Garākā atkārtotā un nepārklājošā apakšvirkne
Izmēģiniet to GfG Practice 
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Ņemot vērā a stīga s uzdevums ir atrast visilgāk atkārtojošā apakšvirkne, kas nepārklājas tajā. Citiem vārdiem sakot, atrast 2 identiskas apakšvirknes no maksimālais garums kas nepārklājas. Atgriež -1 ja šādas virknes nav.
Piezīme: Ir iespējamas vairākas atbildes, bet mums ir jāatgriež apakšvirkne kuru pirmā parādīšanās ir agrāk.
Piemēri:
Ievade: s = 'acdcdcdc'
Izvade: "AC/DC"
Paskaidrojums: Virkne 'acdc' ir s garākā apakšvirkne, kas atkārtojas, bet nepārklājas.Ievade: s = 'geeksforgeeks'
Izvade: 'geeks'
Paskaidrojums: Virkne “geeks” ir s garākā apakšvirkne, kas atkārtojas, bet nepārklājas.
Satura rādītājs
- Brutālā spēka metodes izmantošana - O(n^3) laiks un O(n) telpa
- Izmantojot no augšas uz leju DP (atgādināšana) — O(n^2) laiks un O(n^2) telpa
- Augšupvērsto DP izmantošana (tabulēšana) — O(n^2) laiks un O(n^2) telpa
- Telpas optimizētā DP izmantošana – O(n^2) laiks un O(n) telpa
Brutālā spēka metodes izmantošana - O(n^3) laiks un O(n) telpa
C++Ideja ir, lai ģenerēt viss iespējamās apakšvirknes un pārbaudiet, vai apakšvirkne pastāv atlikušais virkne. Ja apakšvirkne pastāv un tā garums ir lielāks nekā atbildes apakšvirkne, tad iestatiet atbildi uz pašreizējo apakšvirkni.
// C++ program to find longest repeating // and non-overlapping substring // using recursion #include using namespace std ; string longestSubstring ( string & s ) { int n = s . length (); string ans = '' ; int len = 0 ; int i = 0 j = 0 ; while ( i < n && j < n ) { string curr = s . substr ( i j - i + 1 ); // If substring exists compare its length // with ans if ( s . find ( curr j + 1 ) != string :: npos && j - i + 1 > len ) { len = j - i + 1 ; ans = curr ; } // Otherwise increment i else i ++ ; j ++ ; } return len > 0 ? ans : '-1' ; } int main () { string s = 'geeksforgeeks' ; cout < < longestSubstring ( s ) < < endl ; return 0 ; }
Java // Java program to find longest repeating // and non-overlapping substring // using recursion class GfG { static String longestSubstring ( String s ) { int n = s . length (); String ans = '' ; int len = 0 ; int i = 0 j = 0 ; while ( i < n && j < n ) { String curr = s . substring ( i j + 1 ); // If substring exists compare its length // with ans if ( s . indexOf ( curr j + 1 ) != - 1 && j - i + 1 > len ) { len = j - i + 1 ; ans = curr ; } // Otherwise increment i else i ++ ; j ++ ; } return len > 0 ? ans : '-1' ; } public static void main ( String [] args ) { String s = 'geeksforgeeks' ; System . out . println ( longestSubstring ( s )); } }
Python # Python program to find longest repeating # and non-overlapping substring # using recursion def longestSubstring ( s ): n = len ( s ) ans = '' lenAns = 0 i j = 0 0 while i < n and j < n : curr = s [ i : j + 1 ] # If substring exists compare its length # with ans if s . find ( curr j + 1 ) != - 1 and j - i + 1 > lenAns : lenAns = j - i + 1 ans = curr # Otherwise increment i else : i += 1 j += 1 if lenAns > 0 : return ans return '-1' if __name__ == '__main__' : s = 'geeksforgeeks' print ( longestSubstring ( s ))
