\u0131 do\u011frusal bir \u015fekilde artar ki bu, \u00e7o\u011fu uygulama i\u00e7in \u00e7ok daha y\u00f6netilebilir bir durumdur.<\/p>\n
Ayr\u0131ca, bu teknik genellikle O(1) sabit bellek karma\u015f\u0131kl\u0131\u011f\u0131na sahip \u00e7\u00f6z\u00fcmler \u00fcretir. Yani, algoritman\u0131n \u00e7al\u0131\u015fmas\u0131 i\u00e7in ek bellek alan\u0131 girdinin boyutuna ba\u011fl\u0131 olarak artmaz; her zaman sabit ve minimal bir bellek miktar\u0131 kullan\u0131l\u0131r. Bu, \u00f6zellikle bellek k\u0131s\u0131tl\u0131 ortamlarda (\u00f6rne\u011fin g\u00f6m\u00fcl\u00fc sistemler veya eski mobil cihazlar) veya \u00e7ok b\u00fcy\u00fck veri setleriyle \u00e7al\u0131\u015f\u0131rken son derece de\u011ferli bir \u00f6zelliktir. \u0130ki \u0130\u015faret\u00e7i Tekni\u011fi’nin bu kadar geni\u015f bir uygulama yelpazesine sahip olmas\u0131n\u0131n ve bir\u00e7ok standart algoritma m\u00fclakat\u0131nda s\u0131k\u00e7a sorulmas\u0131n\u0131n ana nedenleri de i\u015fte bu \u00fcst\u00fcn performans ve kaynak verimlili\u011fi yetenekleridir.<\/p>\n
Uzman \u0130pucu: \u0130ki \u0130\u015faret\u00e7i Tekni\u011fi, \u00e7o\u011fu zaman s\u0131ral\u0131 (sorted) dizilerde en etkili \u015fekilde \u00e7al\u0131\u015f\u0131r. Verileriniz s\u0131ral\u0131 de\u011filse, \u00f6ncelikle s\u0131ralama algoritmas\u0131 uygulaman\u0131z gerekebilir, ancak bunun getirece\u011fi ek zaman karma\u015f\u0131kl\u0131\u011f\u0131n\u0131 (\u00f6rne\u011fin O(N log N)) g\u00f6z \u00f6n\u00fcnde bulundurun.<\/p>\n<\/div>\n
Bir\u00e7ok programlama problemine ilk yakla\u015ft\u0131\u011f\u0131m\u0131zda akl\u0131m\u0131za gelen en do\u011frudan y\u00f6ntem genellikle “naif \u00e7\u00f6z\u00fcm” olarak adland\u0131r\u0131l\u0131r. Bu \u00e7\u00f6z\u00fcmler, problemin tan\u0131m\u0131n\u0131 do\u011frudan kodlayan ve genellikle basit for d\u00f6ng\u00fcleri veya i\u00e7 i\u00e7e ge\u00e7mi\u015f d\u00f6ng\u00fcler kullanan yakla\u015f\u0131mlard\u0131r. \u00d6rne\u011fin, belirli bir toplam\u0131 veren iki say\u0131y\u0131 bir dizide bulma problemini ele alal\u0131m. Dizideki her say\u0131y\u0131, dizideki di\u011fer her say\u0131yla e\u015fle\u015ftirerek toplamlar\u0131n\u0131 kontrol etmek, naif bir yakla\u015f\u0131md\u0131r. \u0130\u015fte b\u00f6yle bir durum i\u00e7in basit bir Python \u00f6rne\u011fi:<\/p>\n
\ndef naif_iki_sayi_toplami(dizi, hedef_toplam):\n n = len(dizi)\n for i in range(n):\n for j in range(i + 1, n): # Her eleman\u0131 di\u011ferleriyle bir kez kar\u015f\u0131la\u015ft\u0131r\u0131r\n if dizi[i] + dizi[j] == hedef_toplam:\n return [dizi[i], dizi[j]]\n return []\n\n# \u00d6rnek kullan\u0131m\nprint(naif_iki_sayi_toplami([3, 5, 2, 8, 11, 15], 10)) # \u00c7\u0131kt\u0131: [2, 8]\nprint(naif_iki_sayi_toplami([4, 2, 7, 1, 9], 10)) # \u00c7\u0131kt\u0131: [1, 9]\n<\/pre>\n<\/code><\/p>\n
Yukar\u0131daki kod blo\u011funda g\u00f6rd\u00fc\u011f\u00fcn\u00fcz gibi, d\u0131\u015f d\u00f6ng\u00fc N kez, i\u00e7 d\u00f6ng\u00fc ise ortalama N\/2 kez \u00e7al\u0131\u015f\u0131r. Bu durumda, algoritman\u0131n toplam \u00e7al\u0131\u015fma s\u00fcresi N * (N\/2) civar\u0131nda olur, bu da O(N\u00b2) zaman karma\u015f\u0131kl\u0131\u011f\u0131na denk gelir. K\u00fc\u00e7\u00fck diziler i\u00e7in bu kabul edilebilir olabilir. Ancak, dizinin eleman say\u0131s\u0131 (N) 100.000'e ula\u015ft\u0131\u011f\u0131nda, i\u00e7 i\u00e7e d\u00f6ng\u00fcler trilyonlarca i\u015flemi ifade edebilir ve bu da program\u0131n\u0131z\u0131n dakikalarca, hatta saatlerce \u00e7al\u0131\u015fmas\u0131na neden olabilir. Bu senaryolar, ger\u00e7ek d\u00fcnya uygulamalar\u0131nda asla kabul edilemez.<\/p>\n
