Hitta alla strängar som matchar specifika mönster i en ordbok
Prova det på GfG Practice 
#practiceLinkDiv { display: ingen !viktigt; }
#practiceLinkDiv { display: ingen !viktigt; } Med en ordbok med ord, hitta alla strängar som matchar det givna mönstret där varje tecken i mönstret är unikt mappat till ett tecken i ordboken.
Exempel:
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 Matcha specifikt mönster Prova!
Metod 1:
Närma sig: Syftet är att ta reda på om ordet har samma struktur som mönstret. Ett förhållningssätt till detta problem kan vara att göra en hash av ordet och mönstret och jämföra om de är lika eller inte. På ett enkelt språk tilldelar vi olika heltal till de distinkta tecknen i ordet och gör en sträng av heltal (ordets hash) beroende på förekomsten av ett visst tecken i det ordet och jämför sedan det med mönstrets hash.
Exempel:
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
Algoritm:
- Koda mönstret enligt ovanstående tillvägagångssätt och lagra motsvarande hash för mönstret i en strängvariabel hasch .
- Initiera en räknare i=0 som kommer att mappa distinkt karaktär med distinkta heltal.
- Läs strängen och om det aktuella tecknet inte är mappat till ett heltal mappa det till räknarvärdet och öka det.
- Sammanfoga det heltal som är mappat till det aktuella tecknet till hashsträng .
- Läs nu varje ord och gör en hash av det med samma algoritm.
- Om hashen för det aktuella ordet är lika med hashen för mönstret så ingår det ordet i det slutliga svaret.
- Skapa en teckenuppsättning där vi kan mappa tecknen i mönster med motsvarande tecken i ett ord.
- Kontrollera först om längden på ord och mönster är lika eller inte om inga kontrollera sedan nästa ord.
- Om längden är lika, gå igenom mönstret och om det aktuella tecknet i mönstret inte har mappats ännu mappa det till motsvarande tecken i ordet.
- Om det aktuella tecknet är mappat kontrollerar du om tecknet som det har mappats med är lika med det aktuella tecknet i ordet.
- Om inga då följer inte ordet det givna mönstret.
- Om ordet följer mönstret tills sista tecknet skriv ut ordet.
Pseudokod:
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
Komplexitetsanalys:
Här är 'N' antalet ord och 'K' är dess längd. Eftersom vi måste gå igenom varje ord separat för att skapa dess hash.
Användningen av hash_map datastruktur för kartläggning av tecken tar denna mängd utrymme.
Metod 2:
Närma sig: Låt oss nu diskutera ett lite mer konceptuellt tillvägagångssätt som är en ännu bättre tillämpning av kartor. Istället för att göra en hash för varje ord kan vi mappa bokstäverna i själva mönstret med motsvarande bokstav i ordet. Om det aktuella tecknet inte har mappats mappa det till motsvarande tecken i ordet och om det redan har mappats, kontrollera om värdet med vilket det mappades tidigare är detsamma som ordets nuvarande värde eller inte. Exemplet nedan kommer att göra saker lätta att förstå.
Exempel:
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 Algoritm:
Pseudokod:
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
Komplexitetsanalys:
För att korsa varje ord kommer detta att vara tidskravet.
Användningen av hash_map datastruktur för kartläggning av tecken förbrukar N utrymme.
Du Kanske Gillar
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