\n| $S_3$<\/td>\n | 15<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n G\u00f6r\u00fcld\u00fc\u011f\u00fc gibi, $S_i$ dizisinin uzunlu\u011fu $2^{i+1} – 1$ form\u00fcl\u00fc ile ifade edilebilir. Bu, her ad\u0131mda uzunlu\u011fun yakla\u015f\u0131k olarak iki kat\u0131na \u00e7\u0131kt\u0131\u011f\u0131n\u0131 do\u011frular.<\/p>\n 2.3. Karakterlerin \u00d6zellikleri<\/h4>\nDizeleri dikkatle inceledi\u011fimizde baz\u0131 \u00f6nemli \u00f6zellikler fark ederiz:<\/p>\n \n- Her $S_i$ dizisinin tam ortas\u0131nda her zaman ‘1’ karakteri bulunur. (\u00d6rn: $S_1 = “0\\underline{1}1″$, $S_2 = “011\\underline{1}001″$)<\/li>\n
- $S_i$ dizisinin ilk yar\u0131s\u0131, do\u011frudan $S_{i-1}$ dizisidir.<\/li>\n
- $S_i$ dizisinin ikinci yar\u0131s\u0131 (orta ‘1’den sonraki k\u0131s\u0131m), $S_{i-1}$’in ters \u00e7evrilmi\u015f ve tersine \u00e7evrilmi\u015f halidir.<\/li>\n<\/ul>\n
3. \u0130lk Yakla\u015f\u0131m: Brute Force ve Neden Yetersiz?<\/h3>\nDo\u011frudan dizeyi olu\u015fturma fikri, akla gelen ilk yakla\u015f\u0131md\u0131r. Ancak bu, b\u00fcy\u00fck $n$ de\u011ferleri i\u00e7in h\u0131zla kullan\u0131lamaz hale gelir.<\/p>\n 3.1. Basit Bir Sim\u00fclasyon<\/h4>\nE\u011fer $n$ k\u00fc\u00e7\u00fckse (\u00f6rne\u011fin $n < 10$), $S_n$ dizisini yukar\u0131daki kurallara g\u00f6re bir d\u00f6ng\u00fc i\u00e7inde olu\u015fturabilir ve sonra $k$-\u0131nc\u0131 karakteri do\u011frudan alabiliriz. Python'da basit bir implementasyon \u015f\u00f6yle g\u00f6r\u00fcnebilir:<\/p>\n \ndef invert_string(s):\n return \"\".join(['1' if c == '0' else '0' for c in s])\n\ndef find_kth_character_brute_force(n, k):\n s = \"0\"\n for _ in range(n):\n inverted_s = invert_string(s)\n reversed_inverted_s = inverted_s[::-1]\n s = s + '1' + reversed_inverted_s\n return s[k-1] # k 1-indeksli oldu\u011fu i\u00e7in k-1\n\n# \u00d6rnek kullan\u0131m:\n# print(find_kth_character_brute_force(2, 5)) # \u00c7\u0131kt\u0131: 0 (S2 = \"0111001\", 5. karakter '0')\n<\/pre>\n<\/code><\/p>\n 3.2. Zaman ve Bellek S\u0131n\u0131rlamalar\u0131<\/h4>\nBu yakla\u015f\u0131m\u0131n temel sorunu, dizenin uzunlu\u011funun $n$ ile birlikte \u00fcstel olarak b\u00fcy\u00fcmesidir. $S_n$ dizisinin uzunlu\u011fu $2^{n+1} - 1$ oldu\u011fundan, $n=20$ i\u00e7in bile dize uzunlu\u011fu $2^{21}-1 \\approx 2$ milyon karakteri a\u015far. $n=60$ gibi de\u011ferler i\u00e7in dize uzunlu\u011fu $2^{61}-1$ olur ki bu, mevcut bellekte saklanamayacak kadar b\u00fcy\u00fckt\u00fcr. Ayr\u0131ca, her ad\u0131mda dize birle\u015ftirme ve ters \u00e7evirme i\u015flemleri de uzunlukla orant\u0131l\u0131 olarak zaman al\u0131r, bu da zaman karma\u015f\u0131kl\u0131\u011f\u0131n\u0131 $O(2^n)$ yapar.