\u0130\u015fte burada \u0130ki \u0130\u015faret\u00e7i Tekni\u011fi devreye girerek dramatik bir fark yarat\u0131r. Ayn\u0131 problemi, dizi \u00f6nceden s\u0131ralanm\u0131\u015fsa, O(N) zaman karma\u015f\u0131kl\u0131\u011f\u0131 ile \u00e7\u00f6zebiliriz. Tekni\u011fin temelinde, iki i\u015faret\u00e7inin (biri ba\u015ftan, di\u011feri sondan ba\u015flayan) birbirine do\u011fru hareket etmesi yatar. Toplam hedefimizden k\u00fc\u00e7\u00fckse sol i\u015faret\u00e7iyi art\u0131r\u0131r\u0131z, b\u00fcy\u00fckse sa\u011f i\u015faret\u00e7iyi azalt\u0131r\u0131z. B\u00f6ylece, her ad\u0131mda dizideki arama alan\u0131m\u0131z\u0131 daralt\u0131r\u0131z ve her eleman\u0131 sadece bir kez ziyaret etme garantisi veririz. Bu yakla\u015f\u0131m, sadece birka\u00e7 saniye i\u00e7inde devasa veri setlerini i\u015fleyebilen algoritmalar\u0131n kap\u0131lar\u0131n\u0131 aralar. Naif \u00e7\u00f6z\u00fcmlerin aksine, \u0130ki \u0130\u015faret\u00e7i Tekni\u011fi, donan\u0131m kaynaklar\u0131n\u0131 \u00e7ok daha verimli kullanarak h\u0131zl\u0131 ve \u00f6l\u00e7eklenebilir uygulamalar geli\u015ftirmemize olanak tan\u0131r.<\/p>\n
\u0130ki \u0130\u015faret\u00e7i Tekni\u011finin Temel Uygulama Modelleri Nelerdir?<\/h2>\n
\u0130ki \u0130\u015faret\u00e7i Tekni\u011fi, genel olarak iki ana modelde kar\u015f\u0131m\u0131za \u00e7\u0131kar: z\u0131t y\u00f6nl\u00fc (converging) i\u015faret\u00e7iler ve ayn\u0131 y\u00f6nl\u00fc (sliding window veya fast\/slow) i\u015faret\u00e7iler. Her iki modelin de kendine \u00f6zg\u00fc kullan\u0131m alanlar\u0131 ve avantajlar\u0131 bulunmaktad\u0131r. Bu b\u00f6l\u00fcmlerde, bu temel modelleri detayl\u0131 \u00f6rneklerle inceleyelim.<\/p>\n
Z\u0131t Y\u00f6nl\u00fc (Converging) \u0130\u015faret\u00e7ilerle \u00c7\u00f6z\u00fcmler Nas\u0131l Geli\u015ftirilir?<\/h3>\n
Z\u0131t y\u00f6nl\u00fc i\u015faret\u00e7iler modeli, genellikle s\u0131ral\u0131 dizilerde belirli bir \u00f6zelli\u011fe sahip eleman \u00e7iftlerini veya alt dizileri bulmak i\u00e7in kullan\u0131l\u0131r. Bu modelde, bir i\u015faret\u00e7i dizinin ba\u015f\u0131ndan (genellikle left<\/code> veya i<\/code>), di\u011fer i\u015faret\u00e7i ise dizinin sonundan (genellikle right<\/code> veya j<\/code>) ba\u015flar. Her ad\u0131mda, bu iki i\u015faret\u00e7i birbirine do\u011fru ilerler ve aralar\u0131ndaki elemanlar \u00fczerinde i\u015flem yapar.<\/p>\nEn klasik \u00f6rne\u011fi, s\u0131ral\u0131 bir dizide belirli bir hedef_toplam<\/code>\u0131 veren iki say\u0131y\u0131 bulma problemidir. \u0130\u015faret\u00e7iler left = 0<\/code> ve right = len(dizi) - 1<\/code> olarak ba\u015flat\u0131l\u0131r. Her ad\u0131mda, dizi[left]<\/code> ve dizi[right]<\/code> elemanlar\u0131n\u0131n toplam\u0131 kontrol edilir:<\/p>\n\n- E\u011fer toplam
hedef_toplam<\/code>a e\u015fitse, arad\u0131\u011f\u0131m\u0131z \u00e7ifti bulmu\u015fuz demektir.<\/li>\n- E\u011fer toplam
hedef_toplam<\/code>dan k\u00fc\u00e7\u00fckse, toplam\u0131 art\u0131rmak i\u00e7in daha b\u00fcy\u00fck bir say\u0131ya ihtiyac\u0131m\u0131z var demektir. Bu durumda left<\/code> i\u015faret\u00e7isi bir sa\u011fa kayd\u0131r\u0131l\u0131r (left += 1<\/code>).<\/li>\n- E\u011fer toplam
hedef_toplam<\/code>dan b\u00fcy\u00fckse, toplam\u0131 azaltmak i\u00e7in daha k\u00fc\u00e7\u00fck bir say\u0131ya ihtiyac\u0131m\u0131z var demektir. Bu durumda right<\/code> i\u015faret\u00e7isi bir sola kayd\u0131r\u0131l\u0131r (right -= 1<\/code>).<\/li>\n<\/ul>\nBu s\u00fcre\u00e7 left < right<\/code> oldu\u011fu s\u00fcrece devam eder. Bu ak\u0131\u015f, dizinin s\u0131ral\u0131 olmas\u0131n\u0131n sa\u011flad\u0131\u011f\u0131 avantajla her ad\u0131mda do\u011fru y\u00f6nde ilerlememizi garanti eder.<\/p>\n\ndef iki_sayi_toplami_iki_isaretci(dizi, hedef_toplam):\n left = 0\n right = len(dizi) - 1\n\n while left < right:\n mevcut_toplam = dizi[left] + dizi[right]\n\n if mevcut_toplam == hedef_toplam:\n return [dizi[left], dizi[right]]\n elif mevcut_toplam < hedef_toplam:\n left += 1 # Toplam\u0131 art\u0131rmak i\u00e7in sol i\u015faret\u00e7iyi sa\u011fa kayd\u0131r\n else: # mevcut_toplam > hedef_toplam\n right -= 1 # Toplam\u0131 azaltmak i\u00e7in sa\u011f i\u015faret\u00e7iyi sola kayd\u0131r\n \n return [] # \u00c7ift bulunamazsa bo\u015f liste d\u00f6nd\u00fcr\n\n# \u00d6rnek kullan\u0131m (dizinin s\u0131ral\u0131 olmas\u0131 gerekti\u011fini unutmay\u0131n!)\nprint(iki_sayi_toplami_iki_isaretci([2, 3, 5, 8, 11, 15], 10)) # \u00c7\u0131kt\u0131: [2, 8]\nprint(iki_sayi_toplami_iki_isaretci([1, 2, 4, 7, 9], 10)) # \u00c7\u0131kt\u0131: [1, 9]\nprint(iki_sayi_toplami_iki_isaretci([1, 2, 3, 4, 5], 100)) # \u00c7\u0131kt\u0131: []\n<\/pre>\n<\/code><\/p>\n
