Raskite visas eilutes, atitinkančias konkretų šabloną žodyne
Išbandykite GfG praktikoje 
#practiceLinkDiv { display: none !important; }
#practiceLinkDiv { display: none !important; } Žodžių žodyne raskite visas eilutes, atitinkančias nurodytą šabloną, kur kiekvienas rašto simbolis yra unikaliai susietas su žodyno simboliu.
Pavyzdžiai:
Input: dict = ['abb' 'abc' 'xyz' 'xyy']; pattern = 'foo' Output: [xyy abb] xyy and abb have same character at index 1 and 2 like the pattern Input: dict = ['abb' 'abc' 'xyz' 'xyy']; pat = 'mno' Output: [abc xyz] abc and xyz have all distinct characters similar to the pattern. Input: dict = ['abb' 'abc' 'xyz' 'xyy']; pattern = 'aba' Output: [] Pattern has same character at index 0 and 2. No word in dictionary follows the pattern. Input: dict = ['abab' 'aba' 'xyz' 'xyx']; pattern = 'aba' Output: [aba xyx] aba and xyx have same character at index 0 and 2 like the patternRecommended Practice Derinkite konkretų modelį Išbandykite!
1 būdas:
Prieiga: Tikslas yra išsiaiškinti, ar žodžio struktūra tokia pati kaip šablono. Šios problemos sprendimas gali būti žodžio ir šablono maišos sudarymas ir palyginimas, ar jie yra vienodi, ar ne. Paprasta kalba atskiriems žodžio simboliams priskiriame skirtingus sveikuosius skaičius ir sudarome sveikųjų skaičių eilutę (žodžio maišelis) pagal konkretaus simbolio atsiradimą tame žodyje ir palyginkite jį su šablono maiša.
Pavyzdys:
Word='xxyzzaabcdd' Pattern='mmnoopplfmm' For word-: map['x']=1; map['y']=2; map['z']=3; map['a']=4; map['b']=5; map['c']=6; map['d']=7; Hash for Word='11233445677' For Pattern-: map['m']=1; map['n']=2; map['o']=3; map['p']=4; map['l']=5; map['f']=6; Hash for Pattern='11233445611' Therefore in the given example Hash of word is not equal to Hash of pattern so this word is not included in the answer
Algoritmas:
- Užkoduokite šabloną pagal aukščiau pateiktą metodą ir išsaugokite atitinkamą šablono maišą eilutės kintamajame maišos .
- Inicijuoti skaitiklį i=0 kuri atvaizduos atskirą simbolį su skirtingais sveikaisiais skaičiais.
- Perskaitykite eilutę ir, jei dabartinis simbolis nesusietas su sveikuoju skaičiumi, susiekite jį su skaitiklio reikšme ir padidinkite ją.
- Sujunkite sveikąjį skaičių, susietą su dabartiniu simboliu, su maišos eilutė .
- Dabar perskaitykite kiekvieną žodį ir sudarykite maišą naudodami tą patį algoritmą.
- Jei dabartinio žodžio maiša yra lygi šablono maišai, šis žodis įtraukiamas į galutinį atsakymą.
- Sukurkite simbolių masyvą, kuriame galime susieti šablonų simbolius su atitinkamu žodžio simboliu.
- Pirmiausia patikrinkite, ar žodžio ir rašto ilgis yra vienodas, ar ne ne tada patikrinkite kitą žodį.
- Jei ilgis yra lygus, perkelkite šabloną ir jei dabartinis rašto simbolis dar nesusietas su atitinkamu žodžio simboliu.
- Jei dabartinis simbolis susietas, patikrinkite, ar simbolis, su kuriuo jis buvo susietas, yra lygus dabartiniam žodžio simboliui.
- Jeigu ne tada žodis neatitinka pateikto modelio.
- Jei žodis seka šabloną iki paskutinio simbolio, atspausdinkite žodį.