C# // C# program to find longest repeating // and non-overlapping substring // using recursion using System ; class GfG { static string longestSubstring ( string s ) { int n = s . Length ; string ans = '' ; int len = 0 ; int i = 0 j = 0 ; while ( i < n && j < n ) { string curr = s . Substring ( i j - i + 1 ); // If substring exists compare its length // with ans if ( s . IndexOf ( curr j + 1 ) != - 1 && j - i + 1 > len ) { len = j - i + 1 ; ans = curr ; } // Otherwise increment i else i ++ ; j ++ ; } return len > 0 ? ans : '-1' ; } static void Main ( string [] args ) { string s = 'geeksforgeeks' ; Console . WriteLine ( longestSubstring ( s )); } }
JavaScript // JavaScript program to find longest repeating // and non-overlapping substring // using recursion function longestSubstring ( s ) { const n = s . length ; let ans = '' ; let len = 0 ; let i = 0 j = 0 ; while ( i < n && j < n ) { const curr = s . substring ( i j + 1 ); // If substring exists compare its length // with ans if ( s . indexOf ( curr j + 1 ) !== - 1 && j - i + 1 > len ) { len = j - i + 1 ; ans = curr ; } // Otherwise increment i else i ++ ; j ++ ; } return len > 0 ? ans : '-1' ; } const s = 'geeksforgeeks' ; console . log ( longestSubstring ( s ));
Izvade
geeks
Izmantojot no augšas uz leju DP (atgādināšana) — O(n^2) laiks un O(n^2) telpa
Pieeja ir aprēķināt garākais atkārtotais sufikss visiem prefiksiem pāri stīga s . Par indeksiem i un j ja s[i] == s[j] tad rekursīvi aprēķināt sufikss(i+1 j+1) un iestatīt sufikss(i j) kā min(sufikss(i+1 j+1) + 1 j - i - 1) uz novērst pārklāšanos . Ja rakstzīmes nesakrīt iestatīt sufiksu(i j) = 0.
Piezīme:
- Lai izvairītos no pārklāšanās, mums ir jānodrošina, ka garums sufikss ir mazāks par (j-i) jebkurā mirklī.
- Maksimālā vērtība sufikss(i j) nodrošina garākās atkārtotās apakšvirknes garumu, un pašu apakšvirkni var atrast, izmantojot kopīgā sufiksa garumu un sākuma indeksu.
- sufikss(i j) saglabā garākā kopīgā sufiksa garumu starp indeksiem i un j nodrošinot to nepārsniedz j - i - 1 lai izvairītos no pārklāšanās.
// C++ program to find longest repeating // and non-overlapping substring // using memoization #include using namespace std ; int findSuffix ( int i int j string & s vector < vector < int >> & memo ) { // base case if ( j == s . length ()) return 0 ; // return memoized value if ( memo [ i ][ j ] != -1 ) return memo [ i ][ j ]; // if characters match if ( s [ i ] == s [ j ]) { memo [ i ][ j ] = 1 + min ( findSuffix ( i + 1 j + 1 s memo ) j - i - 1 ); } else { memo [ i ][ j ] = 0 ; } return memo [ i ][ j ]; } string longestSubstring ( string s ) { int n = s . length (); vector < vector < int >> memo ( n vector < int > ( n -1 )); // find length of non-overlapping // substrings for all pairs (ij) for ( int i = 0 ; i < n ; i ++ ) { for ( int j = i + 1 ; j < n ; j ++ ) { findSuffix ( i j s memo ); } } string ans = '' ; int ansLen = 0 ; // If length of suffix is greater // than ansLen update ans and ansLen for ( int i = 0 ; i < n ; i ++ ) { for ( int j = i + 1 ; j < n ; j ++ ) { if ( memo [ i ][ j ] > ansLen ) { ansLen = memo [ i ][ j ]; ans = s . substr ( i ansLen ); } } } return ansLen > 0 ? ans : '-1' ; } int main () { string s = 'geeksforgeeks' ; cout < < longestSubstring ( s ) < < endl ; return 0 ; }