<\/p>\n 3.3. Daha Verimli Bir Yola \u0130htiya\u00e7<\/h4>\nBrute force yakla\u015f\u0131m\u0131n\u0131n yetersizli\u011fi, bizi dizeyi fiziksel olarak olu\u015fturmadan $k$-\u0131nc\u0131 karakteri bulman\u0131n yollar\u0131n\u0131 aramaya iter. Bu, genellikle problemin rek\u00fcrsif do\u011fas\u0131n\u0131 ve i\u00e7indeki desenleri kullanmay\u0131 gerektirir.<\/p>\n 4. Desenleri Ke\u015ffetmek: \u00d6zyinelemeli Yap\u0131<\/h3>\nProblemin rek\u00fcrsif tan\u0131m\u0131, \u00e7\u00f6z\u00fcm\u00fc de rek\u00fcrsif bir \u015fekilde d\u00fc\u015f\u00fcnmemiz gerekti\u011fini i\u015faret eder. $S_n$ i\u00e7indeki $k$-\u0131nc\u0131 karakteri bulmak i\u00e7in $S_{n-1}$'e nas\u0131l ba\u015fvurabilece\u011fimizi inceleyelim.<\/p>\n 4.1. Dize Uzunluklar\u0131n\u0131n \u0130li\u015fkisi<\/h4>\n$S_n$ dizisinin uzunlu\u011fu $\\text{len}(S_n) = 2^{n+1} - 1$'dir. Bu, $S_n$ dizisinin orta noktas\u0131n\u0131n indeksinin (1-indeksli) $2^n$'e denk geldi\u011fi anlam\u0131na gelir. (\u00d6rn: $S_2$ uzunlu\u011fu 7, orta nokta $2^2=4$. $S_2 = \"011\\underline{1}001\"$, 4. karakter '1').<\/p>\n 4.2. Simetri ve Ters \u00c7evirme<\/h4>\nDize olu\u015fum kural\u0131 $S_i = S_{i-1} + '1' + \\text{reverse}(\\text{invert}(S_{i-1}))$ bize \u00fc\u00e7 ana b\u00f6l\u00fcm oldu\u011funu g\u00f6sterir:<\/p>\n \n- $S_{i-1}$ (ilk k\u0131s\u0131m)<\/li>\n
- '1' (orta karakter)<\/li>\n
- $\\text{reverse}(\\text{invert}(S_{i-1}))$ (son k\u0131s\u0131m)<\/li>\n<\/ol>\n
Bu yap\u0131, karakterin $S_n$ i\u00e7inde hangi b\u00f6l\u00fcme d\u00fc\u015ft\u00fc\u011f\u00fcne ba\u011fl\u0131 olarak problemin boyutunu k\u00fc\u00e7\u00fcltmemize olanak tan\u0131r.<\/p>\n 4.3. \u00d6zyinelemeli \u0130li\u015fki Kurma<\/h4>\nBir karakterin $S_n$ i\u00e7indeki konumunu $k$ olarak alal\u0131m. \u00dc\u00e7 farkl\u0131 durum s\u00f6z konusudur:<\/p>\n \n- \n E\u011fer $k$ ilk yar\u0131daysa:<\/strong> Yani $1 \\le k < 2^n$. Bu durumda $k$-\u0131nc\u0131 karakter, do\u011frudan $S_{n-1}$ dizisindeki $k$-\u0131nc\u0131 karakterdir. Problemi $f(n, k) \\rightarrow f(n-1, k)$ olarak k\u00fc\u00e7\u00fcltebiliriz.\n <\/li>\n
- \n E\u011fer $k$ orta karakterse:<\/strong> Yani $k = 2^n$. Bu durumda $k$-\u0131nc\u0131 karakter her zaman '1'dir.\n <\/li>\n
- \n E\u011fer $k$ ikinci yar\u0131daysa:<\/strong> Yani $2^n < k \\le 2^{n+1} - 1$. Bu durumda $k$-\u0131nc\u0131 karakter, $\\text{reverse}(\\text{invert}(S_{n-1}))$ k\u0131sm\u0131nda yer al\u0131r.\n \n\n
Bu k\u0131s\u0131mda karakteri bulmak i\u00e7in biraz d\u00f6n\u00fc\u015f\u00fcm yapmam\u0131z gerekir:<\/p>\n \n- \u00d6ncelikle, $k$ de\u011ferini $S_{n-1}$'in sonundan itibaren olan konumuna \u00e7evirmeliyiz. $S_n$'deki $k$ konumu, $S_{n-1}$'in sonundan $(2^{n+1} - 1) - k + 1$ uzakl\u0131ktad\u0131r. Ya da daha basit\u00e7e, $S_{n-1}$'in uzunlu\u011fu $2^n - 1$ oldu\u011fundan, $k$ konumunu orta '1'den sonraki k\u0131sma g\u00f6re ayarlayal\u0131m: $k' = k - (2^n)$. Bu $k'$ de\u011feri, $\\text{reverse}(\\text{invert}(S_{n-1}))$ dizisinin ba\u015flang\u0131c\u0131ndan itibaren olan konumdur.<\/li>\n