Pseudo kodas:
int i=0 Declare map for character in pattern: if(map[character]==map.end()) map[character]=i++; hash_pattern+=to_string(mp[character]) for words in dictionary: i=0; Declare map if(words.length==pattern.length) for character in words: if(map[character]==map.end()) map[character]=i++ hash_word+=to_string(map[character) if(hash_word==hash_pattern) print wordsC++
// C++ program to print all // the strings that match the // given pattern where every // character in the pattern is // uniquely mapped to a character // in the dictionary #include using namespace std ; // Function to encode given string string encodeString ( string str ) { unordered_map < char int > map ; string res = '' ; int i = 0 ; // for each character in given string for ( char ch : str ) { // If the character is occurring // for the first time assign next // unique number to that char if ( map . find ( ch ) == map . end ()) map [ ch ] = i ++ ; // append the number associated // with current character into the // output string res += to_string ( map [ ch ]); } return res ; } // Function to print all the // strings that match the // given pattern where every // character in the pattern is // uniquely mapped to a character // in the dictionary void findMatchedWords ( unordered_set < string > dict string pattern ) { // len is length of the pattern int len = pattern . length (); // Encode the string string hash = encodeString ( pattern ); // for each word in the dictionary for ( string word : dict ) { // If size of pattern is same as // size of current dictionary word // and both pattern and the word // has same hash print the word if ( word . length () == len && encodeString ( word ) == hash ) cout < < word < < ' ' ; } } // Driver code int main () { unordered_set < string > dict = { 'abb' 'abc' 'xyz' 'xyy' }; string pattern = 'foo' ; findMatchedWords ( dict pattern ); return 0 ; }
Java // Java program to print all the // strings that match the // given pattern where every // character in the pattern is // uniquely mapped to a character // in the dictionary import java.io.* ; import java.util.* ; class GFG { // Function to encode given string static String encodeString ( String str ) { HashMap < Character Integer > map = new HashMap <> (); String res = '' ; int i = 0 ; // for each character in given string char ch ; for ( int j = 0 ; j < str . length (); j ++ ) { ch = str . charAt ( j ); // If the character is occurring for the first // time assign next unique number to that char if ( ! map . containsKey ( ch )) map . put ( ch i ++ ); // append the number associated with current // character into the output string res += map . get ( ch ); } return res ; } // Function to print all // the strings that match the // given pattern where every // character in the pattern is // uniquely mapped to a character // in the dictionary static void findMatchedWords ( String [] dict String pattern ) { // len is length of the pattern int len = pattern . length (); // encode the string String hash = encodeString ( pattern ); // for each word in the dictionary array for ( String word : dict ) { // If size of pattern is same // as size of current // dictionary word and both // pattern and the word // has same hash print the word if ( word . length () == len && encodeString ( word ). equals ( hash )) System . out . print ( word + ' ' ); } } // Driver code public static void main ( String args [] ) { String [] dict = { 'abb' 'abc' 'xyz' 'xyy' }; String pattern = 'foo' ; findMatchedWords ( dict pattern ); } // This code is contributed // by rachana soma }
Python3 # Python3 program to print all the # strings that match the # given pattern where every # character in the pattern is # uniquely mapped to a character # in the dictionary # Function to encode # given string def encodeString ( Str ): map = {} res = '' i = 0 # For each character # in given string for ch in Str : # If the character is occurring # for the first time assign next # unique number to that char if ch not in map : map [ ch ] = i i += 1 # Append the number associated # with current character into # the output string res += str ( map [ ch ]) return res # Function to print all # the strings that match the # given pattern where every # character in the pattern is # uniquely mapped to a character # in the dictionary def findMatchedWords ( dict pattern ): # len is length of the # pattern Len = len ( pattern ) # Encode the string hash = encodeString ( pattern ) # For each word in the # dictionary array for word in dict : # If size of pattern is same # as size of current # dictionary word and both # pattern and the word # has same hash print the word if ( len ( word ) == Len and encodeString ( word ) == hash ): print ( word end = ' ' ) # Driver code dict = [ 'abb' 'abc' 'xyz' 'xyy' ] pattern = 'foo' findMatchedWords ( dict pattern ) # This code is contributed by avanitrachhadiya2155