Java // Java program to find longest repeating // and non-overlapping substring // using memoization import java.util.Arrays ; class GfG { static int findSuffix ( int i int j String s int [][] memo ) { // base case if ( j == s . length ()) return 0 ; // return memoized value if ( memo [ i ][ j ] != - 1 ) return memo [ i ][ j ] ; // if characters match if ( s . charAt ( i ) == s . charAt ( j )) { memo [ i ][ j ] = 1 + Math . min ( findSuffix ( i + 1 j + 1 s memo ) j - i - 1 ); } else { memo [ i ][ j ] = 0 ; } return memo [ i ][ j ] ; } static String longestSubstring ( String s ) { int n = s . length (); int [][] memo = new int [ n ][ n ] ; for ( int [] row : memo ) { Arrays . fill ( row - 1 ); } // find length of non-overlapping // substrings for all pairs (i j) for ( int i = 0 ; i < n ; i ++ ) { for ( int j = i + 1 ; j < n ; j ++ ) { findSuffix ( i j s memo ); } } String ans = '' ; int ansLen = 0 ; // If length of suffix is greater // than ansLen update ans and ansLen for ( int i = 0 ; i < n ; i ++ ) { for ( int j = i + 1 ; j < n ; j ++ ) { if ( memo [ i ][ j ] > ansLen ) { ansLen = memo [ i ][ j ] ; ans = s . substring ( i i + ansLen ); } } } return ansLen > 0 ? ans : '-1' ; } public static void main ( String [] args ) { String s = 'geeksforgeeks' ; System . out . println ( longestSubstring ( s )); } }
Python # Python program to find longest repeating # and non-overlapping substring # using memoization def findSuffix ( i j s memo ): # base case if j == len ( s ): return 0 # return memoized value if memo [ i ][ j ] != - 1 : return memo [ i ][ j ] # if characters match if s [ i ] == s [ j ]: memo [ i ][ j ] = 1 + min ( findSuffix ( i + 1 j + 1 s memo ) j - i - 1 ) else : memo [ i ][ j ] = 0 return memo [ i ][ j ] def longestSubstring ( s ): n = len ( s ) memo = [[ - 1 ] * n for _ in range ( n )] # find length of non-overlapping # substrings for all pairs (i j) for i in range ( n ): for j in range ( i + 1 n ): findSuffix ( i j s memo ) ans = '' ansLen = 0 # If length of suffix is greater # than ansLen update ans and ansLen for i in range ( n ): for j in range ( i + 1 n ): if memo [ i ][ j ] > ansLen : ansLen = memo [ i ][ j ] ans = s [ i : i + ansLen ] if ansLen > 0 : return ans return '-1' if __name__ == '__main__' : s = 'geeksforgeeks' print ( longestSubstring ( s ))
C# // C# program to find longest repeating // and non-overlapping substring // using memoization using System ; class GfG { static int findSuffix ( int i int j string s int [ ] memo ) { // base case if ( j == s . Length ) return 0 ; // return memoized value if ( memo [ i j ] != - 1 ) return memo [ i j ]; // if characters match if ( s [ i ] == s [ j ]) { memo [ i j ] = 1 + Math . Min ( findSuffix ( i + 1 j + 1 s memo ) j - i - 1 ); } else { memo [ i j ] = 0 ; } return memo [ i j ]; } static string longestSubstring ( string s ) { int n = s . Length ; int [ ] memo = new int [ n n ]; for ( int i = 0 ; i < n ; i ++ ) { for ( int j = 0 ; j < n ; j ++ ) { memo [ i j ] = - 1 ; } } // find length of non-overlapping // substrings for all pairs (i j) for ( int i = 0 ; i < n ; i ++ ) { for ( int j = i + 1 ; j < n ; j ++ ) { findSuffix ( i j s memo ); } } string ans = '' ; int ansLen = 0 ; // If length of suffix is greater // than ansLen update ans and ansLen for ( int i = 0 ; i < n ; i ++ ) { for ( int j = i + 1 ; j < n ; j ++ ) { if ( memo [ i j ] > ansLen ) { ansLen = memo [ i j ]; ans = s . Substring ( i ansLen ); } } } return ansLen > 0 ? ans : '-1' ; } static void Main ( string [] args ) { string s = 'geeksforgeeks' ; Console . WriteLine ( longestSubstring ( s )); } }