- \u015eimdi, bu $k'$ konumundaki karakteri $\\text{reverse}(\\text{invert}(S_{n-1}))$ i\u00e7inde bulmal\u0131y\u0131z. $\\text{reverse}(X)$ i\u00e7indeki $k'$-\u0131nc\u0131 karakter, $X$ i\u00e7indeki $\\text{len}(X) - k' + 1$-inci karaktere e\u015fittir. Yani, $\\text{invert}(S_{n-1})$ i\u00e7indeki $(2^n - 1) - k' + 1$-inci karaktere.<\/li>\n
- Son olarak, buldu\u011fumuz karakteri ters \u00e7evirmeliyiz (invert).<\/li>\n<\/ul>\n
Bu karma\u015f\u0131k d\u00f6n\u00fc\u015f\u00fcm\u00fc basitle\u015ftirelim: $S_n$'deki $k$-\u0131nc\u0131 karakter, $S_{n-1}$'in ters \u00e7evrilmi\u015f ve tersine \u00e7evrilmi\u015f halindeki bir karaktere kar\u015f\u0131l\u0131k gelir. $S_n$'deki $k$-\u0131nc\u0131 karakterin, $S_{n-1}$'in $\\text{len}(S_{n-1}) - (k - 2^n) + 1$ konumundaki karakterinin tersi oldu\u011funu buluruz. Yani, $k_{new} = \\text{len}(S_{n-1}) - (k - 2^n) + 1 = (2^n - 1) - (k - 2^n) + 1 = 2^{n+1} - k$. Problemi $f(n, k) \\rightarrow \\text{invert}(f(n-1, 2^{n+1} - k))$ olarak k\u00fc\u00e7\u00fcltebiliriz.<\/p>\n<\/li>\n<\/ol>\n 5. Optimal \u00c7\u00f6z\u00fcm: \u0130kili Arama ve Bit Manip\u00fclasyonu<\/h3>\nYukar\u0131daki rek\u00fcrsif ili\u015fkiyi kullanarak, problemi her ad\u0131mda yar\u0131ya indirgeyebiliriz. Ayr\u0131ca, karakterlerin ters \u00e7evrilme (invert) durumunu takip etmek i\u00e7in bit manip\u00fclasyonu veya basit bir saya\u00e7 kullanabiliriz.<\/p>\n 5.1. Problemi K\u00fc\u00e7\u00fcltme Stratejisi<\/h4>\nAna fikir, $k$-\u0131nc\u0131 karakteri bulmak i\u00e7in $n$ de\u011ferini azaltarak temel duruma ($S_0$) ula\u015fmakt\u0131r. Her ad\u0131mda, $k$'n\u0131n $S_n$'in hangi b\u00f6l\u00fcm\u00fcne d\u00fc\u015ft\u00fc\u011f\u00fcne bakar\u0131z:<\/p>\n \n- \n Orta nokta:<\/strong> E\u011fer $k = 2^n$ ise, karakter '1'dir.\n <\/li>\n
- \n Sol taraf:<\/strong> E\u011fer $k < 2^n$ ise, karakter $S_{n-1}$'deki $k$-\u0131nc\u0131 karakterle ayn\u0131d\u0131r. $n$ de\u011ferini $n-1$ yapar\u0131z, $k$ de\u011fi\u015fmez.\n <\/li>\n
- \n Sa\u011f taraf:<\/strong> E\u011fer $k > 2^n$ ise, karakter $S_{n-1}$'deki bir karakterin tersidir. Yeni $k$ de\u011feri $2^{n+1} - k$ olur ve $n$ de\u011ferini $n-1$ yapar\u0131z. Ayr\u0131ca, bu ad\u0131mda bir ters \u00e7evirme (invert) i\u015flemi oldu\u011fu i\u00e7in, nihai sonucu bir kez daha ters \u00e7evirmemiz gerekti\u011fini not etmeliyiz. Bunu bir saya\u00e7 (\u00f6rne\u011fin
