{"id":38127,"date":"2026-01-23T20:00:49","date_gmt":"2026-01-23T17:00:49","guid":{"rendered":"https:\/\/fatihsoysal.com\/blog\/leetcode-1-two-sum-problemi-ve-klasik-hash-map-cozumu-detayli-bir-analiz\/"},"modified":"2026-01-23T20:00:49","modified_gmt":"2026-01-23T17:00:49","slug":"leetcode-1-two-sum-problemi-ve-klasik-hash-map-cozumu-detayli-bir-analiz","status":"publish","type":"post","link":"https:\/\/fatihsoysal.com\/blog\/leetcode-1-two-sum-problemi-ve-klasik-hash-map-cozumu-detayli-bir-analiz\/","title":{"rendered":"LeetCode #1: Two Sum Problemi ve Klasik Hash Map \u00c7\u00f6z\u00fcm\u00fc \u2013 Detayl\u0131 Bir Analiz"},"content":{"rendered":"<h2>LeetCode #1: Two Sum Problemi ve Klasik Hash Map \u00c7\u00f6z\u00fcm\u00fc \u2013 Detayl\u0131 Bir Analiz<\/h2>\n<p>LeetCode&#8217;un ilk problemi olan &#8220;Two Sum&#8221;, algoritmik d\u00fc\u015f\u00fcnmenin temel ta\u015flar\u0131ndan biridir. Bu makalede, Mahdi Shamlou&#8217;nun da \u00fczerinde durdu\u011fu gibi, problemin en verimli \u00e7\u00f6z\u00fcm\u00fc olan klasik hash map yakla\u015f\u0131m\u0131n\u0131 detayl\u0131ca ele alaca\u011f\u0131z. Amac\u0131m\u0131z, \u00e7\u00f6z\u00fcm\u00fcn mant\u0131\u011f\u0131n\u0131, performans\u0131n\u0131 ve \u00f6nemini kapsaml\u0131 bir \u015fekilde a\u00e7\u0131klamakt\u0131r.<\/p>\n<h2>LeetCode ve Algoritma Problemlerine Giri\u015f<\/h2>\n<h3>LeetCode Nedir ve Neden \u00d6nemlidir?<\/h3>\n<p>LeetCode, yaz\u0131l\u0131m geli\u015ftiricilerin algoritmik problem \u00e7\u00f6zme yeteneklerini geli\u015ftirmeleri, veri yap\u0131lar\u0131 bilgilerini peki\u015ftirmeleri ve teknik m\u00fclakatlara haz\u0131rlanmalar\u0131 i\u00e7in tasarlanm\u0131\u015f \u00e7evrimi\u00e7i bir platformdur. Binlerce farkl\u0131 zorluk seviyesinde problem bar\u0131nd\u0131r\u0131r ve kullan\u0131c\u0131lar\u0131n farkl\u0131 programlama dillerinde \u00e7\u00f6z\u00fcmlerini test etmelerine olanak tan\u0131r. LeetCode&#8217;da pratik yapmak, hem teorik bilgiyi peki\u015ftirir hem de ger\u00e7ek d\u00fcnya m\u00fchendislik problemlerine yakla\u015f\u0131m yetene\u011fini art\u0131r\u0131r.<\/p>\n<h3>Algoritmik D\u00fc\u015f\u00fcnmenin Temelleri<\/h3>\n<p>Algoritmik d\u00fc\u015f\u00fcnme, bir problemi ad\u0131m ad\u0131m \u00e7\u00f6zme, verimli y\u00f6ntemler geli\u015ftirme ve bu y\u00f6ntemlerin karma\u015f\u0131kl\u0131\u011f\u0131n\u0131 analiz etme becerisidir. Bu, sadece kod yazmaktan \u00f6te, problemi farkl\u0131 a\u00e7\u0131lardan ele almay\u0131, olas\u0131 \u00e7\u00f6z\u00fcmleri de\u011ferlendirmeyi ve en uygun olan\u0131 se\u00e7meyi i\u00e7erir. Two Sum gibi temel problemler, bu d\u00fc\u015f\u00fcnme bi\u00e7imini geli\u015ftirmek i\u00e7in m\u00fckemmel ba\u015flang\u0131\u00e7 noktalar\u0131d\u0131r.