{"id":2876,"date":"2024-11-04T00:45:10","date_gmt":"2024-11-03T21:45:10","guid":{"rendered":"https:\/\/fatihsoysal.com\/blog\/leetcode-top-interview-150-169-cogunluk-elemani\/"},"modified":"2024-11-04T00:45:10","modified_gmt":"2024-11-03T21:45:10","slug":"leetcode-top-interview-150-169-cogunluk-elemani","status":"publish","type":"post","link":"https:\/\/fatihsoysal.com\/blog\/leetcode-top-interview-150-169-cogunluk-elemani\/","title":{"rendered":"LeetCode Top Interview 150-169: \u00c7o\u011funluk Eleman\u0131"},"content":{"rendered":"

LeetCode Top Interview 150-169: \u00c7o\u011funluk Eleman\u0131<\/p>\n

Merhaba, bug\u00fcn LeetCode’un Top Interview 150-169 serisindeki “\u00c7o\u011funluk Eleman\u0131” problemini ele alaca\u011f\u0131z. Bu problem, bir dizide en \u00e7ok g\u00f6r\u00fcnen eleman\u0131 bulmay\u0131 hedefler. Bir dizi i\u00e7inde, bu eleman toplam eleman say\u0131s\u0131n\u0131n yar\u0131s\u0131ndan fazlas\u0131n\u0131 temsil eder.<\/p>\n

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

Bir dizi nums<\/code> veriliyor. Bu dizide en \u00e7ok g\u00f6r\u00fcnen eleman\u0131 bulun. Bu eleman\u0131n dizi i\u00e7inde toplam eleman say\u0131s\u0131n\u0131n yar\u0131s\u0131ndan fazlas\u0131n\u0131 temsil etti\u011fini varsay\u0131yoruz.<\/p>\n

\u00d6rne\u011fin:<\/p>\n

\nGiri\u015f: nums = [2,2,1,1,1,2,2]\n\u00c7\u0131k\u0131\u015f: 2\n<\/pre>\n
Bu durumda 2 say\u0131s\u0131 toplam eleman say\u0131s\u0131n\u0131n (7) yar\u0131s\u0131ndan fazla (4) g\u00f6r\u00fcnmektedir.<\/p>\n
\u00c7\u00f6z\u00fcmler<\/h2>\n
Bu problemi \u00e7\u00f6zmek i\u00e7in birka\u00e7 farkl\u0131 y\u00f6ntem kullanabiliriz.<\/p>\n
1. Hash Tablosu<\/h3>\n
En basit \u00e7\u00f6z\u00fcmlerden biri hash tablosu kullanmakt\u0131r. Hash tablosunda her eleman\u0131n ka\u00e7 kez g\u00f6r\u00fcnd\u00fc\u011f\u00fcn\u00fc sayabiliriz. Daha sonra, en y\u00fcksek say\u0131ya sahip eleman\u0131 d\u00f6nd\u00fcr\u00fcr\u00fcz.<\/p>\n
“`python
\ndef majorityElement(nums):
\n  counts = {}
\n  for num in nums:
\n    if num in counts:
\n      counts[num] += 1
\n    else:
\n      counts[num] = 1
\n  return max(counts, key=counts.get)<\/p>\n
nums = [2,2,1,1,1,2,2]
\nprint(majorityElement(nums)) # \u00c7\u0131k\u0131\u015f: 2
\n“`<\/p>\n
2. S\u0131ralama<\/h3>\n
Bir di\u011fer yakla\u015f\u0131m ise diziyi s\u0131ralamakt\u0131r. S\u0131ral\u0131 bir dizide, \u00e7o\u011funluk eleman\u0131 ortas\u0131nda yer alacakt\u0131r. Dolay\u0131s\u0131yla diziyi s\u0131ralad\u0131ktan sonra orta eleman\u0131 d\u00f6nd\u00fcrebiliriz.<\/p>\n
“`python
\ndef majorityElement(nums):
\n  nums.sort()
\n  return nums[len(nums) \/\/ 2]<\/p>\n
nums = [2,2,1,1,1,2,2]
\nprint(majorityElement(nums)) # \u00c7\u0131k\u0131\u015f: 2
\n“`<\/p>\n
3. Boyer-Moore Voting Algorithm<\/h3>\n
Boyer-Moore Voting Algorithm, bu problemi en etkili \u015fekilde \u00e7\u00f6zen bir algoritmad\u0131r. Bu algoritma, iki de\u011fi\u015fken kullan\u0131r: candidate<\/code> ve count<\/code>. candidate<\/code>, \u015fu anki \u00e7o\u011funluk aday\u0131n\u0131 temsil ederken, count<\/code>, bu aday\u0131n ka\u00e7 kez g\u00f6r\u00fcnd\u00fc\u011f\u00fcn\u00fc sayar. Her eleman\u0131 dola\u015f\u0131rken, e\u011fer count<\/code> 0 ise candidate<\/code>‘\u0131 mevcut elemana ayarlar\u0131z. E\u011fer mevcut eleman candidate<\/code> ile ayn\u0131ysa count<\/code>‘u artt\u0131r\u0131r, de\u011filse count<\/code>‘u azalt\u0131r\u0131z. Son olarak candidate<\/code>‘\u0131 d\u00f6nd\u00fcr\u00fcr\u00fcz.<\/p>\n
“`python
\ndef majorityElement(nums):
\n  candidate = None
\n  count = 0
\n  for num in nums:
\n    if count == 0:
\n      candidate = num
\n    count += 1 if num == candidate else -1
\n  return candidate<\/p>\n
nums = [2,2,1,1,1,2,2]
\nprint(majorityElement(nums)) # \u00c7\u0131k\u0131\u015f: 2
\n“`<\/p>\n
Karma\u015f\u0131kl\u0131k Analizi<\/h2>\n
Yukar\u0131daki \u00e7\u00f6z\u00fcmlerin karma\u015f\u0131kl\u0131klar\u0131n\u0131 inceleyelim:<\/p>\n
| Y\u00f6ntem | Zaman Karma\u015f\u0131kl\u0131\u011f\u0131 | Uzay Karma\u015f\u0131kl\u0131\u011f\u0131 |
\n|—|—|—|
\n| Hash Tablosu | O(n) | O(n) |
\n| S\u0131ralama | O(n log n) | O(1) |
\n| Boyer-Moore Voting Algorithm | O(n) | O(1) |<\/p>\n
Boyer-Moore Voting Algorithm, zaman ve uzay a\u00e7\u0131s\u0131ndan en verimli \u00e7\u00f6z\u00fcmd\u00fcr.<\/p>\n
Sonu\u00e7<\/h2>\n
Bu makalede, LeetCode’un Top Interview 150-169 serisinden “\u00c7o\u011funluk Eleman\u0131” problemini inceledik. Farkl\u0131 \u00e7\u00f6z\u00fcmlerden bahsettik ve bunlar\u0131n karma\u015f\u0131kl\u0131klar\u0131n\u0131 analiz ettik. Boyer-Moore Voting Algorithm, bu problemi en etkili \u015fekilde \u00e7\u00f6zen y\u00f6ntemdir.<\/p>\n
Daha fazla bilgi i\u00e7in a\u015fa\u011f\u0131daki kaynaklar\u0131 inceleyebilirsiniz:<\/p>\n