flips<\/code> de\u011fi\u015fkeni) ile takip edebiliriz.\n <\/li>\n<\/ul>\n5.2. Orta Nokta ve Karakter Konumu<\/h4>\nHer ad\u0131mda, mevcut $S_n$ dizisinin uzunlu\u011funu $L = 2^{n+1} - 1$ ve orta noktas\u0131n\u0131 $mid = (L+1)\/2 = 2^n$ olarak hesaplar\u0131z. Sonra $k$'y\u0131 bu $mid$ ile kar\u015f\u0131la\u015ft\u0131r\u0131r\u0131z.<\/p>\n 5.3. Bit Manip\u00fclasyonunun Rol\u00fc<\/h4>\nAsl\u0131nda, '0' ve '1' karakterleri ikili say\u0131lar olarak d\u00fc\u015f\u00fcn\u00fclebilir. Bir karakteri ters \u00e7evirmek, o biti flip etmek anlam\u0131na gelir. Ba\u015flang\u0131\u00e7 karakteri $S_0 = '0'$'d\u0131r. Her sa\u011f tarafa ge\u00e7ti\u011fimizde, bir invert i\u015flemi ger\u00e7ekle\u015fir. Bu invert i\u015flemlerinin say\u0131s\u0131n\u0131n tek mi \u00e7ift mi oldu\u011funu takip edebiliriz. E\u011fer toplam invert say\u0131s\u0131 tek ise, nihai karakterin tersini almal\u0131y\u0131z; \u00e7ift ise, karakter oldu\u011fu gibi kal\u0131r.<\/p>\n Alternatif olarak, $S_0 = \"0\"$ oldu\u011fu i\u00e7in, $k$-\u0131nc\u0131 karakterin asl\u0131nda $S_0$ i\u00e7indeki '0'\u0131n ka\u00e7 kez flip edildi\u011fiyle ili\u015fkili oldu\u011funu d\u00fc\u015f\u00fcnebiliriz. $S_0$ tek karakterli bir dizedir ve bu karakter '0'd\u0131r. $S_n$ i\u00e7inde $k$-\u0131nc\u0131 karakterin de\u011ferini belirlemek i\u00e7in, $k-1$ say\u0131s\u0131n\u0131n ikili g\u00f6sterimindeki '1' say\u0131s\u0131n\u0131n (popcount) paritesini kullanabiliriz. E\u011fer $k-1$ i\u00e7indeki '1' say\u0131s\u0131 tekse, karakter '1'dir; \u00e7iftse, karakter '0'd\u0131r. Bu, $S_0 = \"0\"$ ve $S_i = S_{i-1} + '1' + S'_{i-1}$ (burada $S'_{i-1}$ invert edilmi\u015f) yap\u0131s\u0131n\u0131n bir sonucudur. Her invert i\u015flemi, karakterin de\u011ferini de\u011fi\u015ftirir. $k-1$ i\u00e7indeki '1' say\u0131s\u0131, asl\u0131nda o karaktere ula\u015fmak i\u00e7in ka\u00e7 kez \"sa\u011f tarafa\" ge\u00e7ildi\u011fini g\u00f6sterir.<\/p>\n Daha do\u011fru bir ifadeyle, $k$-\u0131nc\u0131 karakterin de\u011ferini bulmak i\u00e7in, $k-1$ say\u0131s\u0131n\u0131n ikili g\u00f6sterimindeki '1' say\u0131s\u0131n\u0131n paritesine bakabiliriz. E\u011fer $S_0 = \"0\"$ olmasayd\u0131, bu do\u011frudan do\u011fru olmazd\u0131. Ancak $S_0 = \"0\"$ oldu\u011fu i\u00e7in, $k-1$ i\u00e7indeki set bit say\u0131s\u0131 (popcount) bize karakterin de\u011ferini verir. \u00d6rne\u011fin, $k=1$ i\u00e7in $k-1=0$, popcount=0, karakter '0'. $k=2$ i\u00e7in $k-1=1$, popcount=1, karakter '1'. $k=3$ i\u00e7in $k-1=2$, popcount=1, karakter '1'. Bu, $S_1 = \"011\"$ ile e\u015fle\u015fir. $k=4$ i\u00e7in $k-1=3$, popcount=2, karakter '0'. Bu, $S_2 = \"0111001\"$ ile e\u015fle\u015fmez. Demek ki bu basit bit manip\u00fclasyonu do\u011frudan \u00e7al\u0131\u015fm\u0131yor. \u00d6nceki rek\u00fcrsif yakla\u015f\u0131m daha sa\u011flam.<\/p>\n Geri d\u00f6nelim rek\u00fcrsif yakla\u015f\u0131ma: $S_0 = \"0\"$. flips<\/code> sayac\u0131 ile ka\u00e7 kez invert edildi\u011fini takip edelim. Ba\u015flang\u0131\u00e7ta flips = 0<\/code>.<\/p>\n\n- E\u011fer $k = 2^n$ ise, karakter '1'dir. Nihai sonu\u00e7
