Subseqüència comuna més llarga amb permutacions permeses
Donades dues cadenes en minúscules, trobeu la cadena més llarga les permutacions de la qual són subseqüències de dues cadenes donades. La cadena de sortida més llarga s'ha d'ordenar.
Exemples:
Input : str1 = 'pink' str2 = 'kite' Output : 'ik' The string 'ik' is the longest sorted string whose one permutation 'ik' is subsequence of 'pink' and another permutation 'ki' is subsequence of 'kite'. Input : str1 = 'working' str2 = 'women' Output : 'now' Input : str1 = 'geeks' str2 = 'cake' Output : 'ek' Input : str1 = 'aaaa' str2 = 'baba' Output : 'aa'Recomanat: si us plau, resol-ho a ' PRÀCTICA primer abans de passar a la solució.
La idea és comptar caràcters en ambdues cadenes.
- calculeu la freqüència de caràcters per a cada cadena i emmagatzemeu-los a les seves respectives matrius de recompte, per exemple count1[] per a str1 i count2[] per a str2.
- Ara tenim matrius de recompte de 26 caràcters. Per tant, travessa count1[] i per a qualsevol índex 'i' afegiu caràcter ('a'+i) a la cadena resultant 'resultat' per min(count1[i] count2[i]) vegades.
- Com que travessem la matriu de recompte en ordre ascendent, els nostres caràcters de cadena finals estaran ordenats.
Implementació:
C++ // C++ program to find LCS with permutations allowed #include using namespace std ; // Function to calculate longest string // str1 --> first string // str2 --> second string // count1[] --> hash array to calculate frequency // of characters in str1 // count[2] --> hash array to calculate frequency // of characters in str2 // result --> resultant longest string whose // permutations are sub-sequence of given two strings void longestString ( string str1 string str2 ) { int count1 [ 26 ] = { 0 } count2 [ 26 ] = { 0 }; // calculate frequency of characters for ( int i = 0 ; i < str1 . length (); i ++ ) count1 [ str1 [ i ] - 'a' ] ++ ; for ( int i = 0 ; i < str2 . length (); i ++ ) count2 [ str2 [ i ] - 'a' ] ++ ; // Now traverse hash array string result ; for ( int i = 0 ; i < 26 ; i ++ ) // append character ('a'+i) in resultant // string 'result' by min(count1[i]count2i]) // times for ( int j = 1 ; j <= min ( count1 [ i ] count2 [ i ]); j ++ ) result . push_back ( 'a' + i ); cout < < result ; } // Driver program to run the case int main () { string str1 = 'geeks' str2 = 'cake' ; longestString ( str1 str2 ); return 0 ; }
Java //Java program to find LCS with permutations allowed class GFG { // Function to calculate longest String // str1 --> first String // str2 --> second String // count1[] --> hash array to calculate frequency // of characters in str1 // count[2] --> hash array to calculate frequency // of characters in str2 // result --> resultant longest String whose // permutations are sub-sequence of given two strings static void longestString ( String str1 String str2 ) { int count1 [] = new int [ 26 ] count2 [] = new int [ 26 ] ; // calculate frequency of characters for ( int i = 0 ; i < str1 . length (); i ++ ) { count1 [ str1 . charAt ( i ) - 'a' ]++ ; } for ( int i = 0 ; i < str2 . length (); i ++ ) { count2 [ str2 . charAt ( i ) - 'a' ]++ ; } // Now traverse hash array String result = '' ; for ( int i = 0 ; i < 26 ; i ++ ) // append character ('a'+i) in resultant // String 'result' by min(count1[i]count2i]) // times { for ( int j = 1 ; j <= Math . min ( count1 [ i ] count2 [ i ] ); j ++ ) { result += ( char )( 'a' + i ); } } System . out . println ( result ); } // Driver program to run the case public static void main ( String [] args ) { String str1 = 'geeks' str2 = 'cake' ; longestString ( str1 str2 ); } } /* This java code is contributed by 29AjayKumar*/
Python3 # Python 3 program to find LCS # with permutations allowed # Function to calculate longest string # str1 --> first string # str2 --> second string # count1[] --> hash array to calculate frequency # of characters in str1 # count[2] --> hash array to calculate frequency # of characters in str2 # result --> resultant longest string whose # permutations are sub-sequence # of given two strings def longestString ( str1 str2 ): count1 = [ 0 ] * 26 count2 = [ 0 ] * 26 # calculate frequency of characters for i in range ( len ( str1 )): count1 [ ord ( str1 [ i ]) - ord ( 'a' )] += 1 for i in range ( len ( str2 )): count2 [ ord ( str2 [ i ]) - ord ( 'a' )] += 1 # Now traverse hash array result = '' for i in range ( 26 ): # append character ('a'+i) in # resultant string 'result' by # min(count1[i]count2i]) times for j in range ( 1 min ( count1 [ i ] count2 [ i ]) + 1 ): result = result + chr ( ord ( 'a' ) + i ) print ( result ) # Driver Code if __name__ == '__main__' : str1 = 'geeks' str2 = 'cake' longestString ( str1 str2 ) # This code is contributed by ita_c