Bu yakla\u015f\u0131m\u0131n zaman karma\u015f\u0131kl\u0131\u011f\u0131 O(N)'dir \u00e7\u00fcnk\u00fc her iki i\u015faret\u00e7i de dizinin ba\u015f\u0131ndan sonuna do\u011fru sadece birer kez ilerler. Bellek karma\u015f\u0131kl\u0131\u011f\u0131 ise O(1)'dir \u00e7\u00fcnk\u00fc ekstra depolama alan\u0131 kullanmay\u0131z. Bu, naif O(N\u00b2) \u00e7\u00f6z\u00fcm\u00fcne k\u0131yasla muazzam bir iyile\u015fmedir.<\/p>\n
Ayn\u0131 Y\u00f6nl\u00fc (Same-Direction) \u0130\u015faret\u00e7ilerle Dizileri Nas\u0131l Optimize Ederiz?<\/h3>\n
Ayn\u0131 y\u00f6nl\u00fc i\u015faret\u00e7iler modeli, genellikle bir dizinin veya ba\u011fl\u0131 listenin belirli ko\u015fullar\u0131 kar\u015f\u0131layan alt k\u0131s\u0131mlar\u0131n\u0131 (alt dizi, alt dize) bulmak veya diziyi yerinde (in-place) d\u00f6n\u00fc\u015ft\u00fcrmek i\u00e7in kullan\u0131l\u0131r. Bu modelde, her iki i\u015faret\u00e7i de ayn\u0131 y\u00f6nde (genellikle ba\u015ftan sona) ilerler, ancak farkl\u0131 h\u0131zlarda veya farkl\u0131 g\u00f6revlerle hareket ederler.<\/p>\n
Bu modelin pop\u00fcler bir kullan\u0131m\u0131, s\u0131ral\u0131 bir dizideki yinelenen elemanlar\u0131 yerinde kald\u0131rmak ve benzersiz elemanlar\u0131 dizinin ba\u015f\u0131na ta\u015f\u0131makt\u0131r. Diyelim ki [1, 1, 2, 2, 3, 4, 4, 5]<\/code> gibi bir dizimiz var ve bunu [1, 2, 3, 4, 5, _, _, _]<\/code> \u015feklinde d\u00f6n\u00fc\u015ft\u00fcrmek istiyoruz, burada alt \u00e7izgiler \u00f6nemsizdir. \u0130ki i\u015faret\u00e7i kullan\u0131r\u0131z: slow<\/code> (yava\u015f) i\u015faret\u00e7i benzersiz elemanlar\u0131n yaz\u0131laca\u011f\u0131 pozisyonu tutar ve fast<\/code> (h\u0131zl\u0131) i\u015faret\u00e7i dizideki t\u00fcm elemanlar\u0131 tarar.<\/p>\n\nslow<\/code> i\u015faret\u00e7isini 0'dan ba\u015flat\u0131r\u0131z.<\/li>\nfast<\/code> i\u015faret\u00e7isini 1'den ba\u015flat\u0131r ve dizinin sonuna kadar ilerletiriz.<\/li>\n- E\u011fer
dizi[fast]<\/code> eleman\u0131 dizi[slow]<\/code> eleman\u0131ndan farkl\u0131ysa, bu yeni bir benzersiz eleman demektir. slow<\/code> i\u015faret\u00e7isini bir art\u0131r\u0131r\u0131z ve dizi[slow]<\/code> pozisyonuna dizi[fast]<\/code> de\u011ferini yazar\u0131z.<\/li>\n- E\u011fer
dizi[fast]<\/code> eleman\u0131 dizi[slow]<\/code> eleman\u0131na e\u015fitse (yani yinelenen bir elemansa), hi\u00e7bir \u015fey yapmay\u0131z ve sadece fast<\/code> i\u015faret\u00e7isini ilerletmeye devam ederiz.<\/li>\n<\/ul>\nBu i\u015flem tamamland\u0131\u011f\u0131nda, slow + 1<\/code> bize benzersiz elemanlar\u0131n say\u0131s\u0131n\u0131 ve ayn\u0131 zamanda benzersiz elemanlardan olu\u015fan yeni dizinin uzunlu\u011funu verecektir.<\/p>\n\ndef yinelenenleri_kaldir_yerinde(dizi):\n if not dizi:\n return 0 # Bo\u015f dizi ise 0 d\u00f6nd\u00fcr\n\n slow = 0 # Benzersiz elemanlar\u0131n yaz\u0131laca\u011f\u0131 indeks\n for fast in range(1, len(dizi)): # Dizideki her eleman\u0131 tarar\n if dizi[fast] != dizi[slow]:\n slow += 1 # slow i\u015faret\u00e7isini ilerlet\n dizi[slow] = dizi[fast] # Yeni benzersiz eleman\u0131 yaz\n \n return slow + 1 # Yeni dizinin uzunlu\u011fu\n\n# \u00d6rnek kullan\u0131m\nmy_list = [1, 1, 2, 2, 3, 4, 4, 5]\nnew_length = yinelenenleri_kaldir_yerinde(my_list)\nprint(f\"Yeni uzunluk: {new_length}\") # \u00c7\u0131kt\u0131: Yeni uzunluk: 5\nprint(f\"D\u00f6n\u00fc\u015ft\u00fcr\u00fclm\u00fc\u015f dizi: {my_list[:new_length]}\") # \u00c7\u0131kt\u0131: D\u00f6n\u00fc\u015ft\u00fcr\u00fclm\u00fc\u015f dizi: [1, 2, 3, 4, 5]\n\nmy_list_2 = [1, 1, 1, 1]\nnew_length_2 = yinelenenleri_kaldir_yerinde(my_list_2)\nprint(f\"Yeni uzunluk: {new_length_2}\") # \u00c7\u0131kt\u0131: Yeni uzunluk: 1\nprint(f\"D\u00f6n\u00fc\u015ft\u00fcr\u00fclm\u00fc\u015f dizi: {my_list_2[:new_length_2]}\") # \u00c7\u0131kt\u0131: D\u00f6n\u00fc\u015ft\u00fcr\u00fclm\u00fc\u015f dizi: [1]\n<\/pre>\n<\/code><\/p>\n