<\/p>\n<h3>\u0130lk Ad\u0131m: Two Sum Problemi<\/h3>\n<p>Two Sum, LeetCode&#8217;un en bilinen ve en s\u0131k kar\u015f\u0131la\u015f\u0131lan problemlerinden biridir. Genellikle m\u00fclakatlarda adaylar\u0131n temel algoritmik yeteneklerini \u00f6l\u00e7mek i\u00e7in kullan\u0131l\u0131r. Problem, basit bir tan\u0131ma sahip olsa da, farkl\u0131 \u00e7\u00f6z\u00fcm yakla\u015f\u0131mlar\u0131n\u0131n performans \u00fczerindeki etkisini net bir \u015fekilde g\u00f6sterir.<\/p>\n<h2>Two Sum Problemini Detayl\u0131 \u0130nceleme<\/h2>\n<h3>Problem Tan\u0131m\u0131 ve K\u0131s\u0131tlamalar<\/h3>\n<p>Two Sum problemi \u015fu \u015fekildedir: Size bir tam say\u0131 dizisi (<code>nums<\/code>) ve bir hedef tam say\u0131 (<code>target<\/code>) verilir. Dizideki hangi iki say\u0131n\u0131n toplam\u0131n\u0131n hedefe e\u015fit oldu\u011funu bulman\u0131z ve bu iki say\u0131n\u0131n dizideki indekslerini d\u00f6nd\u00fcrmeniz istenir. Her giri\u015fin tam olarak bir \u00e7\u00f6z\u00fcm\u00fc oldu\u011fu varsay\u0131l\u0131r ve ayn\u0131 eleman\u0131 iki kez kullanamazs\u0131n\u0131z.<\/p>\n<ul>\n<li><code>nums<\/code> dizisi 2 ile 10<sup>4<\/sup> eleman aras\u0131nda olabilir.<\/li>\n<li><code>-10<sup>9<\/sup> <= nums[i] <= 10<sup>9<\/sup><\/code><\/li>\n<li><code>-10<sup>9<\/sup> <= target <= 10<sup>9<\/sup><\/code><\/li>\n<li>Yaln\u0131zca bir ge\u00e7erli \u00e7\u00f6z\u00fcm vard\u0131r.<\/li>\n<\/ul>\n<h3>\u00d6rnek Senaryolar ve Beklenen \u00c7\u0131kt\u0131lar<\/h3>\n<p>Problemi daha iyi anlamak i\u00e7in birka\u00e7 \u00f6rnek inceleyelim:<\/p>\n<pre><code class=\"language-python\">\n# \u00d6rnek 1:\nnums = [2, 7, 11, 15]\ntarget = 9\n# Beklenen \u00c7\u0131kt\u0131: [0, 1]\n# \u00c7\u00fcnk\u00fc nums[0] + nums[1] == 2 + 7 == 9.\n\n# \u00d6rnek 2:\nnums = [3, 2, 4]\ntarget = 6\n# Beklenen \u00c7\u0131kt\u0131: [1, 2]\n# \u00c7\u00fcnk\u00fc nums[1] + nums[2] == 2 + 4 == 6.\n\n# \u00d6rnek 3:\nnums = [3, 3]\ntarget = 6\n# Beklenen \u00c7\u0131kt\u0131: [0, 1]\n# \u00c7\u00fcnk\u00fc nums[0] + nums[1] == 3 + 3 == 6.\n<\/pre>\n<p><\/code><\/p>\n<h3>Neden Bu Problem Temel Bir K\u00f6\u015fe Ta\u015f\u0131d\u0131r?<\/h3>\n<p>Two Sum, temel arama algoritmalar\u0131, veri yap\u0131lar\u0131 (\u00f6zellikle hash map'ler) ve zaman\/alan karma\u015f\u0131kl\u0131\u011f\u0131 analizi gibi kavramlar\u0131 tan\u0131tmak i\u00e7in idealdir. Bir\u00e7ok karma\u015f\u0131k problemin temelinde yatan prensipleri bar\u0131nd\u0131r\u0131r ve farkl\u0131 \u00e7\u00f6z\u00fcm yakla\u015f\u0131mlar\u0131n\u0131n performans \u00fczerindeki dramatik etkisini a\u00e7\u0131k\u00e7a g\u00f6sterir.<\/p>\n<h2>Naif Yakla\u015f\u0131m: Brute Force ve Performans Dezavantajlar\u0131<\/h2>\n<h3>Brute Force \u00c7\u00f6z\u00fcm\u00fcn\u00fcn Mant\u0131\u011f\u0131<\/h3>\n<p>Bir problemi \u00e7\u00f6zmek i\u00e7in akla gelen ilk ve en basit y\u00f6ntem genellikle \"brute force\" (kaba kuvvet) yakla\u015f\u0131m\u0131d\u0131r. Two Sum i\u00e7in brute force, dizideki her say\u0131y\u0131 al\u0131p, kalan t\u00fcm say\u0131larla toplayarak hedef toplam\u0131 bulmaya \u00e7al\u0131\u015fmakt\u0131r. Bu, i\u00e7 i\u00e7e iki d\u00f6ng\u00fc kullan\u0131larak ger\u00e7ekle\u015ftirilir.