JavaScript // JavaScript program to find longest repeating // and non-overlapping substring // using memoization function findSuffix ( i j s memo ) { // base case if ( j === s . length ) return 0 ; // return memoized value if ( memo [ i ][ j ] !== - 1 ) return memo [ i ][ j ]; // if characters match if ( s [ i ] === s [ j ]) { memo [ i ][ j ] = 1 + Math . min ( findSuffix ( i + 1 j + 1 s memo ) j - i - 1 ); } else { memo [ i ][ j ] = 0 ; } return memo [ i ][ j ]; } function longestSubstring ( s ) { const n = s . length ; const memo = Array . from ({ length : n } () => Array ( n ). fill ( - 1 )); // find length of non-overlapping // substrings for all pairs (i j) for ( let i = 0 ; i < n ; i ++ ) { for ( let j = i + 1 ; j < n ; j ++ ) { findSuffix ( i j s memo ); } } let ans = '' ; let ansLen = 0 ; // If length of suffix is greater // than ansLen update ans and ansLen for ( let i = 0 ; i < n ; i ++ ) { for ( let j = i + 1 ; j < n ; j ++ ) { if ( memo [ i ][ j ] > ansLen ) { ansLen = memo [ i ][ j ]; ans = s . substring ( i i + ansLen ); } } } return ansLen > 0 ? ans : '-1' ; } const s = 'geeksforgeeks' ; console . log ( longestSubstring ( s ));
Izvade
geeks
Augšupvērsto DP izmantošana (tabulēšana) — O(n^2) laiks un O(n^2) telpa
C++Ideja ir, lai izveidot 2D matricu no izmērs (n+1)*(n+1) un aprēķiniet visilgāk atkārtojošos sufiksus visam indeksam pāri (i j) iteratīvi. Mēs sākam no beigas no virknes un strādājiet atpakaļ, lai aizpildītu tabulu. Katram (i j) ja s[i] == s[j] mēs uzstādām sufikss[i][j] līdz min (sufikss[i+1][j+1]+1 j-i-1) lai izvairītos no pārklāšanās; citādi sufikss [i][j] = 0.
// C++ program to find longest repeating // and non-overlapping substring // using tabulation #include using namespace std ; string longestSubstring ( string s ) { int n = s . length (); vector < vector < int >> dp ( n + 1 vector < int > ( n + 1 0 )); string ans = '' ; int ansLen = 0 ; // find length of non-overlapping // substrings for all pairs (ij) for ( int i = n -1 ; i >= 0 ; i -- ) { for ( int j = n -1 ; j > i ; j -- ) { // if characters match set value // and compare with ansLen. if ( s [ i ] == s [ j ]) { dp [ i ][ j ] = 1 + min ( dp [ i + 1 ][ j + 1 ] j - i -1 ); if ( dp [ i ][ j ] >= ansLen ) { ansLen = dp [ i ][ j ]; ans = s . substr ( i ansLen ); } } } } return ansLen > 0 ? ans : '-1' ; } int main () { string s = 'geeksforgeeks' ; cout < < longestSubstring ( s ) < < endl ; return 0 ; }