Bu yine O(N) zaman karma\u015f\u0131kl\u0131\u011f\u0131na ve O(1) sabit bellek karma\u015f\u0131kl\u0131\u011f\u0131na sahip zarif bir \u00e7\u00f6z\u00fcmd\u00fcr. fast<\/code> i\u015faret\u00e7isi dizide bir kez gezinir ve her eleman sabit say\u0131da i\u015flem g\u00f6r\u00fcr. Bu, ayn\u0131 y\u00f6nl\u00fc i\u015faret\u00e7ilerin g\u00fcc\u00fcn\u00fc ve verimlili\u011fini g\u00f6steren harika bir \u00f6rnektir.<\/p>\nGer\u00e7ek D\u00fcnya Senaryolar\u0131nda \u0130ki \u0130\u015faret\u00e7i Tekni\u011fi Vaka Analizleri<\/h2>\n
\u0130ki \u0130\u015faret\u00e7i Tekni\u011fi, teorik problemlerin \u00f6tesinde, ger\u00e7ek d\u00fcnya yaz\u0131l\u0131m geli\u015ftirme s\u00fcre\u00e7lerinde s\u0131kl\u0131kla kar\u015f\u0131m\u0131za \u00e7\u0131kan \u00e7e\u015fitli zorluklar\u0131n \u00fcstesinden gelmek i\u00e7in kullan\u0131labilir. Bu b\u00f6l\u00fcmde, bu tekni\u011fin nas\u0131l pratik uygulamalara d\u00f6n\u00fc\u015ft\u00fc\u011f\u00fcn\u00fc g\u00f6steren iki pop\u00fcler vaka analizine odaklanaca\u011f\u0131z.<\/p>\n
Vaka 1: Ge\u00e7erli Palindrom Kontrol\u00fc Nas\u0131l Yap\u0131l\u0131r?<\/h3>\n
Bir dize (string) palindrom mudur? Palindrom, tersten okundu\u011funda da ayn\u0131 olan bir dizedir (\"madam\", \"kayak\"). Ancak, ger\u00e7ek d\u00fcnya senaryolar\u0131nda, bu kontrol\u00fc yaparken sadece alfan\u00fcmerik karakterleri dikkate almam\u0131z ve b\u00fcy\u00fck\/k\u00fc\u00e7\u00fck harf duyarl\u0131l\u0131\u011f\u0131n\u0131 g\u00f6rmezden gelmemiz gerekebilir. \u00d6rne\u011fin, \"A man, a plan, a canal: Panama\" c\u00fcmlesi, bo\u015fluklar\u0131, noktalama i\u015faretlerini ve harf b\u00fcy\u00fckl\u00fc\u011f\u00fcn\u00fc yok sayd\u0131\u011f\u0131m\u0131zda bir palindromdur. \u0130\u015fte bu problemi iki i\u015faret\u00e7i tekni\u011fiyle etkili bir \u015fekilde \u00e7\u00f6zebiliriz.<\/p>\n
Bu senaryoda, yine z\u0131t y\u00f6nl\u00fc iki i\u015faret\u00e7i kullanaca\u011f\u0131z: biri dizenin ba\u015f\u0131ndan (left<\/code>), di\u011feri dizenin sonundan (right<\/code>). Her ad\u0131mda, left<\/code> i\u015faret\u00e7isi bir alfan\u00fcmerik karaktere gelene kadar sa\u011fa, right<\/code> i\u015faret\u00e7isi bir alfan\u00fcmerik karaktere gelene kadar sola hareket eder. Karakterler bulundu\u011funda, k\u00fc\u00e7\u00fck harfe d\u00f6n\u00fc\u015ft\u00fcr\u00fclerek kar\u015f\u0131la\u015ft\u0131r\u0131l\u0131r. E\u011fer e\u015fle\u015fmezlerse, dize bir palindrom de\u011fildir. E\u015fle\u015firlerse, i\u015faret\u00e7iler bir sonraki alfan\u00fcmerik karakterlere do\u011fru ilerler.<\/p>\n\ndef isValidChar(char):\n # Karakterin alfan\u00fcmerik olup olmad\u0131\u011f\u0131n\u0131 kontrol eder\n return 'a' <= char <= 'z' or '0' <= char <= '9'\n\ndef isPalindrome(s: str) -> bool:\n s = s.lower() # T\u00fcm karakterleri k\u00fc\u00e7\u00fck harfe d\u00f6n\u00fc\u015ft\u00fcr\n left = 0\n right = len(s) - 1\n\n while left < right:\n # Sol i\u015faret\u00e7iyi alfan\u00fcmerik bir karaktere gelene kadar ilerlet\n while left < right and not isValidChar(s[left]):\n left += 1\n # Sa\u011f i\u015faret\u00e7iyi alfan\u00fcmerik bir karaktere gelene kadar geri \u00e7ek\n while left < right and not isValidChar(s[right]):\n right -= 1\n \n # Karakterleri kar\u015f\u0131la\u015ft\u0131r\n if s[left] != s[right]:\n return False # E\u015fle\u015fmezse palindrom de\u011fildir\n \n # \u0130\u015faret\u00e7ileri i\u00e7eri do\u011fru hareket ettir\n left += 1\n right -= 1\n \n return True # T\u00fcm kontroller ba\u015far\u0131l\u0131ysa palindromdur\n\n# \u00d6rnek kullan\u0131m\nprint(isPalindrome(\"A man, a plan, a canal: Panama\")) # \u00c7\u0131kt\u0131: True\nprint(isPalindrome(\"race a car\")) # \u00c7\u0131kt\u0131: False\nprint(isPalindrome(\"madam\")) # \u00c7\u0131kt\u0131: True\nprint(isPalindrome(\"No lemon, no melon\")) # \u00c7\u0131kt\u0131: True\n<\/pre>\n<\/code><\/p>\n
Bu \u00e7\u00f6z\u00fcm, t\u00fcm dizeyi sadece bir kez tarad\u0131\u011f\u0131 i\u00e7in O(N) zaman karma\u015f\u0131kl\u0131\u011f\u0131na sahiptir. Ayr\u0131ca, dizenin bir kopyas\u0131n\u0131 olu\u015fturmad\u0131\u011f\u0131 veya ek veri yap\u0131lar\u0131 kullanmad\u0131\u011f\u0131 i\u00e7in O(1) bellek karma\u015f\u0131kl\u0131\u011f\u0131 sunar.<\/p>\n
Vaka 2: Belirli Bir De\u011feri Dizinin Sonuna Ta\u015f\u0131ma Problemi<\/h3>\n
