{"id":13987,"date":"2025-03-02T10:30:13","date_gmt":"2025-03-02T07:30:13","guid":{"rendered":"https:\/\/fatihsoysal.com\/blog\/en-cok-su-iceren-kap-leetcode-11in-detayli-analizi\/"},"modified":"2025-03-02T10:30:13","modified_gmt":"2025-03-02T07:30:13","slug":"en-cok-su-iceren-kap-leetcode-11in-detayli-analizi","status":"publish","type":"post","link":"https:\/\/fatihsoysal.com\/blog\/en-cok-su-iceren-kap-leetcode-11in-detayli-analizi\/","title":{"rendered":"En \u00c7ok Su \u0130\u00e7eren Kap: LeetCode 11’in Detayl\u0131 Analizi"},"content":{"rendered":"

En \u00c7ok Su \u0130\u00e7eren Kap: LeetCode 11’in Detayl\u0131 Analizi<\/p>\n

LeetCode’un 11. sorusu olan “En \u00c7ok Su \u0130\u00e7eren Kap” problemi, iki i\u015faret\u00e7i tekni\u011finin g\u00fczel bir \u00f6rne\u011fini sergiliyor. Bu problemde, verilen bir y\u00fckseklik dizisiyle, en \u00e7ok suyu tutabilecek kaplar\u0131 bulmam\u0131z gerekiyor. \u00d6ncelikle problemi daha detayl\u0131 inceleyelim, ard\u0131ndan \u00e7\u00f6z\u00fcm algoritmas\u0131n\u0131 ve kodlamas\u0131n\u0131 ad\u0131m ad\u0131m ele alal\u0131m. Sonras\u0131nda ise performans analizi yaparak \u00e7\u00f6z\u00fcm\u00fcn etkinli\u011fini tart\u0131\u015faca\u011f\u0131z.<\/p>\n

Problem Tan\u0131m\u0131<\/h2>\n

Verilen bir height<\/code> dizisi, n tane dikey \u00e7izginin y\u00fcksekliklerini temsil eder. Bu \u00e7izgiler x ekseni \u00fczerinde e\u015fit aral\u0131klarla yer almaktad\u0131r. Herhangi iki \u00e7izgi ve x ekseni aras\u0131nda bir kap olu\u015fur. Hedefimiz, bu kaplar aras\u0131nda en b\u00fcy\u00fck alan\u0131 i\u00e7ereni bulmakt\u0131r. Alan, iki \u00e7izgi aras\u0131ndaki mesafe ile daha k\u0131sa \u00e7izginin y\u00fcksekli\u011finin \u00e7arp\u0131m\u0131yla hesaplan\u0131r. \u00d6rne\u011fin, height = [1,8,6,2,5,4,8,3,7]<\/code> i\u00e7in en b\u00fcy\u00fck alan 49’dur.<\/p>\n

\u00c7\u00f6z\u00fcm Algoritmas\u0131: \u0130ki \u0130\u015faret\u00e7i Tekni\u011fi<\/h2>\n

Bu problemi \u00e7\u00f6zmek i\u00e7in en etkin y\u00f6ntem, iki i\u015faret\u00e7i tekni\u011fini kullanmakt\u0131r. \u0130ki i\u015faret\u00e7i, dizinin ba\u015flang\u0131c\u0131 ve sonuna yerle\u015ftirilir. Her ad\u0131mda, daha k\u00fc\u00e7\u00fck y\u00fcksekli\u011fe sahip i\u015faret\u00e7i i\u00e7eri do\u011fru hareket ettirilir. Bunun nedeni, daha k\u00fc\u00e7\u00fck y\u00fcksekli\u011fin alan s\u0131n\u0131rlay\u0131c\u0131s\u0131 olmas\u0131d\u0131r. Daha b\u00fcy\u00fck y\u00fcksekli\u011fi de\u011fi\u015ftirerek daha b\u00fcy\u00fck bir alan elde etme olas\u0131l\u0131\u011f\u0131m\u0131z daha y\u00fcksektir. Bu i\u015flem, i\u015faret\u00e7iler \u00e7ak\u0131\u015fana kadar devam eder. Her ad\u0131mda hesaplanan alan, \u015fimdiye kadar bulunan en b\u00fcy\u00fck alan ile kar\u015f\u0131la\u015ft\u0131r\u0131l\u0131r ve en b\u00fcy\u00fck alan g\u00fcncellenir.<\/p>\n

\u00d6rne\u011fin, height = [1,8,6,2,5,4,8,3,7]<\/code> i\u00e7in:<\/p>\n

    \n
  1. Ba\u015flang\u0131\u00e7ta sol i\u015faret\u00e7i 1’de, sa\u011f i\u015faret\u00e7i 7’dedir. Alan = min(1,7) * (7-1) = 6<\/li>\n
  2. Sol i\u015faret\u00e7i hareket eder. Alan = min(8,7) * (7-1) = 42<\/li>\n
  3. Sa\u011f i\u015faret\u00e7i hareket eder. Alan = min(8,3) * (7-2) = 15<\/li>\n
  4. ve b\u00f6yle devam eder…<\/li>\n<\/ol>\n