<\/p>\n<h3>Zaman Karma\u015f\u0131kl\u0131\u011f\u0131 Analizi (O(n^2))<\/h3>\n<p>Brute force \u00e7\u00f6z\u00fcm\u00fcnde, d\u0131\u015f d\u00f6ng\u00fc dizideki her eleman i\u00e7in bir kez \u00e7al\u0131\u015f\u0131rken (n kez), i\u00e7 d\u00f6ng\u00fc de her d\u0131\u015f d\u00f6ng\u00fc iterasyonunda kalan elemanlar i\u00e7in (yakla\u015f\u0131k n kez) \u00e7al\u0131\u015f\u0131r. Bu durumda toplam i\u015flem say\u0131s\u0131 yakla\u015f\u0131k n * n = n<sup>2<\/sup> olur. B\u00fcy\u00fck O notasyonu ile bu, O(n<sup>2<\/sup>) olarak ifade edilir. Bu, dizi boyutu (n) artt\u0131k\u00e7a \u00e7\u00f6z\u00fcm s\u00fcresinin katlanarak artaca\u011f\u0131 anlam\u0131na gelir. \u00d6rne\u011fin, 10.000 elemanl\u0131 bir dizi i\u00e7in yakla\u015f\u0131k 100 milyon i\u015flem gerekebilir, bu da modern sistemlerde bile kabul edilemez derecede yava\u015f olabilir.<\/p>\n<h3>Alan Karma\u015f\u0131kl\u0131\u011f\u0131<\/h3>\n<p>Brute force yakla\u015f\u0131m\u0131nda, ek bir veri yap\u0131s\u0131 kullanmad\u0131\u011f\u0131m\u0131z i\u00e7in alan karma\u015f\u0131kl\u0131\u011f\u0131 O(1) veya O(n) (e\u011fer sonu\u00e7 dizisi d\u00f6nd\u00fcrme maliyetini sayarsak) d\u00fczeyindedir. Ancak, bu d\u00fc\u015f\u00fck alan maliyeti, y\u00fcksek zaman maliyeti nedeniyle genellikle tercih edilmez.<\/p>\n<h2>Hash Map (Karma Tablo) Veri Yap\u0131s\u0131n\u0131n G\u00fcc\u00fc<\/h2>\n<h3>Hash Map Nedir ve Nas\u0131l \u00c7al\u0131\u015f\u0131r?<\/h3>\n<p>Hash map (karma tablo veya s\u00f6zl\u00fck), anahtar-de\u011fer \u00e7iftlerini depolayan, verimli arama, ekleme ve silme i\u015flemleri sunan bir veri yap\u0131s\u0131d\u0131r. Bir anahtar verildi\u011finde, bu anahtar\u0131 bir \"hash fonksiyonu\" kullanarak bir dizin de\u011ferine d\u00f6n\u00fc\u015ft\u00fcr\u00fcr ve de\u011feri o dizinde saklar. Bu sayede, bir de\u011feri anahtar\u0131 arac\u0131l\u0131\u011f\u0131yla neredeyse sabit zamanda (ortalama O(1)) bulmak m\u00fcmk\u00fcn olur.<\/p>\n<h3>Temel Operasyonlar: Ekleme, Arama, Silme<\/h3>\n<p>Hash map'lerin en b\u00fcy\u00fck avantaj\u0131, temel operasyonlar\u0131n\u0131n ortalama olarak O(1) zaman karma\u015f\u0131kl\u0131\u011f\u0131na sahip olmas\u0131d\u0131r:<\/p>\n<ul>\n<li><strong>Ekleme (Insertion):<\/strong> Yeni bir anahtar-de\u011fer \u00e7iftini tabloya eklemek.<\/li>\n<li><strong>Arama (Lookup):<\/strong> Bir anahtara kar\u015f\u0131l\u0131k gelen de\u011feri bulmak.<\/li>\n<li><strong>Silme (Deletion):<\/strong> Bir anahtar-de\u011fer \u00e7iftini tablodan kald\u0131rmak.<\/li>\n<\/ul>\n<p>Bu h\u0131z, \u00f6zellikle b\u00fcy\u00fck veri k\u00fcmelerinde arama i\u015flemlerinin kritik oldu\u011fu durumlarda hash map'leri vazge\u00e7ilmez k\u0131lar.