Bazen bir dizideki belirli bir de\u011feri (\u00f6rne\u011fin s\u0131f\u0131rlar\u0131) dizinin sonuna ta\u015f\u0131mak isteyebiliriz, ancak di\u011fer elemanlar\u0131n g\u00f6receli s\u0131ralamas\u0131n\u0131 bozmamal\u0131y\u0131z. \u00d6rne\u011fin, [0, 1, 0, 3, 12]<\/code> dizisini [1, 3, 12, 0, 0]<\/code> haline getirmek. Bu da ayn\u0131 y\u00f6nl\u00fc iki i\u015faret\u00e7i ile kolayca \u00e7\u00f6z\u00fclebilir bir problemdir.<\/p>\nBurada yine slow<\/code> ve fast<\/code> ad\u0131nda iki i\u015faret\u00e7i kullanaca\u011f\u0131z. slow<\/code> i\u015faret\u00e7isi, hedef de\u011fere sahip olmayan elemanlar\u0131n yaz\u0131laca\u011f\u0131 mevcut pozisyonu g\u00f6sterirken, fast<\/code> i\u015faret\u00e7isi dizinin t\u00fcm elemanlar\u0131n\u0131 tarayacak. Hedef de\u011ferimiz, \u00f6rne\u011fin 0 olsun.<\/p>\n\nslow<\/code> i\u015faret\u00e7isi 0'dan ba\u015flar.<\/li>\nfast<\/code> i\u015faret\u00e7isi 0'dan ba\u015flar ve dizinin sonuna kadar ilerler.<\/li>\n- E\u011fer
dizi[fast]<\/code> hedef de\u011fer (0) de\u011filse, bu eleman\u0131 dizi[slow]<\/code> pozisyonuna yazar\u0131z ve slow<\/code> i\u015faret\u00e7isini bir art\u0131r\u0131r\u0131z. Bu, hedef olmayan elemanlar\u0131 dizinin ba\u015f\u0131na toplamam\u0131z\u0131 sa\u011flar.<\/li>\n- E\u011fer
dizi[fast]<\/code> hedef de\u011fer (0) ise, hi\u00e7bir \u015fey yapmay\u0131z ve sadece fast<\/code> i\u015faret\u00e7isini ilerletiriz. Bu eleman\u0131 \u015fimdilik g\u00f6z ard\u0131 ederiz, \u00e7\u00fcnk\u00fc dizinin sonuna ta\u015f\u0131nacak.<\/li>\n<\/ul>\nD\u00f6ng\u00fc bittikten sonra, slow<\/code> i\u015faret\u00e7isinden ba\u015flayarak dizinin sonuna kadar olan t\u00fcm elemanlar\u0131 hedef de\u011fer (0) ile doldururuz. Bu, di\u011fer elemanlar\u0131n g\u00f6receli s\u0131ras\u0131n\u0131 korurken t\u00fcm s\u0131f\u0131rlar\u0131 dizinin sonuna ta\u015f\u0131m\u0131\u015f olur.<\/p>\n\ndef belirli_degeri_sona_tasi(dizi, hedef_deger):\n slow = 0 # Hedef olmayan elemanlar\u0131n yaz\u0131laca\u011f\u0131 pozisyon\n\n # Diziyi fast i\u015faret\u00e7isi ile tarar\n for fast in range(len(dizi)):\n if dizi[fast] != hedef_deger:\n dizi[slow] = dizi[fast] # Hedef olmayan eleman\u0131 slow pozisyonuna yaz\n slow += 1 # slow i\u015faret\u00e7isini ilerlet\n\n # slow'dan sonraki t\u00fcm pozisyonlar\u0131 hedef de\u011fer ile doldur\n while slow < len(dizi):\n dizi[slow] = hedef_deger\n slow += 1\n \n return dizi\n\n# \u00d6rnek kullan\u0131m\narr1 = [0, 1, 0, 3, 12]\nprint(belirli_degeri_sona_tasi(arr1, 0)) # \u00c7\u0131kt\u0131: [1, 3, 12, 0, 0]\n\narr2 = [2, 0, 0, 1, 3, 0]\nprint(belirli_degeri_sona_tasi(arr2, 0)) # \u00c7\u0131kt\u0131: [2, 1, 3, 0, 0, 0]\n\narr3 = [1, 2, 3, 4, 5]\nprint(belirli_degeri_sona_tasi(arr3, 0)) # \u00c7\u0131kt\u0131: [1, 2, 3, 4, 5]\n<\/pre>\n<\/code><\/p>\n
Yine, bu \u00e7\u00f6z\u00fcm O(N) zaman karma\u015f\u0131kl\u0131\u011f\u0131na ve O(1) bellek karma\u015f\u0131kl\u0131\u011f\u0131na sahiptir, \u00e7\u00fcnk\u00fc dizi \u00fczerinde tek bir ge\u00e7i\u015f yapar\u0131z ve yerinde de\u011fi\u015fiklikler uygular\u0131z. Bu \u00f6rnekler, iki i\u015faret\u00e7i tekni\u011finin sadece teorik de\u011fil, ayn\u0131 zamanda g\u00fcnl\u00fck programlama g\u00f6revlerinde de ne kadar g\u00fc\u00e7l\u00fc ve kullan\u0131\u015fl\u0131 olabilece\u011fini a\u00e7\u0131k\u00e7a g\u00f6stermektedir.<\/p>\n
\u0130leri D\u00fczey \u0130ki \u0130\u015faret\u00e7i Stratejileri ve Performans \u0130pu\u00e7lar\u0131<\/h2>\n
\u0130ki \u0130\u015faret\u00e7i Tekni\u011fi'nin temellerini ve pop\u00fcler uygulama alanlar\u0131n\u0131 ele ald\u0131k. Ancak bu tekni\u011fin potansiyeli, basit \u00e7ift bulma veya eleman ta\u015f\u0131ma i\u015flemlerinin \u00e7ok \u00f6tesine uzan\u0131r. Daha karma\u015f\u0131k problemlerde de bu temel deseni farkl\u0131 stratejilerle birle\u015ftirerek g\u00fc\u00e7l\u00fc \u00e7\u00f6z\u00fcmler \u00fcretebiliriz.<\/p>\n
Kayar Pencere (Sliding Window) Deseni<\/h3>\n