C# // C# program to print all the strings // that match the given pattern where // every character in the pattern is // uniquely mapped to a character in the dictionary using System ; using System.Collections.Generic ; public class GFG { // Function to encode given string static String encodeString ( String str ) { Dictionary < char int > map = new Dictionary < char int > (); String res = '' ; int i = 0 ; // for each character in given string char ch ; for ( int j = 0 ; j < str . Length ; j ++ ) { ch = str [ j ]; // If the character is occurring for the first // time assign next unique number to that char if ( ! map . ContainsKey ( ch )) map . Add ( ch i ++ ); // append the number associated with current // character into the output string res += map [ ch ]; } return res ; } // Function to print all the // strings that match the // given pattern where every // character in the pattern is // uniquely mapped to a character // in the dictionary static void findMatchedWords ( String [] dict String pattern ) { // len is length of the pattern int len = pattern . Length ; // encode the string String hash = encodeString ( pattern ); // for each word in the dictionary array foreach ( String word in dict ) { // If size of pattern is same as // size of current dictionary word // and both pattern and the word // has same hash print the word if ( word . Length == len && encodeString ( word ). Equals ( hash )) Console . Write ( word + ' ' ); } } // Driver code public static void Main ( String [] args ) { String [] dict = { 'abb' 'abc' 'xyz' 'xyy' }; String pattern = 'foo' ; findMatchedWords ( dict pattern ); } } // This code is contributed by 29AjayKumar
JavaScript < script > // Javascript program to print all the // strings that match the // given pattern where every // character in the pattern is // uniquely mapped to a character // in the dictionary // Function to encode given string function encodeString ( str ) { let map = new Map (); let res = '' ; let i = 0 ; // for each character in given string let ch ; for ( let j = 0 ; j < str . length ; j ++ ) { ch = str [ j ]; // If the character is occurring for the first // time assign next unique number to that char if ( ! map . has ( ch )) map . set ( ch i ++ ); // append the number associated with current // character into the output string res += map . get ( ch ); } return res ; } // Function to print all // the strings that match the // given pattern where every // character in the pattern is // uniquely mapped to a character // in the dictionary function findMatchedWords ( dict pattern ) { // len is length of the pattern let len = pattern . length ; // encode the string let hash = encodeString ( pattern ); // for each word in the dictionary array for ( let word = 0 ; word < dict . length ; word ++ ) { // If size of pattern is same // as size of current // dictionary word and both // pattern and the word // has same hash print the word if ( dict [ word ]. length == len && encodeString ( dict [ word ]) == ( hash )) document . write ( dict [ word ] + ' ' ); } } // Driver code let dict = [ 'abb' 'abc' 'xyz' 'xyy' ]; let pattern = 'foo' ; findMatchedWords ( dict pattern ); // This code is contributed by unknown2108 < /script>
Išvestis
xyy abb
Sudėtingumo analizė:
Čia „N“ yra žodžių skaičius, o „K“ yra jo ilgis. Kadangi turime perbraukti kiekvieną žodį atskirai, kad sukurtume maišą.
Naudojimas hash_map duomenų struktūra simbolių atvaizdavimui užima tiek vietos.
2 būdas:
Prieiga: Dabar aptarkime šiek tiek konceptualesnį požiūrį, kuris yra dar geresnis žemėlapių pritaikymas. Užuot darę maišą kiekvienam žodžiui, galime susieti paties modelio raides su atitinkama žodžio raide. Jei dabartinis simbolis nebuvo susietas, susiekite jį su atitinkamu žodžio simboliu ir, jei jis jau buvo susietas, patikrinkite, ar reikšmė, su kuria jis buvo susietas anksčiau, yra tokia pati kaip dabartinė žodžio reikšmė, ar ne. Toliau pateiktame pavyzdyje viskas bus lengviau suprasti.