Java // Java program to find longest repeating // and non-overlapping substring // using tabulation class GfG { static String longestSubstring ( String s ) { int n = s . length (); int [][] dp = new int [ n + 1 ][ n + 1 ] ; String ans = '' ; int ansLen = 0 ; // find length of non-overlapping // substrings for all pairs (i j) for ( int i = n - 1 ; i >= 0 ; i -- ) { for ( int j = n - 1 ; j > i ; j -- ) { // if characters match set value // and compare with ansLen. if ( s . charAt ( i ) == s . charAt ( j )) { dp [ i ][ j ] = 1 + Math . min ( dp [ i + 1 ][ j + 1 ] j - i - 1 ); if ( dp [ i ][ j ] >= ansLen ) { ansLen = dp [ i ][ j ] ; ans = s . substring ( i i + ansLen ); } } } } return ansLen > 0 ? ans : '-1' ; } public static void main ( String [] args ) { String s = 'geeksforgeeks' ; System . out . println ( longestSubstring ( s )); } }
Python # Python program to find longest repeating # and non-overlapping substring # using tabulation def longestSubstring ( s ): n = len ( s ) dp = [[ 0 ] * ( n + 1 ) for _ in range ( n + 1 )] ans = '' ansLen = 0 # find length of non-overlapping # substrings for all pairs (i j) for i in range ( n - 1 - 1 - 1 ): for j in range ( n - 1 i - 1 ): # if characters match set value # and compare with ansLen. if s [ i ] == s [ j ]: dp [ i ][ j ] = 1 + min ( dp [ i + 1 ][ j + 1 ] j - i - 1 ) if dp [ i ][ j ] >= ansLen : ansLen = dp [ i ][ j ] ans = s [ i : i + ansLen ] return ans if ansLen > 0 else '-1' if __name__ == '__main__' : s = 'geeksforgeeks' print ( longestSubstring ( s ))
C# // C# program to find longest repeating // and non-overlapping substring // using tabulation using System ; class GfG { static string longestSubstring ( string s ) { int n = s . Length ; int [] dp = new int [ n + 1 n + 1 ]; string ans = '' ; int ansLen = 0 ; // find length of non-overlapping // substrings for all pairs (i j) for ( int i = n - 1 ; i >= 0 ; i -- ) { for ( int j = n - 1 ; j > i ; j -- ) { // if characters match set value // and compare with ansLen. if ( s [ i ] == s [ j ]) { dp [ i j ] = 1 + Math . Min ( dp [ i + 1 j + 1 ] j - i - 1 ); if ( dp [ i j ] >= ansLen ) { ansLen = dp [ i j ]; ans = s . Substring ( i ansLen ); } } } } return ansLen > 0 ? ans : '-1' ; } static void Main ( string [] args ) { string s = 'geeksforgeeks' ; Console . WriteLine ( longestSubstring ( s )); } }
JavaScript // JavaScript program to find longest repeating // and non-overlapping substring // using tabulation function longestSubstring ( s ) { const n = s . length ; const dp = Array . from ({ length : n + 1 } () => Array ( n + 1 ). fill ( 0 )); let ans = '' ; let ansLen = 0 ; // find length of non-overlapping // substrings for all pairs (i j) for ( let i = n - 1 ; i >= 0 ; i -- ) { for ( let j = n - 1 ; j > i ; j -- ) { // if characters match set value // and compare with ansLen. if ( s [ i ] === s [ j ]) { dp [ i ][ j ] = 1 + Math . min ( dp [ i + 1 ][ j + 1 ] j - i - 1 ); if ( dp [ i ][ j ] >= ansLen ) { ansLen = dp [ i ][ j ]; ans = s . substring ( i i + ansLen ); } } } } return ansLen > 0 ? ans : '-1' ; } const s = 'geeksforgeeks' ; console . log ( longestSubstring ( s ));
Izvade
geeks
Telpas optimizētā DP izmantošana – O(n^2) laiks un O(n) telpa
C++Ideja ir izmantot a viens 1D masīvs a vietā 2D matrica sekojot tikai 'nākamā rinda' aprēķināšanai nepieciešamās vērtības sufikss[i][j]. Tā kā katra vērtība s piedēklis[i][j] atkarīgs tikai no sufikss[i+1][j+1] zemāk esošajā rindā mēs varam saglabāt iepriekšējās rindas vērtības 1D masīvā un atjaunināt tās iteratīvi katrai rindai.
