İki dize (küçük harflerden oluşan) verildiğinde, bir desen ve bir dize. Görev, dizeleri desen tarafından tanımlanan sıraya göre sıralamaktır. Desenin dizedeki tüm karakterleri içerdiği ve desendeki tüm karakterlerin yalnızca bir kez göründüğü varsayılabilir.
Örnekler:
Input : pat = 'bca' str = 'abc' Output : str = 'bca' Input : pat = 'bxyzca' str = 'abcabcabc' Output : str = 'bbbcccaaa' Input : pat = 'wcyuogmlrdfphitxjakqvzbnes' str = 'jcdokai' Output : str = 'codijak'
Yaklaşım 1: Fikir ilk önce str'deki tüm karakterlerin oluşumlarını saymak ve bu sayıları bir sayım dizisinde saklamaktır. Daha sonra deseni soldan sağa doğru hareket ettirin ve her pat[i] karakteri için count dizisinde kaç kez göründüğüne bakın ve bu karakteri bu kadar çok kez str'ye kopyalayın.
Aşağıda yukarıdaki fikrin uygulanması yer almaktadır.
Uygulama:
C++// C++ program to sort a string according to the // order defined by a pattern string #include using namespace std; const int MAX_CHAR = 26; // Sort str according to the order defined by pattern. void sortByPattern(string& str string pat) { // Create a count array store count of characters in str. int count[MAX_CHAR] = { 0 }; // Count number of occurrences of each character // in str. for (int i = 0; i < str.length(); i++) count[str[i] - 'a']++; // Traverse the pattern and print every characters // same number of times as it appears in str. This // loop takes O(m + n) time where m is length of // pattern and n is length of str. int index = 0; for (int i = 0; i < pat.length(); i++) for (int j = 0; j < count[pat[i] - 'a']; j++) str[index++] = pat[i]; } // Driver code int main() { string pat = 'bca'; string str = 'abc'; sortByPattern(str pat); cout << str; return 0; }
Java // Java program to sort a string according to the // order defined by a pattern string class GFG { static int MAX_CHAR = 26; // Sort str according to the order defined by pattern. static void sortByPattern(char[] str char[] pat) { // Create a count array store // count of characters in str. int count[] = new int[MAX_CHAR]; // Count number of occurrences of // each character in str. for (int i = 0; i < str.length; i++) { count[str[i] - 'a']++; } // Traverse the pattern and print every characters // same number of times as it appears in str. This // loop takes O(m + n) time where m is length of // pattern and n is length of str. int index = 0; for (int i = 0; i < pat.length; i++) { for (int j = 0; j < count[pat[i] - 'a']; j++) { str[index++] = pat[i]; } } } // Driver code public static void main(String args[]) { char[] pat = 'bca'.toCharArray(); char[] str = 'abc'.toCharArray(); sortByPattern(str pat); System.out.println(String.valueOf(str)); } } // This code has been contributed by 29AjayKumar
Python3 # Python3 program to sort a string according to # the order defined by a pattern string MAX_CHAR = 26 # Sort str according to the order defined by pattern. def sortByPattern(str pat): global MAX_CHAR # Create a count array store count # of characters in str. count = [0] * MAX_CHAR # Count number of occurrences of # each character in str. for i in range (0 len(str)): count[ord(str[i]) - 97] += 1 # Traverse the pattern and print every characters # same number of times as it appears in str. This # loop takes O(m + n) time where m is length of # pattern and n is length of str. index = 0; str = '' for i in range (0 len(pat)): j = 0 while(j < count[ord(pat[i]) - ord('a')]): str += pat[i] j = j + 1 index += 1 return str # Driver code pat = 'bca' str = 'abc' print(sortByPattern(str pat)) # This code is contributed by ihritik
C# // C# program to sort a string according to the // order defined by a pattern string using System; class GFG { static int MAX_CHAR = 26; // Sort str according to the order defined by pattern. static void sortByPattern(char[] str char[] pat) { // Create a count array store // count of characters in str. int[] count = new int[MAX_CHAR]; // Count number of occurrences of // each character in str. for (int i = 0; i < str.Length; i++) { count[str[i] - 'a']++; } // Traverse the pattern and print every characters // same number of times as it appears in str. This // loop takes O(m + n) time where m is length of // pattern and n is length of str. int index = 0; for (int i = 0; i < pat.Length; i++) { for (int j = 0; j < count[pat[i] - 'a']; j++) { str[index++] = pat[i]; } } } // Driver code public static void Main(String[] args) { char[] pat = 'bca'.ToCharArray(); char[] str = 'abc'.ToCharArray(); sortByPattern(str pat); Console.WriteLine(String.Join('' str)); } } /* This code contributed by PrinciRaj1992 */
JavaScript <script> // Javascript program to sort a string according to the // order defined by a pattern string let MAX_CHAR = 26; // Sort str according to the order defined by pattern. function sortByPattern(strpat) { // Create a count array stor // count of characters in str. let count = new Array(MAX_CHAR); for(let i = 0; i < MAX_CHAR; i++) { count[i] = 0; } // Count number of occurrences of // each character in str. for (let i = 0; i < str.length; i++) { count[str[i].charCodeAt(0) - 'a'.charCodeAt(0)]++; } // Traverse the pattern and print every characters // same number of times as it appears in str. This // loop takes O(m + n) time where m is length of // pattern and n is length of str. let index = 0; for (let i = 0; i < pat.length; i++) { for (let j = 0; j < count[pat[i].charCodeAt(0) - 'a'.charCodeAt(0)]; j++) { str[index++] = pat[i]; } } } // Driver code let pat = 'bca'.split(''); let str = 'abc'.split(''); sortByPattern(str pat); document.write((str).join('')); // This code is contributed by rag2127 </script>
Çıkış
bca
Zaman karmaşıklığı: O(m+n) burada m desenin uzunluğu ve n str'nin uzunluğudur.
