Find alle strenge, der matcher et bestemt mønster i en ordbog
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#practiceLinkDiv { display: ingen !important; }
#practiceLinkDiv { display: ingen !important; } Givet en ordbog med ord, find alle strenge, der matcher det givne mønster, hvor hvert tegn i mønsteret er unikt knyttet til et tegn i ordbogen.
Eksempler:
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 Match specifikt mønster Prøv det!
Metode 1:
Nærme sig: Målet er at finde ud af, om ordet har samme struktur som mønsteret. En tilgang til dette problem kan være at lave en hash af ordet og mønsteret og sammenligne, om de er ens eller ej. I simpelt sprog tildeler vi forskellige heltal til ordets distinkte tegn og laver en streng af heltal (ordets hash) i henhold til forekomsten af et bestemt tegn i det ord, og sammenlign det derefter med mønsterets hash.
Eksempel:
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
Algoritme:
- Indkode mønsteret i overensstemmelse med ovenstående tilgang og gem den tilsvarende hash af mønsteret i en strengvariabel hash .
- Initialiser en tæller i=0 som vil kortlægge distinkt karakter med distinkte heltal.
- Læs strengen, og hvis det aktuelle tegn ikke er afbildet til et heltal, kortlæg det til tællerværdien og forøg det.
- Sammenkæd det heltal, der er knyttet til det aktuelle tegn, til hash streng .
- Læs nu hvert ord og lav en hash af det ved hjælp af den samme algoritme.
- Hvis hashen af det aktuelle ord er lig med hashen af mønsteret, er det ord inkluderet i det endelige svar.
- Opret et tegnarray, hvor vi kan kortlægge tegnene i mønstre med en tilsvarende karakter af et ord.
- Tjek først om længden af ord og mønster er ens eller ej hvis ingen tjek derefter det næste ord.
- Hvis længden er ens, skal du krydse mønsteret, og hvis det aktuelle tegn i mønsteret ikke er blevet afbildet, skal du tilknytte det til det tilsvarende tegn i ordet.
- Hvis det aktuelle tegn er afbildet, skal du kontrollere, om tegnet, som det er blevet afbildet med, er lig med det aktuelle tegn i ordet.
- Hvis ingen så følger ordet ikke det givne mønster.
- Hvis ordet følger mønsteret indtil det sidste tegn, så udskriv ordet.
Pseudokode:
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>
Produktion
xyy abb
Kompleksitetsanalyse:
Her er 'N' antallet af ord og 'K' er dets længde. Da vi er nødt til at krydse hvert ord separat for at skabe dets hash.
Brugen af hash_map datastruktur til kortlægning af tegn tager denne mængde plads.
Metode 2:
Nærme sig: Lad os nu diskutere en lidt mere konceptuel tilgang, som er en endnu bedre anvendelse af kort. I stedet for at lave en hash for hvert ord kan vi kortlægge bogstaverne i selve mønsteret med det tilsvarende bogstav i ordet. Hvis det aktuelle tegn ikke er blevet tilknyttet, skal det tilknyttes det tilsvarende tegn i ordet, og hvis det allerede er blevet afbildet, skal du kontrollere, om værdien, som det tidligere blev kortlagt med, er den samme som ordets aktuelle værdi eller ej. Eksemplet nedenfor vil gøre tingene nemme at forstå.
Eksempel:
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 Algoritme:
Pseudokode:
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>
Produktion
xyy abb
Kompleksitetsanalyse:
For at krydse hvert ord vil dette være tidskravet.
Brugen af hash_map datastruktur til kortlægning af tegn bruger N plads.
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