<\/p>\n<h3>Two Sum \u0130\u00e7in Hash Map'in Avantajlar\u0131<\/h3>\n<p>Two Sum probleminde, bir say\u0131y\u0131 bulduktan sonra hedefi tamamlayacak di\u011fer say\u0131y\u0131 h\u0131zl\u0131ca bulmam\u0131z gerekir. Hash map'ler, bu \"tamamlay\u0131c\u0131\" say\u0131y\u0131 dizinin geri kalan\u0131nda tekrar aramak yerine, O(1) s\u00fcrede mevcut olup olmad\u0131\u011f\u0131n\u0131 kontrol etme yetene\u011fi sunar. Bu, i\u00e7 i\u00e7e d\u00f6ng\u00fc ihtiyac\u0131n\u0131 ortadan kald\u0131rarak zaman karma\u015f\u0131kl\u0131\u011f\u0131n\u0131 \u00f6nemli \u00f6l\u00e7\u00fcde iyile\u015ftirir.<\/p>\n<h2>Klasik Hash Map \u00c7\u00f6z\u00fcm\u00fcn\u00fcn Ad\u0131m Ad\u0131m Uygulanmas\u0131<\/h2>\n<h3>Algoritma Mant\u0131\u011f\u0131 ve Ak\u0131\u015f Diyagram\u0131<\/h3>\n<p>Hash map \u00e7\u00f6z\u00fcm\u00fc, dizide tek bir ge\u00e7i\u015f yaparak \u00e7al\u0131\u015f\u0131r. Her eleman\u0131 i\u015flerken, hedefi tamamlayacak \"tamamlay\u0131c\u0131\" say\u0131n\u0131n daha \u00f6nce g\u00f6r\u00fcl\u00fcp g\u00f6r\u00fclmedi\u011fini kontrol ederiz. E\u011fer g\u00f6r\u00fclm\u00fc\u015fse, \u00e7\u00f6z\u00fcm bulunmu\u015ftur. G\u00f6r\u00fclmemi\u015fse, mevcut say\u0131y\u0131 ve indeksini hash map'e ekleriz.<\/p>\n<h3>Ad\u0131m 1: Hash Map'i Ba\u015flatma<\/h3>\n<p>Bo\u015f bir hash map (Python'da s\u00f6zl\u00fck, Java'da HashMap) olu\u015fturulur. Bu map, anahtar olarak say\u0131lar\u0131 ve de\u011fer olarak da o say\u0131n\u0131n dizideki indeksini saklayacakt\u0131r.<\/p>\n<h3>Ad\u0131m 2: Dizide Tek Ge\u00e7i\u015f<\/h3>\n<p>Verilen <code>nums<\/code> dizisi \u00fczerinde ba\u015ftan sona tek bir d\u00f6ng\u00fc ba\u015flat\u0131l\u0131r. Her iterasyonda, mevcut say\u0131ya (<code>num<\/code>) ve onun indeksine (<code>i<\/code>) eri\u015filir.<\/p>\n<h3>Ad\u0131m 3: Tamamlay\u0131c\u0131 Say\u0131y\u0131 Arama<\/h3>\n<p>Mevcut <code>num<\/code> i\u00e7in, hedefi tamamlayacak olan say\u0131y\u0131 (<code>complement = target - num<\/code>) hesaplar\u0131z. Daha sonra bu <code>complement<\/code> say\u0131s\u0131n\u0131n hash map'te zaten mevcut olup olmad\u0131\u011f\u0131n\u0131 kontrol ederiz.<\/p>\n<h3>Ad\u0131m 4: E\u015fle\u015fme Bulundu\u011funda<\/h3>\n<p>E\u011fer <code>complement<\/code> hash map'te bulunuyorsa, bu demektir ki daha \u00f6nce bu say\u0131y\u0131 g\u00f6rm\u00fc\u015f ve indeksini kaydetmi\u015fiz. Bu durumda, <code>complement<\/code>'in hash map'teki indeksi ile mevcut <code>num<\/code>'un indeksi (<code>i<\/code>) aranan cevapt\u0131r. Bu iki indeksi i\u00e7eren bir liste d\u00f6nd\u00fcr\u00fcl\u00fcr ve algoritma sona erer.<\/p>\n<h3>Ad\u0131m 5: E\u015fle\u015fme Bulunamazsa<\/h3>\n<p>E\u011fer <code>complement<\/code> hash map'te bulunam\u0131yorsa, bu, mevcut <code>num<\/code>'un hen\u00fcz bir e\u015fle\u015fme bulamad\u0131\u011f\u0131 anlam\u0131na gelir. Bu durumda, <code>num<\/code>'u ve onun indeksini (<code>i<\/code>) hash map'e ekleriz. B\u00f6ylece, dizinin ilerleyen elemanlar\u0131 i\u00e7in <code>num<\/code> bir <code>complement<\/code> olarak i\u015flev g\u00f6rebilir.