Zoek of de string K-Palindroom is of niet | Stel 2 in
Gegeven een string, zoek uit of de string K-Palindroom is of niet. Een K-palindroomreeks verandert in een palindroom als er maximaal k tekens uit worden verwijderd.
Voorbeelden:
Input : String - abcdecba k = 1 Output : Yes String can become palindrome by removing 1 character i.e. either d or e Input : String - abcdeca K = 2 Output : Yes Can become palindrome by removing 2 characters b and e (or b and d). Input : String - acdcb K = 1 Output : No String can not become palindrome by removing only one character.
We hebben een DP-oplossing besproken in vorig post waarin we zagen dat het probleem in feite een variatie is op Afstand bewerken probleem. In dit bericht wordt een andere interessante DP-oplossing besproken.
Het idee is om de langste palindroomdeelreeks van de gegeven reeks te vinden. Als het verschil tussen de langste palindroomdeelreeks en de originele string kleiner is dan gelijk aan k, dan is de string een k-palindroom, anders is het geen k-palindroom.
Bijvoorbeeld de langste palindrome deelreeks van een string abcdeca is accdca (of aceca). De karakters die niet bijdragen aan de langste palindroomreeks van de string moeten worden verwijderd om de string een palindroom te maken. Dus als u b en d (of e) uit de abcdeca-reeks verwijdert, verandert deze in een palindroom.
De langste palindroomdeelreeks van een string kan eenvoudig worden gevonden met behulp van LCS . Hieronder volgt de tweestapsoplossing voor het vinden van de langste palindrome deelreeks die LCS gebruikt.
- Keer de gegeven reeks om en sla het omgekeerde op in een andere array, bijvoorbeeld rev[0..n-1]
- LCS van de gegeven reeks en rev[] zal de langste palindroomreeks zijn.
Hieronder vindt u de implementatie van het bovenstaande idee -
// C++ program to find if given string is K-Palindrome // or not #include using namespace std ; /* Returns length of LCS for X[0..m-1] Y[0..n-1] */ int lcs ( string X string Y int m int n ) { int L [ m + 1 ][ n + 1 ]; /* Following steps build L[m+1][n+1] in bottom up fashion. Note that L[i][j] contains length of LCS of X[0..i-1] and Y[0..j-1] */ for ( int i = 0 ; i <= m ; i ++ ) { for ( int j = 0 ; j <= n ; j ++ ) { if ( i == 0 || j == 0 ) L [ i ][ j ] = 0 ; else if ( X [ i - 1 ] == Y [ j - 1 ]) L [ i ][ j ] = L [ i - 1 ][ j - 1 ] + 1 ; else L [ i ][ j ] = max ( L [ i - 1 ][ j ] L [ i ][ j - 1 ]); } } // L[m][n] contains length of LCS for X and Y return L [ m ][ n ]; } // find if given string is K-Palindrome or not bool isKPal ( string str int k ) { int n = str . length (); // Find reverse of string string revStr = str ; reverse ( revStr . begin () revStr . end ()); // find longest palindromic subsequence of // given string int lps = lcs ( str revStr n n ); // If the difference between longest palindromic // subsequence and the original string is less // than equal to k then the string is k-palindrome return ( n - lps <= k ); } // Driver program int main () { string str = 'abcdeca' ; int k = 2 ; isKPal ( str k ) ? cout < < 'Yes' : cout < < 'No' ; return 0 ; }
Java // Java program to find if given // String is K-Palindrome or not import java.util.* ; import java.io.* ; class GFG { /* Returns length of LCS for X[0..m-1] Y[0..n-1] */ static int lcs ( String X String Y int m int n ) { int L [][] = new int [ m + 1 ][ n + 1 ] ; /* Following steps build L[m+1][n+1] in bottom up fashion. Note that L[i][j] contains length of LCS of X[0..i-1] and Y[0..j-1] */ for ( int i = 0 ; i <= m ; i ++ ) { for ( int j = 0 ; j <= n ; j ++ ) { if ( i == 0 || j == 0 ) { L [ i ][ j ] = 0 ; } else if ( X . charAt ( i - 1 ) == Y . charAt ( j - 1 )) { L [ i ][ j ] = L [ i - 1 ][ j - 1 ] + 1 ; } else { L [ i ][ j ] = Math . max ( L [ i - 1 ][ j ] L [ i ][ j - 1 ] ); } } } // L[m][n] contains length // of LCS for X and Y return L [ m ][ n ] ; } // find if given String is // K-Palindrome or not static boolean isKPal ( String str int k ) { int n = str . length (); // Find reverse of String StringBuilder revStr = new StringBuilder ( str ); revStr = revStr . reverse (); // find longest palindromic // subsequence of given String int lps = lcs ( str revStr . toString () n n ); // If the difference between longest // palindromic subsequence and the // original String is less than equal // to k then the String is k-palindrome return ( n - lps <= k ); } // Driver code public static void main ( String [] args ) { String str = 'abcdeca' ; int k = 2 ; if ( isKPal ( str k )) { System . out . println ( 'Yes' ); } else System . out . println ( 'No' ); } } // This code is contributed by Rajput-JI
Python3 # Python program to find # if given string is K-Palindrome # or not # Returns length of LCS # for X[0..m-1] Y[0..n-1] def lcs ( X Y m n ): L = [[ 0 ] * ( n + 1 ) for _ in range ( m + 1 )] # Following steps build # L[m+1][n+1] in bottom up # fashion. Note that L[i][j] # contains length of # LCS of X[0..i-1] and Y[0..j-1] for i in range ( m + 1 ): for j in range ( n + 1 ): if not i or not j : L [ i ][ j ] = 0 elif X [ i - 1 ] == Y [ j - 1 ]: L [ i ][ j ] = L [ i - 1 ][ j - 1 ] + 1 else : L [ i ][ j ] = max ( L [ i - 1 ][ j ] L [ i ][ j - 1 ]) # L[m][n] contains length # of LCS for X and Y return L [ m ][ n ] # find if given string is # K-Palindrome or not def isKPal ( string k ): n = len ( string ) # Find reverse of string revStr = string [:: - 1 ] # find longest palindromic # subsequence of # given string lps = lcs ( string revStr n n ) # If the difference between # longest palindromic # subsequence and the original # string is less # than equal to k then # the string is k-palindrome return ( n - lps <= k ) # Driver program string = 'abcdeca' k = 2 print ( 'Yes' if isKPal ( string k ) else 'No' ) # This code is contributed # by Ansu Kumari.