(1 + flips) % 2<\/code> olur.<\/li>\n- E\u011fer $k < 2^n$ ise, $n \\leftarrow n-1$, $k$ ayn\u0131 kal\u0131r.<\/li>\n
- E\u011fer $k > 2^n$ ise, $n \\leftarrow n-1$, $k \\leftarrow 2^{n+1} - k$,
flips \\leftarrow flips + 1<\/code>.<\/li>\n<\/ul>\nTemel durum $n=0, k=1$ oldu\u011funda $S_0 = \"0\"$'d\u0131r. Bu durumda nihai karakter (0 + flips) % 2<\/code> olur.<\/p>\n6. Algoritman\u0131n Ad\u0131m Ad\u0131m \u0130ncelenmesi<\/h3>\n\u015eimdi bu rek\u00fcrsif algoritmay\u0131 daha somut bir \u015fekilde inceleyelim.<\/p>\n 6.1. Temel Durumlar<\/h4>\nRek\u00fcrsif fonksiyonun durma ko\u015fulu, $n=0$ ve $k=1$ oldu\u011funda ger\u00e7ekle\u015fir. Bu durumda, $S_0$ dizisinin tek karakteri olan '0'\u0131 d\u00f6nd\u00fcr\u00fcr\u00fcz. Ancak, bu '0' karakteri, yol boyunca ka\u00e7 kez ters \u00e7evrildi\u011fine ba\u011fl\u0131 olarak de\u011fi\u015febilir. Bu y\u00fczden, flips<\/code> sayac\u0131n\u0131 da hesaba katmal\u0131y\u0131z.<\/p>\n\ndef find_kth_char(n, k):\n # flips, karakterin ka\u00e7 kez ters \u00e7evrildi\u011fini tutar.\n # Ba\u015flang\u0131\u00e7ta 0.\n flips = 0\n\n while n > 0:\n length_sn_minus_1 = (1 << n) - 1 # S_{n-1} uzunlu\u011fu\n mid_point = (1 << n) # S_n'deki orta '1'in konumu (1-indeksli)\n\n if k == mid_point:\n # Orta karakter her zaman '1'dir.\n # Ancak yol boyunca flip edilmi\u015f olabilir.\n return str((1 + flips) % 2)\n elif k < mid_point:\n # Karakter S_{n-1} i\u00e7inde. n'i azalt, k ayn\u0131 kal\u0131r.\n n -= 1\n else:\n # Karakter S_{n-1}'in ters \u00e7evrilmi\u015f ve tersine \u00e7evrilmi\u015f k\u0131sm\u0131nda.\n # Yeni k, S_{n-1} i\u00e7indeki simetrik konumudur.\n # Bir flip i\u015flemi ger\u00e7ekle\u015fti.\n k = length_sn_minus_1 - (k - mid_point) + 1\n n -= 1\n flips += 1\n \n # n=0 durumuna ula\u015f\u0131ld\u0131. S_0 = \"0\".\n # flips say\u0131s\u0131na g\u00f6re nihai karakteri d\u00f6nd\u00fcr.\n return str((0 + flips) % 2)\n\n<\/pre>\n<\/code><\/p>\n 6.2. \u00d6zyineleme Ad\u0131mlar\u0131<\/h4>\nYukar\u0131daki d\u00f6ng\u00fc tabanl\u0131 \u00e7\u00f6z\u00fcm, asl\u0131nda rek\u00fcrsif bir fonksiyonun yinelemeli halidir. Her ad\u0131mda $n$ de\u011feri azal\u0131r, bu da logaritmik say\u0131da ad\u0131mda sonuca ula\u015faca\u011f\u0131m\u0131z anlam\u0131na gelir.<\/p>\n 6.3. \u00d6rnek \u00c7al\u0131\u015fma (n=3, k=12 i\u00e7in)<\/h4>\nHedef: $S_3$ dizisindeki 12. karakter.<\/p>\n \n- \n Ba\u015flang\u0131\u00e7:<\/strong> $n=3, k=12$,
flips=0<\/code>.\n <\/li>\n- \n Ad\u0131m 1 ($n=3$):<\/strong><\/p>\n
\n- $\\text{len}(S_2) = 2^3 - 1 = 7$.<\/li>\n
- $\\text{mid_point} = 2^3 = 8$.<\/li>\n