<\/p>\n<h2>Kod \u00d6rne\u011fi ve Detayl\u0131 A\u00e7\u0131klamalar<\/h2>\n<h3>Python ile \u00c7\u00f6z\u00fcm Kodu<\/h3>\n<pre><code class=\"language-python\">\nclass Solution:\n    def twoSum(self, nums: list[int], target: int) -> list[int]:\n        # 1. Bo\u015f bir hash map (s\u00f6zl\u00fck) olu\u015ftur\n        num_map = {}  # {say\u0131: indeks}\n\n        # 2. Dizide tek ge\u00e7i\u015f yap\n        for i, num in enumerate(nums):\n            # 3. Tamamlay\u0131c\u0131 say\u0131y\u0131 hesapla\n            complement = target - num\n\n            # 4. Tamamlay\u0131c\u0131 say\u0131 map'te var m\u0131 kontrol et\n            if complement in num_map:\n                # Varsa, \u00e7\u00f6z\u00fcm bulundu.\n                # Tamamlay\u0131c\u0131n\u0131n indeksi ve mevcut say\u0131n\u0131n indeksi d\u00f6nd\u00fcr\u00fcl\u00fcr.\n                return [num_map[complement], i]\n            \n            # 5. Tamamlay\u0131c\u0131 say\u0131 bulunamazsa, mevcut say\u0131y\u0131 map'e ekle\n            num_map[num] = i\n        \n        # E\u011fer d\u00f6ng\u00fc biter ve \u00e7\u00f6z\u00fcm bulunamazsa (problem tan\u0131m\u0131na g\u00f6re bu durum olmaz)\n        return []\n<\/pre>\n<p><\/code><\/p>\n<h3>Kodun Sat\u0131r Sat\u0131r Analizi<\/h3>\n<ul>\n<li><code>num_map = {}<\/code>: Bo\u015f bir Python s\u00f6zl\u00fc\u011f\u00fc (hash map) ba\u015flat\u0131l\u0131r. Anahtarlar say\u0131lar, de\u011ferler ise o say\u0131lar\u0131n dizideki indeksleri olacakt\u0131r.<\/li>\n<li><code>for i, num in enumerate(nums):<\/code>: <code>enumerate<\/code> fonksiyonu, dizideki her eleman\u0131 hem indeksi (<code>i<\/code>) hem de de\u011feri (<code>num<\/code>) ile birlikte d\u00f6ng\u00fcye sokar.<\/li>\n<li><code>complement = target - num<\/code>: Mevcut <code>num<\/code> i\u00e7in hedefi tamamlayacak say\u0131y\u0131 hesaplar\u0131z.<\/li>\n<li><code>if complement in num_map:<\/code>: Hesaplanan <code>complement<\/code> say\u0131s\u0131n\u0131n daha \u00f6nce hash map'e eklenip eklenmedi\u011fini kontrol ederiz. Bu kontrol ortalama O(1) s\u00fcrede ger\u00e7ekle\u015fir.<\/li>\n<li><code>return [num_map[complement], i]<\/code>: E\u011fer <code>complement<\/code> map'te bulunursa, \u00e7\u00f6z\u00fcm bulunmu\u015ftur. <code>num_map[complement]<\/code> bize <code>complement<\/code>'in indeksini verirken, <code>i<\/code> de mevcut <code>num<\/code>'un indeksidir. Bu iki indeks d\u00f6nd\u00fcr\u00fcl\u00fcr.<\/li>\n<li><code>num_map[num] = i<\/code>: E\u011fer <code>complement<\/code> bulunamazsa, mevcut <code>num<\/code>'u ve onun indeksini hash map'e ekleriz. Bu, gelecekteki elemanlar i\u00e7in <code>num<\/code>'un bir <code>complement<\/code> olarak bulunabilmesini sa\u011flar.