Pavyzdys:
Word='xxyzzaa' Pattern='mmnoopp' Step 1-: map['m'] = x Step 2-: 'm' is already mapped to some value check whether that value is equal to current character of word-:YES ('m' is mapped to x). Step 3-: map['n'] = y Step 4-: map['o'] = z Step 5-: 'o' is already mapped to some value check whether that value is equal to current character of word-:YES ('o' is mapped to z). Step 6-: map['p'] = a Step 7-: 'p' is already mapped to some value check whether that value is equal to current character of word-: YES ('p' is mapped to a). No contradiction so current word matches the pattern Algoritmas:
Pseudo kodas:
for words in dictionary: char arr_map[128]=0 char map_word[128]=0 if(words.length==pattern.length) for 0 to length of pattern: if(arr_map[character in pattern]==0 && map_word[character in word]==0) arr_map[character in pattern]=word[character in word] map_word[character in word]=pattern[character in pattern] else if(arr_map[character]!=word[character] ||map_word[character]!=pattern[character] ) break the loop If above loop runs successfully Print(words)C++
// C++ program to print all // the strings that match the // given pattern where every // character in the pattern is // uniquely mapped to a character // in the dictionary #include using namespace std ; bool check ( string pattern string word ) { if ( pattern . length () != word . length ()) return false ; char ch [ 128 ] = { 0 }; char map_word [ 128 ] = { 0 }; int len = word . length (); for ( int i = 0 ; i < len ; i ++ ) { if ( ch [ pattern [ i ]] == 0 && map_word [ word [ i ] ] == 0 ) { ch [ pattern [ i ]] = word [ i ]; map_word [ word [ i ] ] = pattern [ i ]; } else if ( ch [ pattern [ i ]] != word [ i ] || map_word [ word [ i ] ] != pattern [ i ]) return false ; } return true ; } // Function to print all the // strings that match the // given pattern where every // character in the pattern is // uniquely mapped to a character // in the dictionary void findMatchedWords ( unordered_set < string > dict string pattern ) { // len is length of the pattern int len = pattern . length (); // for each word in the dictionary for ( string word : dict ) { if ( check ( pattern word )) cout < < word < < ' ' ; } } // Driver code int main () { unordered_set < string > dict = { 'abb' 'abc' 'xyz' 'xyy' 'bbb' }; string pattern = 'foo' ; findMatchedWords ( dict pattern ); return 0 ; } // This code is contributed by Ankur Goel And Priobrata Malik
Java // Java program to print all // the strings that match the // given pattern where every // character in the pattern is // uniquely mapped to a character // in the dictionary import java.util.* ; class GFG { static boolean check ( String pattern String word ) { if ( pattern . length () != word . length ()) return false ; int [] ch = new int [ 128 ] ; int Len = word . length (); for ( int i = 0 ; i < Len ; i ++ ) { if ( ch [ ( int ) pattern . charAt ( i ) ] == 0 ) { ch [ ( int ) pattern . charAt ( i ) ] = word . charAt ( i ); } else if ( ch [ ( int ) pattern . charAt ( i ) ] != word . charAt ( i )) { return false ; } } return true ; } // Function to print all the // strings that match the // given pattern where every // character in the pattern is // uniquely mapped to a character // in the dictionary static void findMatchedWords ( HashSet < String > dict String pattern ) { // len is length of the pattern int Len = pattern . length (); // For each word in the dictionary String result = ' ' ; for ( String word : dict ) { if ( check ( pattern word )) { result = word + ' ' + result ; } } System . out . print ( result ); } // Driver code public static void main ( String [] args ) { HashSet < String > dict = new HashSet < String > (); dict . add ( 'abb' ); dict . add ( 'abc' ); dict . add ( 'xyz' ); dict . add ( 'xyy' ); String pattern = 'foo' ; findMatchedWords ( dict pattern ); } } // This code is contributed by divyeshrabadiya07