Kayar Pencere deseni, \u0130ki \u0130\u015faret\u00e7i Tekni\u011fi'nin daha soyut ve g\u00fc\u00e7l\u00fc bir uygulamas\u0131d\u0131r. Bu desen, genellikle bir dizi veya dize i\u00e7inde belirli ko\u015fullar\u0131 kar\u015f\u0131layan en uzun, en k\u0131sa veya en \u00e7ok\/az tekrarlayan alt diziyi (veya alt dizeyi) bulmak i\u00e7in kullan\u0131l\u0131r. \u0130ki i\u015faret\u00e7i (start<\/code> ve end<\/code>), dizide bir \"pencere\" tan\u0131mlar. Bu pencere, end<\/code> i\u015faret\u00e7isi ilerledik\u00e7e geni\u015fler ve start<\/code> i\u015faret\u00e7isi belirli ko\u015fullar kar\u015f\u0131land\u0131\u011f\u0131nda veya a\u015f\u0131ld\u0131\u011f\u0131nda daral\u0131r.<\/p>\n\u00d6rne\u011fin, bir dizide belirli bir toplam\u0131 a\u015fmayan en uzun alt diziyi bulma problemi. end<\/code> i\u015faret\u00e7isiyle pencereyi sa\u011fa do\u011fru geni\u015fletiriz, mevcut toplam\u0131 g\u00fcncelleriz. E\u011fer mevcut toplam hedefi a\u015farsa, start<\/code> i\u015faret\u00e7isini sa\u011fa kayd\u0131rarak pencereyi daralt\u0131r\u0131z ve start<\/code> eleman\u0131n\u0131 toplamdan \u00e7\u0131kar\u0131r\u0131z. Bu s\u00fcre\u00e7te en uzun ge\u00e7erli pencereyi takip ederiz. Kayar pencere, genellikle O(N) zaman karma\u015f\u0131kl\u0131\u011f\u0131 sunar \u00e7\u00fcnk\u00fc her eleman pencereye bir kez girer ve bir kez \u00e7\u0131kar.<\/p>\n\n# Kayar Pencere \u00d6rne\u011fi: En uzun K ay\u0131rt edici karaktere sahip alt dize\n# Basit bir kod blo\u011fu (tam \u00e7\u00f6z\u00fcmden ziyade desen vurgusu i\u00e7in)\ndef en_uzun_kayar_pencere(s, k):\n char_map = {}\n window_start = 0\n max_length = 0\n\n for window_end in range(len(s)):\n right_char = s[window_end]\n char_map[right_char] = char_map.get(right_char, 0) + 1\n\n # K ay\u0131rt edici karakter s\u0131n\u0131r\u0131n\u0131 a\u015ft\u0131ysa pencereyi k\u00fc\u00e7\u00fclt\n while len(char_map) > k:\n left_char = s[window_start]\n char_map[left_char] -= 1\n if char_map[left_char] == 0:\n del char_map[left_char]\n window_start += 1\n \n # Ge\u00e7erli pencere uzunlu\u011funu g\u00fcncelle\n max_length = max(max_length, window_end - window_start + 1)\n \n return max_length\n\n# print(en_uzun_kayar_pencere(\"araaci\", 2)) # \u00c7\u0131kt\u0131: 4 (araa)\n<\/pre>\n<\/code><\/p>\n
H\u0131zl\u0131 ve Yava\u015f \u0130\u015faret\u00e7iler (Fast and Slow Pointers)<\/h3>\n
Bu \u00f6zel ayn\u0131 y\u00f6nl\u00fc i\u015faret\u00e7i t\u00fcr\u00fc, \u00f6zellikle ba\u011fl\u0131 listelerde d\u00f6ng\u00fcleri tespit etmek veya ortadaki eleman\u0131 bulmak gibi problemlerde kullan\u0131l\u0131r. H\u0131zl\u0131 i\u015faret\u00e7i, yava\u015f i\u015faret\u00e7iden daha h\u0131zl\u0131 ilerler (\u00f6rne\u011fin, h\u0131zl\u0131 i\u015faret\u00e7i her ad\u0131mda iki, yava\u015f i\u015faret\u00e7i bir ad\u0131m ilerler). E\u011fer bir d\u00f6ng\u00fc varsa, h\u0131zl\u0131 i\u015faret\u00e7i sonunda yava\u015f i\u015faret\u00e7iye yeti\u015fecektir. Bu teknik, Floyd'un D\u00f6ng\u00fc Bulma Algoritmas\u0131 (Tortoise and Hare algorithm) olarak da bilinir ve O(N) zaman, O(1) bellek karma\u015f\u0131kl\u0131\u011f\u0131na sahiptir.<\/p>\n
\u00c7oklu \u0130\u015faret\u00e7iler (Multiple Pointers)<\/h3>\n
Baz\u0131 problemler ikiden fazla i\u015faret\u00e7i gerektirebilir. \u00d6rne\u011fin, \"\u00dc\u00e7 Say\u0131n\u0131n Toplam\u0131 S\u0131f\u0131r\" (3Sum) problemi. Bu problemde, s\u0131ral\u0131 bir dizide toplamlar\u0131 s\u0131f\u0131r olan \u00fc\u00e7l\u00fcleri bulmak i\u00e7in d\u0131\u015fta bir d\u00f6ng\u00fc ve i\u00e7te iki i\u015faret\u00e7ili bir yakla\u015f\u0131m (z\u0131t y\u00f6nl\u00fc) kullan\u0131r\u0131z. Bu t\u00fcr yakla\u015f\u0131mlar genellikle O(N\u00b2) zaman karma\u015f\u0131kl\u0131\u011f\u0131na sahip olsa da, naif O(N\u00b3) \u00e7\u00f6z\u00fcmlere g\u00f6re \u00f6nemli bir iyile\u015fme sa\u011flar.<\/p>\n
\nUzman \u0130pucu: \u0130ki i\u015faret\u00e7i kullan\u0131rken, d\u00f6ng\u00fc sonland\u0131rma ko\u015fullar\u0131na (left < right<\/code>, fast < len(dizi)<\/code> vb.) ve i\u015faret\u00e7i ba\u015flang\u0131\u00e7 pozisyonlar\u0131na \u00f6zellikle dikkat edin. K\u00f6\u015fe durumlar\u0131 (bo\u015f dizi, tek elemanl\u0131 dizi) i\u00e7in test yapmay\u0131 unutmay\u0131n!<\/p>\n<\/div>\nAlgoritmik Verimlilik ve Web Performans\u0131: Mobil Uyumlu Yakla\u015f\u0131mlar<\/h2>\n