<\/li>\n<\/ul>\n<h3>Edge Case'ler ve Dikkat Edilmesi Gerekenler<\/h3>\n<p>Problem tan\u0131m\u0131na g\u00f6re her zaman tam olarak bir \u00e7\u00f6z\u00fcm oldu\u011fu varsay\u0131ld\u0131\u011f\u0131 i\u00e7in, kodun d\u00f6ng\u00fcden \u00e7\u0131kmadan bir \u00e7\u00f6z\u00fcm bulmas\u0131 garanti edilir. Bu nedenle, d\u00f6ng\u00fcden sonra bo\u015f bir liste d\u00f6nd\u00fcrme durumu pratikte ger\u00e7ekle\u015fmez, ancak iyi bir uygulama olarak eklenebilir. Ayn\u0131 eleman\u0131 iki kez kullanamama k\u0131s\u0131tlamas\u0131, hash map'e ekleme stratejimizle otomatik olarak ele al\u0131n\u0131r, \u00e7\u00fcnk\u00fc mevcut eleman\u0131 ancak tamamlay\u0131c\u0131s\u0131 bulunamazsa ekleriz. Bu, ayn\u0131 eleman\u0131n kendisiyle e\u015fle\u015fmesini engeller.<\/p>\n<h2>Performans Analizi ve Optimizasyon<\/h2>\n<h3>Zaman Karma\u015f\u0131kl\u0131\u011f\u0131 (O(n))<\/h3>\n<p>Hash map \u00e7\u00f6z\u00fcm\u00fcnde, dizide sadece tek bir ge\u00e7i\u015f yapar\u0131z. Her eleman i\u00e7in bir \u00e7\u0131karma i\u015flemi (O(1)), hash map'te bir arama i\u015flemi (ortalama O(1)) ve potansiyel olarak bir ekleme i\u015flemi (ortalama O(1)) ger\u00e7ekle\u015ftiririz. Bu nedenle, toplam zaman karma\u015f\u0131kl\u0131\u011f\u0131 dizinin boyutuyla do\u011fru orant\u0131l\u0131d\u0131r, yani O(n) olur. Bu, brute force \u00e7\u00f6z\u00fcm\u00fcn\u00fcn O(n<sup>2<\/sup>) karma\u015f\u0131kl\u0131\u011f\u0131na g\u00f6re \u00e7ok daha verimlidir.<\/p>\n<h3>Alan Karma\u015f\u0131kl\u0131\u011f\u0131 (O(n))<\/h3>\n<p>Bu \u00e7\u00f6z\u00fcmde, en k\u00f6t\u00fc durumda dizideki t\u00fcm elemanlar\u0131 hash map'e eklememiz gerekebilir (\u00e7\u00f6z\u00fcm son elemanla bulunursa). Bu durumda hash map'in boyutu dizinin boyutuyla (n) orant\u0131l\u0131 olacakt\u0131r. Dolay\u0131s\u0131yla, alan karma\u015f\u0131kl\u0131\u011f\u0131 O(n) olarak kabul edilir. Bu, brute force \u00e7\u00f6z\u00fcm\u00fcne g\u00f6re daha fazla bellek kullan\u0131m\u0131 anlam\u0131na gelir, ancak zaman performans\u0131ndaki kazan\u00e7 genellikle bu maliyeti hakl\u0131 \u00e7\u0131kar\u0131r.<\/p>\n<h3>Brute Force ile Kar\u015f\u0131la\u015ft\u0131rma<\/h3>\n<p>A\u015fa\u011f\u0131daki tablo, iki \u00e7\u00f6z\u00fcm yakla\u015f\u0131m\u0131n\u0131n performans\u0131n\u0131 \u00f6zetlemektedir:<\/p>\n<table>\n<thead>\n<tr>\n<th>Yakla\u015f\u0131m<\/th>\n<th>Zaman Karma\u015f\u0131kl\u0131\u011f\u0131<\/th>\n<th>Alan Karma\u015f\u0131kl\u0131\u011f\u0131<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>Brute Force<\/td>\n<td>O(n<sup>2<\/sup>)<\/td>\n<td>O(1)<\/td>\n<\/tr>\n<tr>\n<td>Hash Map<\/td>\n<td>O(n)<\/td>\n<td>O(n)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>G\u00f6r\u00fcld\u00fc\u011f\u00fc \u00fczere, hash map \u00e7\u00f6z\u00fcm\u00fc zaman karma\u015f\u0131kl\u0131\u011f\u0131 a\u00e7\u0131s\u0131ndan \u00e7ok daha \u00fcst\u00fcnd\u00fcr, \u00f6zellikle b\u00fcy\u00fck veri setlerinde bu fark hayati \u00f6nem ta\u015f\u0131r.