Python3 # Python3 program to print all # the strings that match the # given pattern where every # character in the pattern is # uniquely mapped to a character # in the dictionary def check ( pattern word ): if ( len ( pattern ) != len ( word )): return False ch = [ 0 for i in range ( 128 )] Len = len ( word ) for i in range ( Len ): if ( ch [ ord ( pattern [ i ])] == 0 ): ch [ ord ( pattern [ i ])] = word [ i ] else if ( ch [ ord ( pattern [ i ])] != word [ i ]): return False return True # Function to print all the # strings that match the # given pattern where every # character in the pattern is # uniquely mapped to a character # in the dictionary def findMatchedWords ( Dict pattern ): # len is length of the pattern Len = len ( pattern ) # For each word in the dictionary for word in range ( len ( Dict ) - 1 - 1 - 1 ): if ( check ( pattern Dict [ word ])): print ( Dict [ word ] end = ' ' ) # Driver code Dict = [ 'abb' 'abc' 'xyz' 'xyy' ] pattern = 'foo' findMatchedWords ( Dict pattern ) # This code is contributed by rag2127
C# // C# program to print all // the strings that match the // given pattern where every // character in the pattern is // uniquely mapped to a character // in the dictionary using System ; using System.Collections ; using System.Collections.Generic ; class GFG { static bool check ( string pattern string word ) { if ( pattern . Length != word . Length ) return false ; int [] ch = new int [ 128 ]; int Len = word . Length ; for ( int i = 0 ; i < Len ; i ++ ) { if ( ch [( int ) pattern [ i ]] == 0 ) { ch [( int ) pattern [ i ]] = word [ i ]; } else if ( ch [( int ) pattern [ i ]] != word [ i ]) { return false ; } } return true ; } // Function to print all the // strings that match the // given pattern where every // character in the pattern is // uniquely mapped to a character // in the dictionary static void findMatchedWords ( HashSet < string > dict string pattern ) { // len is length of the pattern int Len = pattern . Length ; // For each word in the dictionary string result = ' ' ; foreach ( string word in dict ) { if ( check ( pattern word )) { result = word + ' ' + result ; } } Console . Write ( result ); } // Driver Code static void Main () { HashSet < string > dict = new HashSet < string > ( new string []{ 'abb' 'abc' 'xyz' 'xyy' }); string pattern = 'foo' ; findMatchedWords ( dict pattern ); } } // This code is contributed by divyesh072019
JavaScript < script > // Javascript program to print all // the strings that match the // given pattern where every // character in the pattern is // uniquely mapped to a character // in the dictionary function check ( pattern word ) { if ( pattern . length != word . length ) return false ; let ch = new Array ( 128 ); for ( let i = 0 ; i < 128 ; i ++ ) { ch [ i ] = 0 ; } let Len = word . length ; for ( let i = 0 ; i < Len ; i ++ ) { if ( ch [ pattern [ i ]. charCodeAt ( 0 )] == 0 ) { ch [ pattern [ i ]. charCodeAt ( 0 )] = word [ i ]; } else if ( ch [ pattern [ i ]. charCodeAt ( 0 )] != word [ i ]) { return false ; } } return true ; } // Function to print all the // strings that match the // given pattern where every // character in the pattern is // uniquely mapped to a character // in the dictionary function findMatchedWords ( dict pattern ) { // len is length of the pattern let Len = pattern . length ; // For each word in the dictionary let result = ' ' ; for ( let word of dict . values ()) { if ( check ( pattern word )) { result = word + ' ' + result ; } } document . write ( result ); } // Driver code let dict = new Set (); dict . add ( 'abb' ); dict . add ( 'abc' ); dict . add ( 'xyz' ); dict . add ( 'xyy' ); let pattern = 'foo' ; findMatchedWords ( dict pattern ); // This code is contributed by patel2127 < /script>
Išvestis
xyy abb
Sudėtingumo analizė:
Norint perskaityti kiekvieną žodį, reikės laiko.
Naudojimas hash_map simbolių atvaizdavimo duomenų struktūra užima N vietos.
Jums Gali Patikti
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