C# // C# program to find LCS with // permutations allowed using System ; class GFG { // Function to calculate longest String // str1 --> first String // str2 --> second String // count1[] --> hash array to calculate // frequency of characters in str1 // count[2] --> hash array to calculate // frequency of characters in str2 // result --> resultant longest String whose // permutations are sub-sequence of // given two strings static void longestString ( String str1 String str2 ) { int [] count1 = new int [ 26 ]; int [] count2 = new int [ 26 ]; // calculate frequency of characters for ( int i = 0 ; i < str1 . Length ; i ++ ) { count1 [ str1 [ i ] - 'a' ] ++ ; } for ( int i = 0 ; i < str2 . Length ; i ++ ) { count2 [ str2 [ i ] - 'a' ] ++ ; } // Now traverse hash array String result = '' ; for ( int i = 0 ; i < 26 ; i ++ ) // append character ('a'+i) in resultant // String 'result' by min(count1[i]count2i]) // times { for ( int j = 1 ; j <= Math . Min ( count1 [ i ] count2 [ i ]); j ++ ) { result += ( char )( 'a' + i ); } } Console . Write ( result ); } // Driver Code public static void Main () { String str1 = 'geeks' str2 = 'cake' ; longestString ( str1 str2 ); } } // This code is contributed // by PrinciRaj1992
PHP // PHP program to find LCS with // permutations allowed // Function to calculate longest string // str1 --> first string // str2 --> second string // count1[] --> hash array to calculate frequency // of characters in str1 // count[2] --> hash array to calculate frequency // of characters in str2 // result --> resultant longest string whose // permutations are sub-sequence of given two strings function longestString ( $str1 $str2 ) { $count1 = array_fill ( 0 26 NULL ); $count2 = array_fill ( 0 26 NULL ); // calculate frequency of characters for ( $i = 0 ; $i < strlen ( $str1 ); $i ++ ) $count1 [ ord ( $str1 [ $i ]) - ord ( 'a' )] ++ ; for ( $i = 0 ; $i < strlen ( $str2 ); $i ++ ) $count2 [ ord ( $str2 [ $i ]) - ord ( 'a' )] ++ ; // Now traverse hash array $result = '' ; for ( $i = 0 ; $i < 26 ; $i ++ ) // append character ('a'+i) in resultant // string 'result' by min(count1[$i] // count2[$i]) times for ( $j = 1 ; $j <= min ( $count1 [ $i ] $count2 [ $i ]); $j ++ ) $result = $result . chr ( ord ( 'a' ) + $i ); echo $result ; } // Driver Code $str1 = 'geeks' ; $str2 = 'cake' ; longestString ( $str1 $str2 ); // This code is contributed by ita_c ?>
JavaScript < script > // Javascript program to find LCS with permutations allowed function min ( a b ) { if ( a < b ) return a ; else return b ; } // Function to calculate longest String // str1 --> first String // str2 --> second String // count1[] --> hash array to calculate frequency // of characters in str1 // count[2] --> hash array to calculate frequency // of characters in str2 // result --> resultant longest String whose // permutations are sub-sequence of given two strings function longestString ( str1 str2 ) { var count1 = new Array ( 26 ); var count2 = new Array ( 26 ); count1 . fill ( 0 ); count2 . fill ( 0 ); // calculate frequency of characters for ( var i = 0 ; i < str1 . length ; i ++ ) { count1 [ str1 . charCodeAt ( i ) - 97 ] ++ ; } for ( var i = 0 ; i < str2 . length ; i ++ ) { count2 [ str2 . charCodeAt ( i ) - 97 ] ++ ; } // Now traverse hash array var result = '' ; for ( var i = 0 ; i < 26 ; i ++ ) // append character ('a'+i) in resultant // String 'result' by min(count1[i]count2i]) // times { for ( var j = 1 ; j <= min ( count1 [ i ] count2 [ i ]); j ++ ) { result += String . fromCharCode ( 97 + i ); } } document . write ( result ); } var str1 = 'geeks' ; var str2 = 'cake' ; longestString ( str1 str2 ); // This code is contributed by akshitsaxenaa09. < /script>
Sortida
ek
Complexitat temporal: O(m + n) on m i n són longituds de cadenes d'entrada.
Espai auxiliar: O(1)
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