    Kod Uygulamas\u0131 (Python)<\/h2>\n
    \ndef maxArea(height):\n    max_area = 0\n    left = 0\n    right = len(height) - 1\n    while left < right:\n        area = min(height[left], height[right]) * (right - left)\n        max_area = max(max_area, area)\n        if height[left] < height[right]:\n            left += 1\n        else:\n            right -= 1\n    return max_area\n\n<\/code><\/pre>\n
    Performans Analizi<\/h2>\n
    \u0130ki i\u015faret\u00e7i tekni\u011fi, bu problemi O(n) zaman karma\u015f\u0131kl\u0131\u011f\u0131yla \u00e7\u00f6zer.  Yani, dizi boyutu artt\u0131k\u00e7a \u00e7\u00f6z\u00fcm s\u00fcresi lineer olarak artar.  Uzay karma\u015f\u0131kl\u0131\u011f\u0131 ise O(1)'dir \u00e7\u00fcnk\u00fc yaln\u0131zca sabit miktarda ek uzay kullan\u0131l\u0131r. Bu, olduk\u00e7a verimli bir \u00e7\u00f6z\u00fcmd\u00fcr.<\/p>\n
    Sonu\u00e7<\/h2>\n
    LeetCode 11 problemi, iki i\u015faret\u00e7i tekni\u011finin problem \u00e7\u00f6z\u00fcm\u00fcnde ne kadar etkili oldu\u011funu g\u00f6steren g\u00fczel bir \u00f6rnektir.  Bu teknik, do\u011fru uyguland\u0131\u011f\u0131nda, karma\u015f\u0131k problemleri verimli bir \u015fekilde \u00e7\u00f6zmemizi sa\u011flar.  Umar\u0131m bu detayl\u0131 a\u00e7\u0131klama, problemi anlaman\u0131za ve \u00e7\u00f6zmenize yard\u0131mc\u0131 olmu\u015ftur.  Daha fazla algoritma ve veri yap\u0131s\u0131 \u00f6rne\u011fi i\u00e7in fatihsoysal.com<\/a> adresini ziyaret edebilirsiniz.<\/p>\n
    Ayr\u0131ca,  konu hakk\u0131nda daha fazla bilgi edinmek i\u00e7in buray\u0131<\/a> inceleyebilirsiniz.<\/p>\n
    #Etiketler:<\/strong> LeetCode, Algoritma, Veri Yap\u0131lar\u0131, \u0130ki \u0130\u015faret\u00e7i Tekni\u011fi, En \u00c7ok Su \u0130\u00e7eren Kap, Problem \u00c7\u00f6zme, Python, Zaman Karma\u015f\u0131kl\u0131\u011f\u0131, Uzay Karma\u015f\u0131kl\u0131\u011f\u0131, Programlama<\/p>\n","protected":false},"excerpt":{"rendered":"En \u00c7ok Su \u0130\u00e7eren Kap: LeetCode 11’in Detayl\u0131 Analizi LeetCode’un 11. sorusu olan “En \u00c7ok Su \u0130\u00e7eren Kap”…","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-13987","1":"post","2":"type-post","3":"status-publish","4":"format-standard","6":"category-genel","7":"cs-entry","8":"cs-video-wrap"},"yoast_head":"\nEn \u00c7ok Su \u0130\u00e7eren Kap: LeetCode 11'in Detayl\u0131 Analizi<\/title>\n<meta name=\"description\" content=\"LeetCode'un 11. sorusu olan "En \u00c7ok Su \u0130\u00e7eren Kap" problemi, iki i\u015faret\u00e7i tekni\u011finin g\u00fczel bir \u00f6rne\u011fini sergiliyor. Bu problemde, verilen bir y\u00fckseklik dizisiyle, en \u00e7ok suyu tutabilecek kaplar\u0131 bulmam\u0131z gerekiyor. \u00d6ncelikle problemi daha detayl\u0131 inceleyelim, ard\u0131ndan \u00e7\u00f6z\u00fcm algoritmas\u0131n\u0131 ve kodlamas\u0131n\u0131 ad\u0131m ad\u0131m ele alal\u0131m. Sonras\u0131nda ise performans analizi yaparak \u00e7\u00f6z\u00fcm\u00fcn etkinli\u011fini tart\u0131\u015faca\u011f\u0131z.\" \/>\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\/en-cok-su-iceren-kap-leetcode-11in-detayli-analizi\/\" \/>\n<meta property=\"og:locale\" content=\"tr_TR\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"En \u00c7ok Su \u0130\u00e7eren Kap: LeetCode 11'in Detayl\u0131 Analizi\" \/>\n<meta property=\"og:description\" content=\"LeetCode'un 11. sorusu olan "En \u00c7ok Su \u0130\u00e7eren Kap" problemi, iki i\u015faret\u00e7i tekni\u011finin g\u00fczel bir \u00f6rne\u011fini sergiliyor. 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