C# // C# program to find if given // String is K-Palindrome or not using System ; class GFG { /* Returns length of LCS for X[0..m-1] Y[0..n-1] */ static int lcs ( String X String Y int m int n ) { int [] L = new int [ m + 1 n + 1 ]; /* Following steps build L[m+1n+1] in bottom up fashion. Note that L[ij] contains length of LCS of X[0..i-1] and Y[0..j-1] */ for ( int i = 0 ; i <= m ; i ++ ) { for ( int j = 0 ; j <= n ; j ++ ) { if ( i == 0 || j == 0 ) { L [ i j ] = 0 ; } else if ( X [ i - 1 ] == Y [ j - 1 ]) { L [ i j ] = L [ i - 1 j - 1 ] + 1 ; } else { L [ i j ] = Math . Max ( L [ i - 1 j ] L [ i j - 1 ]); } } } // L[mn] contains length // of LCS for X and Y return L [ m n ]; } // find if given String is // K-Palindrome or not static bool isKPal ( String str int k ) { int n = str . Length ; // Find reverse of String str = reverse ( str ); // find longest palindromic // subsequence of given String int lps = lcs ( str str n n ); // If the difference between longest // palindromic subsequence and the // original String is less than equal // to k then the String is k-palindrome return ( n - lps <= k ); } static String reverse ( String input ) { char [] temparray = input . ToCharArray (); int left right = 0 ; right = temparray . Length - 1 ; for ( left = 0 ; left < right ; left ++ right -- ) { // Swap values of left and right char temp = temparray [ left ]; temparray [ left ] = temparray [ right ]; temparray [ right ] = temp ; } return String . Join ( '' temparray ); } // Driver code public static void Main ( String [] args ) { String str = 'abcdeca' ; int k = 2 ; if ( isKPal ( str k )) { Console . WriteLine ( 'Yes' ); } else Console . WriteLine ( 'No' ); } } // This code is contributed by PrinciRaj1992
JavaScript < script > // JavaScript program to find // if given string is K-Palindrome // or not // Returns length of LCS // for X[0..m-1] Y[0..n-1] function lcs ( X Y m n ){ let L = new Array ( m + 1 ); for ( let i = 0 ; i < m + 1 ; i ++ ){ L [ i ] = new Array ( n + 1 ). fill ( 0 ); } // Following steps build // L[m+1][n+1] in bottom up // fashion. Note that L[i][j] // contains length of // LCS of X[0..i-1] and Y[0..j-1] for ( let i = 0 ; i < m + 1 ; i ++ ) { for ( let j = 0 ; j < n + 1 ; j ++ ) { if ( ! i || ! j ) L [ i ][ j ] = 0 else if ( X [ i - 1 ] == Y [ j - 1 ]) L [ i ][ j ] = L [ i - 1 ][ j - 1 ] + 1 else L [ i ][ j ] = Math . max ( L [ i - 1 ][ j ] L [ i ][ j - 1 ]) } } // L[m][n] contains length // of LCS for X and Y return L [ m ][ n ] } // find if given string is // K-Palindrome or not function isKPal ( string k ){ let n = string . length // Find reverse of string let revStr = string . split ( '' ). reverse (). join ( '' ) // find longest palindromic // subsequence of // given string let lps = lcs ( string revStr n n ) // If the difference between // longest palindromic // subsequence and the original // string is less // than equal to k then // the string is k-palindrome return ( n - lps <= k ) } // Driver program let string = 'abcdeca' let k = 2 document . write ( isKPal ( string k ) ? 'Yes' : 'No' ) // This code is contributed by shinjanpatra < /script>
Uitvoer
Yes
Tijdcomplexiteit van bovenstaande oplossing is O(n 2 ).
Hulpruimte gebruikt door het programma is O(n 2 ). Het kan verder worden gereduceerd tot O(n) door gebruik te maken van Ruimtegeoptimaliseerde oplossing van LCS .
Dankzij Het ravijn dat jij hebt versmald voor het voorstellen van bovenstaande oplossing.