<\/p>\n<h2>Sonu\u00e7 ve \u0130leri Ad\u0131mlar<\/h2>\n<h3>Two Sum \u00c7\u00f6z\u00fcm\u00fcn\u00fcn \u00d6nemi<\/h3>\n<p>Two Sum problemi, basit bir ba\u015flang\u0131\u00e7 noktas\u0131 olmas\u0131na ra\u011fmen, algoritmik d\u00fc\u015f\u00fcnme, veri yap\u0131lar\u0131 se\u00e7imi ve performans analizi gibi temel yaz\u0131l\u0131m m\u00fchendisli\u011fi kavramlar\u0131n\u0131 peki\u015ftirmek i\u00e7in m\u00fckemmel bir ara\u00e7t\u0131r. Hash map \u00e7\u00f6z\u00fcm\u00fcn\u00fcn anla\u015f\u0131lmas\u0131, daha karma\u015f\u0131k LeetCode problemlerini \u00e7\u00f6zmek i\u00e7in sa\u011flam bir temel olu\u015fturur.<\/p>\n<h3>\u00d6\u011frenilen Dersler<\/h3>\n<p>Bu problemden \u00e7\u0131kar\u0131lacak en \u00f6nemli ders, do\u011fru veri yap\u0131s\u0131 se\u00e7iminin bir algoritman\u0131n performans\u0131 \u00fczerindeki dramatik etkisidir. Hash map'ler, h\u0131zl\u0131 arama gerektiren bir\u00e7ok senaryoda vazge\u00e7ilmez bir ara\u00e7t\u0131r ve bu \u00e7\u00f6z\u00fcm, onlar\u0131n g\u00fcc\u00fcn\u00fc a\u00e7\u0131k\u00e7a ortaya koymaktad\u0131r.<\/p>\n<h3>Gelecek Problemlere Haz\u0131rl\u0131k<\/h3>\n<p>Two Sum'\u0131 hash map ile \u00e7\u00f6zmek, LeetCode yolculu\u011funuzda att\u0131\u011f\u0131n\u0131z ilk \u00f6nemli ad\u0131md\u0131r. Bu deneyim, dinamik programlama, grafik algoritmalar\u0131 ve di\u011fer geli\u015fmi\u015f veri yap\u0131lar\u0131 gibi daha karma\u015f\u0131k konulara ge\u00e7i\u015f yaparken size rehberlik edecektir. Her zaman problemi anlamaya, farkl\u0131 yakla\u015f\u0131mlar\u0131 d\u00fc\u015f\u00fcnmeye ve performans analizini yapmaya odaklan\u0131n.<\/p>\n<h2>S\u0131k\u00e7a Sorulan Sorular (SSS)<\/h2>\n<h3>Two Sum i\u00e7in ba\u015fka \u00e7\u00f6z\u00fcm yollar\u0131 var m\u0131?<\/h3>\n<p>Evet, diziyi \u00f6nce s\u0131ralay\u0131p sonra iki i\u015faret\u00e7i (two-pointer) tekni\u011fiyle de \u00e7\u00f6z\u00fclebilir. Ancak bu durumda, orijinal indeksleri korumak i\u00e7in ek ad\u0131mlar gerekebilir ve s\u0131ralama i\u015flemi O(n log n) zaman karma\u015f\u0131kl\u0131\u011f\u0131na sahiptir. Hash map \u00e7\u00f6z\u00fcm\u00fc genellikle en optimal ve pratik y\u00f6ntem olarak kabul edilir.<\/p>\n<h3>Hash Map yerine ba\u015fka hangi veri yap\u0131lar\u0131 kullan\u0131labilir?<\/h3>\n<p>E\u011fer say\u0131lar belirli bir aral\u0131kta ve pozitif ise, bir boolean dizisi veya frekans dizisi (array) de kullan\u0131labilir. Ancak bu, say\u0131lar\u0131n aral\u0131\u011f\u0131 \u00e7ok geni\u015f oldu\u011funda veya negatif say\u0131lar i\u00e7erdi\u011finde pratik de\u011fildir. Genel durumlar i\u00e7in hash map en esnek ve verimli \u00e7\u00f6z\u00fcmd\u00fcr.<\/p>\n<h3>Bo\u015f dizi veya tek elemanl\u0131 dizi gibi \u00f6zel durumlar nas\u0131l ele al\u0131n\u0131r?