- $k=12 > \\text{mid_point}=8$. Sa\u011f tarafa d\u00fc\u015ft\u00fc.<\/li>\n
- Yeni $k = \\text{len}(S_2) - (k - \\text{mid_point}) + 1 = 7 - (12 - 8) + 1 = 7 - 4 + 1 = 4$.<\/li>\n
- $n \\leftarrow 2$.<\/li>\n
flips \\leftarrow 1<\/code>.<\/li>\n<\/ul>\nDurum: $n=2, k=4$, flips=1<\/code>.<\/p>\n<\/li>\n- \n Ad\u0131m 2 ($n=2$):<\/strong><\/p>\n
\n- $\\text{len}(S_1) = 2^2 - 1 = 3$.<\/li>\n
- $\\text{mid_point} = 2^2 = 4$.<\/li>\n
- $k=4 == \\text{mid_point}=4$. Orta karaktere ula\u015f\u0131ld\u0131.<\/li>\n
- D\u00f6nd\u00fcr: $\\text{str}((1 + \\text{flips}) \\% 2) = \\text{str}((1 + 1) \\% 2) = \\text{str}(2 \\% 2) = \\text{str}(0)$.<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n
Sonu\u00e7: '0'. Do\u011frulayal\u0131m: $S_3 = \"011100110110001\"$. 12. karakter ger\u00e7ekten '0'd\u0131r.<\/p>\n 7. Karma\u015f\u0131kl\u0131k Analizi<\/h3>\nOptimal \u00e7\u00f6z\u00fcm\u00fcn performans\u0131n\u0131 anlamak i\u00e7in zaman ve uzay karma\u015f\u0131kl\u0131klar\u0131n\u0131 inceleyelim.<\/p>\n 7.1. Zaman Karma\u015f\u0131kl\u0131\u011f\u0131<\/h4>\nHer ad\u0131mda $n$ de\u011feri 1 azal\u0131r ve $k$ de\u011feri ya ayn\u0131 kal\u0131r ya da yeni bir konuma ayarlan\u0131r. $n$ de\u011feri 0 olana kadar bu i\u015flem devam eder. Bu, $n$ kez tekrarlanan bir d\u00f6ng\u00fc veya rek\u00fcrsif \u00e7a\u011fr\u0131 anlam\u0131na gelir. Dolay\u0131s\u0131yla, zaman karma\u015f\u0131kl\u0131\u011f\u0131 $O(n)$'dir. $n$ genellikle 60'a kadar olabilece\u011finden, bu \u00e7ok verimli bir \u00e7\u00f6z\u00fcmd\u00fcr.<\/p>\n 7.2. Uzay Karma\u015f\u0131kl\u0131\u011f\u0131<\/h4>\n\u00c7\u00f6z\u00fcm, sabit say\u0131da de\u011fi\u015fken (n, k, flips) kullan\u0131r ve rek\u00fcrsif \u00e7a\u011fr\u0131 y<\/p>\n","protected":false},"excerpt":{"rendered":"Problem, $S_0 = “0”$ ile ba\u015flayan bir dize serisi tan\u0131mlar. Her bir sonraki dize $S_i$, $S_{i-1}$ dizisinden t\u00fcretilir. Kural \u015fu \u015fekildedir:","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-39280","1":"post","2":"type-post","3":"status-publish","4":"format-standard","6":"category-genel","7":"cs-entry","8":"cs-video-wrap"},"yoast_head":"\n LeetCode 3304: Dize Oyununda K'\u0131nc\u0131 Karakteri Anlamak (Ad\u0131m Ad\u0131m K\u0131lavuz) - 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-3304-dize-oyununda-kinci-karakteri-anlamak-adim-adim-kilavuz\/\" \/>\n<meta property=\"og:locale\" content=\"tr_TR\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"LeetCode 3304: Dize Oyununda K'\u0131nc\u0131 Karakteri Anlamak (Ad\u0131m Ad\u0131m K\u0131lavuz)\" \/>\n<meta property=\"og:description\" content=\"Problem, $S_0 = "0"$ ile ba\u015flayan bir dize serisi tan\u0131mlar. 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