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Comproveu si la cadena segueix l'ordre dels caràcters definit per un patró o no | Set 2

Donada una cadena d'entrada i un patró, comproveu si els caràcters de la cadena d'entrada segueixen el mateix ordre que el determinat pels caràcters presents al patró. Suposem que no hi haurà caràcters duplicats al patró.
Es publica una altra solució al mateix problema aquí .
Exemples:  
 

Input: string = 'engineers rock' pattern = 'er'; Output: true All 'e' in the input string are before all 'r'. Input: string = 'engineers rock' pattern = 'egr'; Output: false There are two 'e' after 'g' in the input string. Input: string = 'engineers rock' pattern = 'gsr'; Output: false There are one 'r' before 's' in the input string.


 



.següent java


La idea aquí és reduir la cadena donada al patró donat. Per als caràcters indicats al patró només conservem els caràcters corresponents a la cadena. A la cadena nova suprimim els caràcters repetits continus. Aleshores, la cadena modificada hauria de ser igual al patró donat. Finalment, comparem la cadena modificada amb el patró donat i tornem true o false en conseqüència.
Il·lustració: 
 

str = 'bfbaeadeacc' pat[] = 'bac' 1) Remove extra characters from str (characters that are not present in pat[] str = 'bbaaacc' [f e and d are removed] 3) Removed consecutive repeating occurrences of characters str = 'bac' 4) Since str is same as pat[] we return true


A continuació es mostra la implementació dels passos anteriors.
 