G\u00fcn\u00fcm\u00fcz\u00fcn dijital d\u00fcnyas\u0131nda web uygulamalar\u0131n\u0131n ve sitelerin mobil cihazlarda sorunsuz \u00e7al\u0131\u015fmas\u0131, kullan\u0131c\u0131 deneyimi a\u00e7\u0131s\u0131ndan olmazsa olmaz bir gerekliliktir. Algoritmik verimlilik, yaln\u0131zca b\u00fcy\u00fck veri i\u015fleme veya karma\u015f\u0131k hesaplamalar i\u00e7in de\u011fil, ayn\u0131 zamanda son kullan\u0131c\u0131n\u0131n kar\u015f\u0131la\u015ft\u0131\u011f\u0131 web ve mobil uygulama performans\u0131n\u0131 do\u011frudan etkileyen kritik bir fakt\u00f6rd\u00fcr. O(N) gibi verimli algoritmalar, sunucu taraf\u0131nda veri i\u015fleme veya istemci taraf\u0131nda b\u00fcy\u00fck listeleri i\u015fleme gibi g\u00f6revlerde kullan\u0131ld\u0131\u011f\u0131nda, sayfa y\u00fckleme s\u00fcrelerini azalt\u0131r ve uygulaman\u0131n genel yan\u0131t h\u0131z\u0131n\u0131 art\u0131r\u0131r.<\/p>\n
D\u00fc\u015f\u00fcn\u00fcn ki bir e-ticaret sitesinde binlerce \u00fcr\u00fcn\u00fcn filtrelenmesi veya aranmas\u0131 gerekiyor. E\u011fer bu i\u015flemler i\u00e7in naif O(N\u00b2) algoritmalar kullan\u0131l\u0131rsa, mobil cihazlarda ba\u011flant\u0131 h\u0131z\u0131n\u0131n ve i\u015flem g\u00fcc\u00fcn\u00fcn daha k\u0131s\u0131tl\u0131 oldu\u011fu durumlarda, kullan\u0131c\u0131 saniyelerce hatta dakikalarca beklemek zorunda kalabilir. O(N) optimizasyonlar sayesinde ise bu i\u015flemler saniyeler i\u00e7inde tamamlanarak ak\u0131c\u0131 bir deneyim sunulur. Dolay\u0131s\u0131yla, \u0130ki \u0130\u015faret\u00e7i Tekni\u011fi gibi algoritmik yakla\u015f\u0131mlar, web uygulamas\u0131n\u0131n arka planda ne kadar verimli \u00e7al\u0131\u015ft\u0131\u011f\u0131n\u0131 belirler.<\/p>\n
Web performans\u0131 sadece algoritmik verimlilikle s\u0131n\u0131rl\u0131 de\u011fildir; ayn\u0131 zamanda \u00f6n u\u00e7 geli\u015ftirme pratikleriyle de yak\u0131ndan ili\u015fkilidir. Mobil uyumlu bir HTML yap\u0131s\u0131 olu\u015fturmak, sayfa boyutunu k\u00fc\u00e7\u00fcltmek ve taray\u0131c\u0131n\u0131n i\u015fleme y\u00fck\u00fcn\u00fc azaltmak, algoritmalar\u0131n sa\u011flad\u0131\u011f\u0131 h\u0131z\u0131 destekleyen unsurlard\u0131r. \u00d6rne\u011fin, gereksiz DOM elemanlar\u0131ndan ka\u00e7\u0131nmak, resimleri optimize etmek (lazy loading), CSS ve JavaScript dosyalar\u0131n\u0131 k\u00fc\u00e7\u00fcltmek (minification) ve HTTP isteklerini azaltmak gibi teknikler, mobil cihazlarda daha h\u0131zl\u0131 y\u00fckleme s\u00fcreleri sa\u011flar.<\/p>\n
Ayr\u0131ca, duyarl\u0131 tasar\u0131m (responsive design) ilkeleri, farkl\u0131 ekran boyutlar\u0131na ve \u00e7\u00f6z\u00fcn\u00fcrl\u00fcklerine uyum sa\u011flayarak mobil kullan\u0131c\u0131lar\u0131n sitenizi sorunsuz bir \u015fekilde g\u00f6r\u00fcnt\u00fclemesini ve etkile\u015fimde bulunmas\u0131n\u0131 sa\u011flar. CSS medya sorgular\u0131 (media queries) kullanarak cihaz \u00f6zelliklerine g\u00f6re farkl\u0131 stiller uygulamak, bu duyarl\u0131 tasar\u0131m\u0131n temelini olu\u015fturur. \u00d6rne\u011fin, k\u00fc\u00e7\u00fck ekranlar i\u00e7in daha basit bir d\u00fczen sunmak veya belirli \u00f6\u011feleri gizlemek, hem g\u00f6rsel deneyimi iyile\u015ftirir hem de sayfan\u0131n daha h\u0131zl\u0131 i\u015flenmesine yard\u0131mc\u0131 olur. Bu ba\u011flamda, arka planda \u00e7al\u0131\u015fan algoritmalar\u0131n verimlili\u011fi ile \u00f6n y\u00fczde uygulanan performans optimizasyonlar\u0131, mobil cihazlarda en iyi kullan\u0131c\u0131 deneyimini sunmak i\u00e7in el ele gider.<\/p>\n
Sonu\u00e7<\/h2>\n
\u0130ki \u0130\u015faret\u00e7i Tekni\u011fi, algoritmik problem \u00e7\u00f6zme repertuvar\u0131n\u0131zda mutlaka bulunmas\u0131 gereken temel ve g\u00fc\u00e7l\u00fc bir ara\u00e7t\u0131r. Bu makalede, bu tekni\u011fin ne oldu\u011funu, naif yakla\u015f\u0131mlara g\u00f6re neden daha \u00fcst\u00fcn oldu\u011funu ve z\u0131t y\u00f6nl\u00fc ile ayn\u0131 y\u00f6nl\u00fc olmak \u00fczere iki ana uygulama modelini detayl\u0131 kod \u00f6rnekleriyle inceledik. Palindrom kontrol\u00fc ve belirli bir de\u011feri dizinin sonuna ta\u015f\u0131ma gibi ger\u00e7ek d\u00fcnya senaryolar\u0131nda nas\u0131l kullan\u0131labilece\u011fini g\u00f6sterdik. Ayr\u0131ca, kayar pencere ve h\u0131zl\u0131-yava\u015f i\u015faret\u00e7iler gibi ileri d\u00fczey stratejilere de\u011findik ve mobil uyumlu web uygulamalar\u0131n\u0131n performans\u0131nda algoritmik verimlili\u011fin \u00f6nemini vurgulad\u0131k.<\/p>\n