// C++ program to find longest repeating // and non-overlapping substring // using space optimised #include using namespace std ; string longestSubstring ( string s ) { int n = s . length (); vector < int > dp ( n + 1 0 ); string ans = '' ; int ansLen = 0 ; // find length of non-overlapping // substrings for all pairs (ij) for ( int i = n -1 ; i >= 0 ; i -- ) { for ( int j = i ; j < n ; j ++ ) { // if characters match set value // and compare with ansLen. if ( s [ i ] == s [ j ]) { dp [ j ] = 1 + min ( dp [ j + 1 ] j - i -1 ); if ( dp [ j ] >= ansLen ) { ansLen = dp [ j ]; ans = s . substr ( i ansLen ); } } else dp [ j ] = 0 ; } } return ansLen > 0 ? ans : '-1' ; } int main () { string s = 'geeksforgeeks' ; cout < < longestSubstring ( s ) < < endl ; return 0 ; }
Java // Java program to find longest repeating // and non-overlapping substring // using space optimised class GfG { static String longestSubstring ( String s ) { int n = s . length (); int [] dp = new int [ n + 1 ] ; String ans = '' ; int ansLen = 0 ; // find length of non-overlapping // substrings for all pairs (i j) for ( int i = n - 1 ; i >= 0 ; i -- ) { for ( int j = i ; j < n ; j ++ ) { // if characters match set value // and compare with ansLen. if ( s . charAt ( i ) == s . charAt ( j )) { dp [ j ] = 1 + Math . min ( dp [ j + 1 ] j - i - 1 ); if ( dp [ j ] >= ansLen ) { ansLen = dp [ j ] ; ans = s . substring ( i i + ansLen ); } } else { dp [ j ] = 0 ; } } } return ansLen > 0 ? ans : '-1' ; } public static void main ( String [] args ) { String s = 'geeksforgeeks' ; System . out . println ( longestSubstring ( s )); } }
Python # Python program to find longest repeating # and non-overlapping substring # using space optimised def longestSubstring ( s ): n = len ( s ) dp = [ 0 ] * ( n + 1 ) ans = '' ansLen = 0 # find length of non-overlapping # substrings for all pairs (i j) for i in range ( n - 1 - 1 - 1 ): for j in range ( i n ): # if characters match set value # and compare with ansLen. if s [ i ] == s [ j ]: dp [ j ] = 1 + min ( dp [ j + 1 ] j - i - 1 ) if dp [ j ] >= ansLen : ansLen = dp [ j ] ans = s [ i : i + ansLen ] else : dp [ j ] = 0 return ans if ansLen > 0 else '-1' if __name__ == '__main__' : s = 'geeksforgeeks' print ( longestSubstring ( s ))
C# // C# program to find longest repeating // and non-overlapping substring // using space optimised using System ; class GfG { static string longestSubstring ( string s ) { int n = s . Length ; int [] dp = new int [ n + 1 ]; string ans = '' ; int ansLen = 0 ; // find length of non-overlapping // substrings for all pairs (i j) for ( int i = n - 1 ; i >= 0 ; i -- ) { for ( int j = i ; j < n ; j ++ ) { // if characters match set value // and compare with ansLen. if ( s [ i ] == s [ j ]) { dp [ j ] = 1 + Math . Min ( dp [ j + 1 ] j - i - 1 ); if ( dp [ j ] >= ansLen ) { ansLen = dp [ j ]; ans = s . Substring ( i ansLen ); } } else { dp [ j ] = 0 ; } } } return ansLen > 0 ? ans : '-1' ; } static void Main ( string [] args ) { string s = 'geeksforgeeks' ; Console . WriteLine ( longestSubstring ( s )); } }
JavaScript // JavaScript program to find longest repeating // and non-overlapping substring // using space optimised function longestSubstring ( s ) { const n = s . length ; const dp = new Array ( n + 1 ). fill ( 0 ); let ans = '' ; let ansLen = 0 ; // find length of non-overlapping // substrings for all pairs (i j) for ( let i = n - 1 ; i >= 0 ; i -- ) { for ( let j = i ; j < n ; j ++ ) { // if characters match set value // and compare with ansLen. if ( s [ i ] === s [ j ]) { dp [ j ] = 1 + Math . min ( dp [ j + 1 ] j - i - 1 ); if ( dp [ j ] >= ansLen ) { ansLen = dp [ j ]; ans = s . substring ( i i + ansLen ); } } else { dp [ j ] = 0 ; } } } return ansLen > 0 ? ans : '-1' ; } const s = 'geeksforgeeks' ; console . log ( longestSubstring ( s ));
Izvade
geeks
Saistītie raksti:
- Garākā kopējā apakšvirkne