C++ 
// C++ code for the above approach #include    #include  using namespace std; bool followsPattern(string str string pattern) {  // Insert all characters of pattern in a hash set  unordered_set<char> patternSet;  for (int i = 0; i < pattern.length(); i++) {  patternSet.insert(pattern[i]);  }  // Build modified string (string with characters only from pattern are taken)  string modifiedStr = str;  for (int i = str.length() - 1; i >= 0; i--) {  if (patternSet.find(str[i]) == patternSet.end()) {  modifiedStr.erase(i 1);  }  }  // Remove more than one consecutive occurrences of pattern characters from modified string  for (int i = modifiedStr.length() - 1; i > 0; i--) {  if (modifiedStr[i] == modifiedStr[i - 1]) {  modifiedStr.erase(i 1);  }  }  // After above modifications the length of modified string must be same as pattern length  if (pattern.length() != modifiedStr.length()) {  return false;  }  // And pattern characters must also be same as modified string characters  for (int i = 0; i < pattern.length(); i++) {  if (pattern[i] != modifiedStr[i]) {  return false;  }  }  return true; } int main() {  string str = 'engineers rock';  string pattern = 'er';  cout << 'Expected: true Actual: ' << followsPattern(str pattern) << endl;  str = 'engineers rock';  pattern = 'egr';  cout << 'Expected: false Actual: ' << followsPattern(str pattern) << endl;  str = 'engineers rock';  pattern = 'gsr';  cout << 'Expected: false Actual: ' << followsPattern(str pattern) << endl;  str = 'engineers rock';  pattern = 'eger';  cout << 'Expected: true Actual: ' << followsPattern(str pattern) << endl;  return 0; } // This code is contributed by adityashatmfh
Java 
// Java program to check if characters of a string follow // pattern defined by given pattern. import java.util.*; public class OrderOfCharactersForPattern {  public static boolean followsPattern(String str String pattern)  {  // Insert all characters of pattern in a hash set  Set<Character> patternSet = neHashSet<>();  for (int i=0; i<pattern.length(); i++)  patternSet.add(pattern.charAt(i));  // Build modified string (string with characters only from  // pattern are taken)  StringBuilder modifiedString = new StringBuilder(str);  for (int i=str.length()-1; i>=0; i--)  if (!patternSet.contains(modifiedString.charAt(i)))  modifiedString.deleteCharAt(i);  // Remove more than one consecutive occurrences of pattern  // characters from modified string.  for (int i=modifiedString.length()-1; i>0; i--)  if (modifiedString.charAt(i) == modifiedString.charAt(i-1))  modifiedString.deleteCharAt(i);  // After above modifications the length of modified string  // must be same as pattern length  if (pattern.length() != modifiedString.length())  return false;  // And pattern characters must also be same as modified string  // characters  for (int i=0; i<pattern.length(); i++)  if (pattern.charAt(i) != modifiedString.charAt(i))  return false;  return true;  }  // Driver program  int main()  {  String str = 'engineers rock';  String pattern = 'er';  System.out.println('Expected: true Actual: ' +  followsPattern(str pattern));  str = 'engineers rock';  pattern = 'egr';  System.out.println('Expected: false Actual: ' +  followsPattern(str pattern));  str = 'engineers rock';  pattern = 'gsr';  System.out.println('Expected: false Actual: ' +  followsPattern(str pattern));  str = 'engineers rock';  pattern = 'eger';  System.out.println('Expected: true Actual: ' +  followsPattern(str pattern));  return 0;  } }
Python3 
# Python3 program to check if characters of  # a string follow pattern defined by given pattern. def followsPattern(string pattern): # Insert all characters of pattern in a hash set patternSet = set() for i in range(len(pattern)): patternSet.add(pattern[i]) # Build modified string (string with characters  # only from pattern are taken) modifiedString = string for i in range(len(string) - 1 -1 -1): if not modifiedString[i] in patternSet: modifiedString = modifiedString[:i] +  modifiedString[i + 1:] # Remove more than one consecutive occurrences  # of pattern characters from modified string. for i in range(len(modifiedString) - 1 0 -1): if modifiedString[i] == modifiedString[i - 1]: modifiedString = modifiedString[:i] +  modifiedString[i + 1:] # After above modifications the length of  # modified string must be same as pattern length if len(pattern) != len(modifiedString): return False # And pattern characters must also be same  # as modified string characters for i in range(len(pattern)): if pattern[i] != modifiedString[i]: return False return True # Driver Code if __name__ == '__main__': string = 'engineers rock' pattern = 'er' print('Expected: true Actual:' followsPattern(string pattern)) string = 'engineers rock' pattern = 'egr' print('Expected: false Actual:' followsPattern(string pattern)) string = 'engineers rock' pattern = 'gsr' print('Expected: false Actual:' followsPattern(string pattern)) string = 'engineers rock' pattern = 'eger' print('Expected: true Actual:' followsPattern(string pattern)) # This code is contributed by # sanjeev2552
C# 
// C# program to check if characters of a string follow // pattern defined by given pattern. using System; using System.Collections.Generic; using System.Text; class GFG {  public static bool followsPattern(String str String pattern)  {  // Insert all characters of pattern in a hash set  HashSet<char> patternSet = new HashSet<char>();  for (int i = 0; i < pattern.Length; i++)  patternSet.Add(pattern[i]);  // Build modified string (string with characters   // only from pattern are taken)  StringBuilder modifiedString = new StringBuilder(str);  for (int i = str.Length - 1; i >= 0; i--)  if (!patternSet.Contains(modifiedString[i]))  modifiedString.Remove(i 1);  // Remove more than one consecutive occurrences of pattern  // characters from modified string.  for (int i = modifiedString.Length - 1; i > 0; i--)  if (modifiedString[i] == modifiedString[i - 1])  modifiedString.Remove(i 1);  // After above modifications the length of modified string  // must be same as pattern length  if (pattern.Length != modifiedString.Length)  return false;  // And pattern characters must also be same   // as modified string characters  for (int i = 0; i < pattern.Length; i++)  if (pattern[i] != modifiedString[i])  return false;  return true;  }  // Driver program  public static void Main(String[] args)  {  String str = 'engineers rock';  String pattern = 'er';  Console.WriteLine('Expected: true Actual: ' +  followsPattern(str pattern));  str = 'engineers rock';  pattern = 'egr';  Console.WriteLine('Expected: false Actual: ' +  followsPattern(str pattern));  str = 'engineers rock';  pattern = 'gsr';  Console.WriteLine('Expected: false Actual: ' +  followsPattern(str pattern));  str = 'engineers rock';  pattern = 'eger';  Console.WriteLine('Expected: true Actual: ' +  followsPattern(str pattern));  } } // This code is contributed by 29AjayKumar
JavaScript 
<script> // Javascript program to check if characters of a string follow // pattern defined by given pattern. function followsPattern(str pattern) {  // Insert all characters of pattern in a hash set  let patternSet = new Set();  for (let i=0; i<pattern.length; i++)  patternSet.add(pattern[i]);    // Build modified string (string with characters only from  // pattern are taken)  let modifiedString = (str).split('');  for (let i=str.length-1; i>=0; i--)  if (!patternSet.has(modifiedString[i]))  modifiedString.splice(i1);    // Remove more than one consecutive occurrences of pattern  // characters from modified string.  for (let i=modifiedString.length-1; i>0; i--)  if (modifiedString[i] == modifiedString[i-1])  modifiedString.splice(i1);    // After above modifications the length of modified string  // must be same as pattern length  if (pattern.length != modifiedString.length)  return false;    // And pattern characters must also be same as modified string  // characters  for (let i=0; i<pattern.length; i++)  if (pattern[i] != modifiedString[i])  return false;    return true; } // Driver program let str = 'engineers rock'; let pattern = 'er'; document.write('Expected: true Actual: ' +  followsPattern(str pattern)+'  
'); str = 'engineers rock'; pattern = 'egr'; document.write('Expected: false Actual: ' +  followsPattern(str pattern)+'  
'); str = 'engineers rock'; pattern = 'gsr'; document.write('Expected: false Actual: ' +  followsPattern(str pattern)+'  
'); str = 'engineers rock'; pattern = 'eger'; document.write('Expected: true Actual: ' +  followsPattern(str pattern)+'  
'); // This code is contributed by rag2127 </script>

Sortida:  
 



Expected: true Actual: true Expected: false Actual: false Expected: false Actual: false Expected: true Actual: true


Complexitat temporal: La complexitat temporal de les implementacions anteriors és en realitat O(mn + n^2) ja que fem servir deleteCharAt() per eliminar caràcters. Podem optimitzar la solució anterior per treballar en temps lineal. En lloc d'utilitzar deleteCharAr() podem crear una cadena buida i afegir-hi només els caràcters necessaris.
StringBuilder s'utilitza per operar amb una cadena d'entrada. Això es deu al fet que StringBuilder és mutable mentre que String és un objecte immutable. Per crear una cadena nova es necessita espai O(n), de manera que l'espai addicional és O(n).
Hem comentat dos enfocaments més per resoldre aquest problema. 
Comproveu si la cadena segueix l'ordre dels caràcters definit per un patró o no | Set 1  
Comproveu si la cadena segueix l'ordre dels caràcters definit per un patró o no | Set 3
 

convertir la cadena en data