<\/h3>\n<p>Problem k\u0131s\u0131tlamalar\u0131na g\u00f6re <code>nums<\/code> dizisi en az iki eleman i\u00e7erecektir. Bu nedenle, bo\u015f veya tek elemanl\u0131 dizi gibi durumlar\u0131 \u00f6zel olarak ele almaya gerek kalmaz. Ancak genel bir \u00e7\u00f6z\u00fcm yazarken bu t\u00fcr \"edge case\"leri d\u00fc\u015f\u00fcnmek iyi bir al\u0131\u015fkanl\u0131kt\u0131r.<\/p>\n<h3>Hash Map \u00e7\u00f6z\u00fcm\u00fcn\u00fcn dezavantajlar\u0131 nelerdir?<\/h3>\n<p>En b\u00fcy\u00fck dezavantaj\u0131, ek bellek kullan\u0131m\u0131d\u0131r (O(n) alan karma\u015f\u0131kl\u0131\u011f\u0131). \u00c7ok k\u0131s\u0131tl\u0131 bellek ortamlar\u0131nda bu bir sorun olabilir. Ayr\u0131ca, hash \u00e7arp\u0131\u015fmalar\u0131 (collision) nedeniyle en k\u00f6t\u00fc durumda arama ve ekleme i\u015flemleri O(n) olabilir, ancak bu durum modern hash map uygulamalar\u0131nda nadiren g\u00f6r\u00fcl\u00fcr ve ortalama performans O(1)'dir.<\/p>\n<h3>Bu problemde neden iki ge\u00e7i\u015f yerine tek ge\u00e7i\u015f tercih edilir?<\/h3>\n<p>Tek ge\u00e7i\u015fli \u00e7\u00f6z\u00fcm, diziyi sadece bir kez tarayarak hem mevcut say\u0131y\u0131 i\u015fler hem de tamamlay\u0131c\u0131s\u0131n\u0131 arar. Bu, iki ayr\u0131 ge\u00e7i\u015f yapmaktan (bir ge\u00e7i\u015fle t\u00fcm say\u0131lar\u0131 map'e ekleyip, ikinci ge\u00e7i\u015fle tamamlay\u0131c\u0131lar\u0131 aramak) daha verimlidir \u00e7\u00fcnk\u00fc her eleman\u0131 sadece bir kez ziyaret eder. \u0130ki ge\u00e7i\u015f de O(n) zaman karma\u015f\u0131kl\u0131\u011f\u0131na sahip olsa da, tek ge\u00e7i\u015f pratik uygulamada daha az sabit fakt\u00f6rle daha h\u0131zl\u0131d\u0131r.<\/p>\n","protected":false},"excerpt":{"rendered":"LeetCode&#8217;un ilk problemi olan &#8220;Two Sum&#8221;, algoritmik d\u00fc\u015f\u00fcnmenin temel ta\u015flar\u0131ndan biridir. Bu makalede, Mahdi Shamlou&#8217;nun da \u00fczerinde durdu\u011fu gibi, pro&#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-38127","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) - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>LeetCode #1: Two Sum Problemi ve Klasik Hash Map \u00c7\u00f6z\u00fcm\u00fc \u2013 Detayl\u0131 Bir Analiz - Kodlar\u0131n Gizemli D\u00fcnyas\u0131<\/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:\/\/fatihsoysal.com\/blog\/leetcode-1-two-sum-problemi-ve-klasik-hash-map-cozumu-detayli-bir-analiz\/\" \/>\n<meta property=\"og:locale\" content=\"tr_TR\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"LeetCode #1: Two Sum Problemi ve Klasik Hash Map \u00c7\u00f6z\u00fcm\u00fc \u2013 Detayl\u0131 Bir Analiz\" \/>\n<meta property=\"og:description\" content=\"LeetCode&#039;un ilk problemi olan &quot;Two Sum&quot;, algoritmik d\u00fc\u015f\u00fcnmenin temel ta\u015flar\u0131ndan biridir. 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