순열이 허용되는 가장 긴 공통 부분 수열
소문자로 된 두 문자열이 주어지면 순열이 주어진 두 문자열의 하위 시퀀스인 가장 긴 문자열을 찾습니다. 출력되는 가장 긴 문자열을 정렬해야 합니다.
예:
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'추천 : '에서 해결해주세요 관행 ' 먼저 솔루션으로 넘어가기 전에.
아이디어는 두 문자열 모두에서 문자 수를 계산하는 것입니다.
- 각 문자열의 문자 빈도를 계산하고 이를 각각의 카운트 배열에 저장합니다(예: str1의 경우 count1[], str2의 경우 count2[]).
- 이제 26자에 대한 개수 배열이 생겼습니다. 따라서 count1[]을 순회하고 모든 인덱스 'i'에 대해 결과 문자열 'result'에 문자('a'+i)를 min(count1[i] count2[i])번만큼 추가합니다.
- count 배열을 오름차순으로 순회하므로 최종 문자열 문자는 정렬된 순서로 표시됩니다.
구현:
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>
산출
ek
시간 복잡도: O(m + n) 여기서 m과 n은 입력 문자열의 길이입니다.
보조 공간: O(1)
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