Yardımcı Alan: O(1)
Yaklaşım 2:
alfabe ve sayılar
sort() fonksiyonuna bir karşılaştırıcı aktarabilir ve dizeyi desene göre sıralayabiliriz.
C++#include using namespace std; // Declaring a vector globally that stores which character // is occurring first vector<int> position(26 -1); //Comparator function bool cmp(char& char1 char& char2) { return position[char1 - 'a'] < position[char2 - 'a']; } int main() { // Pattern string pat = 'wcyuogmlrdfphitxjakqvzbnes'; for (int i = 0; i < pat.length(); i++) { if (position[pat[i] - 'a'] == -1) position[pat[i] - 'a'] = i; } // String to be sorted string str = 'jcdokai'; // Passing a comparator to sort function sort(str.begin() str.end() cmp); cout << str; }
Java import java.util.*; class Main { // Declaring a list globally that stores which character is occurring first static List<Integer> position = new ArrayList<>(Collections.nCopies(26 -1)); // Comparator function static int cmp(char char1 char char2) { if (position.get(char1 - 'a') < position.get(char2 - 'a')) { return -1; } else if (position.get(char1 - 'a') > position.get(char2 - 'a')) { return 1; } else { return 0; } } public static void main(String[] args) { // Pattern String pat = 'wcyuogmlrdfphitxjakqvzbnes'; for (int i = 0; i < pat.length(); i++) { if (position.get(pat.charAt(i) - 'a') == -1) { position.set(pat.charAt(i) - 'a' i); } } // String to be sorted String str = 'jcdokai'; // Passing a comparator to the sorted function char[] charArr = str.toCharArray(); Arrays.sort(charArr new Comparator<Character>() { public int compare(Character c1 Character c2) { return cmp(c1 c2); } }); String sortedStr = new String(charArr); System.out.println(sortedStr); } }
Python3 from typing import List from functools import cmp_to_key # Declaring a list globally that stores which character is occurring first position: List[int] = [-1] * 26 # Comparator function def cmp(char1: str char2: str) -> int: if position[ord(char1) - ord('a')] < position[ord(char2) - ord('a')]: return -1 elif position[ord(char1) - ord('a')] > position[ord(char2) - ord('a')]: return 1 else: return 0 if __name__ == '__main__': # Pattern pat = 'wcyuogmlrdfphitxjakqvzbnes' for i in range(len(pat)): if position[ord(pat[i]) - ord('a')] == -1: position[ord(pat[i]) - ord('a')] = i # String to be sorted str = 'jcdokai' # Passing a comparator to the sorted function sorted_str = sorted(str key=cmp_to_key(cmp)) print(''.join(sorted_str)) # This code is contributed by adityashatmfh
JavaScript <script> // Declaring a vector globally that stores which character // is occurring first let position = new Array(26).fill(-1); //Comparator function function cmp(char1 char2) { return position[char1.charCodeAt(0) - 'a'.charCodeAt(0)] - position[char2.charCodeAt(0) - 'a'.charCodeAt(0)]; } // driver code // Pattern let pat = 'wcyuogmlrdfphitxjakqvzbnes'; for (let i = 0; i <br pat.length; i++) { if (position[pat.charCodeAt(i) - 'a'.charCodeAt(0)] == -1) position[pat.charCodeAt(i) - 'a'.charCodeAt(0)] = i; } // String to be sorted let str = 'jcdokai'; // Passing a comparator to sort function str = str.split('').sort(cmp).join(''); document.write(str''); // This code is contributed by Shinjan Patra </script>
C# // C# program for the above approach using System; using System.Collections.Generic; using System.Linq; class GFG { // Declaring a list globally that stores which character is occurring first static List<int> position = Enumerable.Repeat(-1 26).ToList(); // Comparator function static int Cmp(char char1 char char2) { if (position[char1 - 'a'] < position[char2 - 'a']) { return -1; } else if (position[char1 - 'a'] > position[char2 - 'a']) { return 1; } else { return 0; } } public static void Main() { // Pattern string pat = 'wcyuogmlrdfphitxjakqvzbnes'; for (int i = 0; i < pat.Length; i++) { if (position[pat[i] - 'a'] == -1) { position[pat[i] - 'a'] = i; } } // String to be sorted string str = 'jcdokai'; // Passing a comparator to the sorted function char[] charArr = str.ToCharArray(); Array.Sort(charArr new Comparison<char>(Cmp)); string sortedStr = new string(charArr); Console.WriteLine(sortedStr); } } // This code is contributed by sdeadityasharma
Çıkış
codijak
Zaman karmaşıklığı: O(m+nlogn ) burada m desenin uzunluğu ve n str'nin uzunluğudur.
Yardımcı Alan: O(1)
Egzersiz yapmak : Yukarıdaki çözümde desenin tüm str karakterlerine sahip olduğu varsayılmaktadır. Desenin tüm karakterlere sahip olamayabileceği ve görevin kalan tüm karakterleri (dizgede ancak desende değil) sona koymak olduğu değiştirilmiş bir versiyonunu düşünün. Kalan karakterlerin alfabetik olarak sıralanması gerekir. İpucu: İkinci döngüde indeksi arttırırken ve karakteri str'ye koyarken o andaki sayıyı da azaltabiliriz. Ve son olarak geri kalan karakterleri alfabetik olarak sıralamak için count dizisini dolaşıyoruz.
Test Oluştur