Bu tekni\u011fi \u00f6\u011frenmek ve uygulamak, kodunuzu sadece daha h\u0131zl\u0131 ve verimli hale getirmekle kalmaz, ayn\u0131 zamanda daha elegant ve okunabilir \u00e7\u00f6z\u00fcmler \u00fcretmenize de yard\u0131mc\u0131 olur. Unutmay\u0131n ki pratik yapmak, bu ve benzeri algoritmik desenlerde ustala\u015fman\u0131n anahtar\u0131d\u0131r. Farkl\u0131 problem t\u00fcrlerinde bu tekni\u011fi uygulamaya \u00e7al\u0131\u015farak, problem \u00e7\u00f6zme kaslar\u0131n\u0131z\u0131 g\u00fc\u00e7lendirecek ve daha karma\u015f\u0131k algoritmik zorluklar\u0131n \u00fcstesinden gelebilecek bir yetkinlik kazanacaks\u0131n\u0131z. Algoritma d\u00fcnyas\u0131n\u0131n kap\u0131lar\u0131n\u0131 aralayan bu yolculukta, \u0130ki \u0130\u015faret\u00e7i Tekni\u011fi sizin sa\u011flam bir temeliniz olacakt\u0131r.<\/p>\n
S\u0131k\u00e7a Sorulan Sorular<\/h3>\n\n- \n \u0130ki \u0130\u015faret\u00e7i tekni\u011fi her zaman uygulanabilir mi?<\/strong>
\n Hay\u0131r, \u0130ki \u0130\u015faret\u00e7i Tekni\u011fi her zaman uygulanamaz. Genellikle, s\u0131ral\u0131 veri yap\u0131lar\u0131 (diziler, ba\u011fl\u0131 listeler) \u00fczerinde veya belirli bir mant\u0131ksal s\u0131raya sahip problemlerde en etkili \u015fekilde kullan\u0131l\u0131r. Verinin s\u0131ral\u0131 olmad\u0131\u011f\u0131 durumlarda, \u00f6ncelikle s\u0131ralama i\u015flemi yapmak gerekebilir, ancak bu ek bir zaman karma\u015f\u0131kl\u0131\u011f\u0131 getirecektir.\n <\/li>\n- \n Hangi veri yap\u0131lar\u0131nda en etkilidir?<\/strong>
\n Bu teknik, ba\u015fta diziler (array) ve dizeler (string) olmak \u00fczere, elemanlara indeks veya referans arac\u0131l\u0131\u011f\u0131yla h\u0131zl\u0131 eri\u015fim sa\u011flanabilen do\u011frusal veri yap\u0131lar\u0131nda son derece etkilidir. Ba\u011fl\u0131 listelerde de h\u0131zl\u0131 ve yava\u015f i\u015faret\u00e7i deseniyle s\u0131k\u00e7a kullan\u0131l\u0131r.\n <\/li>\n- \n Zaman karma\u015f\u0131kl\u0131\u011f\u0131 neden genellikle O(N)'dir?<\/strong>
\n \u0130ki \u0130\u015faret\u00e7i Tekni\u011fi'nin zaman karma\u015f\u0131kl\u0131\u011f\u0131 genellikle O(N)'dir \u00e7\u00fcnk\u00fc her i\u015faret\u00e7i, veri yap\u0131s\u0131 \u00fczerinde en fazla bir kez ba\u015ftan sona (veya sondan ba\u015fa) hareket eder. Bu, her eleman\u0131n sabit say\u0131da i\u015flem g\u00f6rmesini sa\u011flar ve girdinin boyutuna (N) do\u011frusal olarak orant\u0131l\u0131 bir \u00e7al\u0131\u015fma s\u00fcresi anlam\u0131na gelir gelir.\n <\/li>\n- \n \u0130ki \u0130\u015faret\u00e7i tekni\u011fi ile Kayar Pencere (Sliding Window) aras\u0131ndaki fark nedir?<\/strong>
\n \u0130ki \u0130\u015faret\u00e7i Tekni\u011fi, iki i\u015faret\u00e7i kullanarak bir veri yap\u0131s\u0131n\u0131 i\u015fleme genel bir algoritma desenidir. Kayar Pencere (Sliding Window) ise, genellikle iki i\u015faret\u00e7iyi (pencerenin ba\u015flang\u0131c\u0131 ve sonu) kullanarak bir alt diziyi veya alt dizeyi dinamik olarak tan\u0131mlayan ve boyutunu ayarlayan \u00f6zel bir \u0130ki \u0130\u015faret\u00e7i uygulamas\u0131d\u0131r. Yani, Kayar Pencere, \u0130ki \u0130\u015faret\u00e7i Tekni\u011fi'nin belirli bir problem s\u0131n\u0131f\u0131 i\u00e7in \u00f6zelle\u015fmi\u015f bir t\u00fcr\u00fcd\u00fcr.\n <\/li>\n<\/ol>\n<\/body><\/p>\n","protected":false},"excerpt":{"rendered":"\u0130ki \u0130\u015faret\u00e7i Tekni\u011fi, algoritmik verimlilik arayan yaz\u0131l\u0131mc\u0131lar i\u00e7in vazge\u00e7ilmez bir ara\u00e7t\u0131r. Naif O(N\u00b2) \u00e7\u00f6z\u00fcmlerden zarif O(N) optimizasyonlara ge\u00e7i\u015f…","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":[1342],"tags":[],"class_list":{"0":"post-34081","1":"post","2":"type-post","3":"status-publish","4":"format-standard","6":"category-ai","7":"cs-entry","8":"cs-video-wrap"},"yoast_head":"\n
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