Patikrinkite, ar duotą eilutę galima padalyti į keturias skirtingas eilutes
Išbandykite GfG praktikoje 
Išvestis
Pateikus eilutę, užduotis yra patikrinti, ar galime ją padalinti į 4 eilutes taip, kad kiekviena eilutė būtų netuščia ir skirtųsi nuo kitos.
Pavyzdžiai:
Įvestis : str[] = 'geeksforgeeks'
Išvestis : Taip
„geeks“ „for“ „gee“ „ks“ yra keturios skirtingos eilutės, kurios gali susidaryti iš nurodytos eilutės.
Įvestis : str[] = 'aaabb'
Išvestis : Ne
Stebėkite, jei eilutės ilgis yra didesnis arba lygus 10, tada kiekvienas laikas gali būti padalintas į keturias dalis. Tarkime, kad ilgis yra 10, tada galima sudaryti 1 2 3 4 ilgio eilutę.
Jei eilutės ilgis yra mažesnis nei 10, galime naudoti brutalią jėgą, t. y. kartoti visus įmanomus eilutės padalijimo būdus ir patikrinti kiekvieną.
Jeigu ilgis yra didesnis nei 10
grąžinti tiesa
Kitaip (Jei ilgis mažesnis nei 10)
Naudokite Brute Force metodą, kad patikrintumėte, ar galime jį sulaužyti į keturias skirtingas eilutes.
Žemiau yra aukščiau pateiktos idėjos įgyvendinimas.
C++ // C++ program to check if we can break a string // into four distinct strings. #include using namespace std ; // Return if the given string can be split or not. bool check ( string s ) { // We can always break a string of size 10 or // more into four distinct strings. if ( s . size () >= 10 ) return true ; // Brute Force for ( int i = 1 ; i < s . size (); i ++ ) { for ( int j = i + 1 ; j < s . size (); j ++ ) { for ( int k = j + 1 ; k < s . size (); k ++ ) { // Making 4 string from the given string string s1 = s . substr ( 0 i ); string s2 = s . substr ( i j - i ); string s3 = s . substr ( j k - j ); string s4 = s . substr ( k s . size () - k ); // Checking if they are distinct or not. if ( s1 != s2 && s1 != s3 && s1 != s4 && s2 != s3 && s2 != s4 && s3 != s4 ) return true ; } } } return false ; } // Driven Program int main () { string str = 'aaabb' ; ( check ( str )) ? ( cout < < 'Yes' < < endl ) : ( cout < < 'No' < < endl ); return 0 ; }
Java // Java program to check if we can break a string // into four distinct strings. class GFG { // Return true if both strings are equal public static boolean strcheck ( String s1 String s2 ) { if ( s1 != s2 ) return false ; return true ; } // Return if the given string can be split or not. public static boolean check ( String s ) { if ( s . length () >= 10 ) return true ; // Brute Force for ( int i = 1 ; i < s . length (); i ++ ) { for ( int j = i + 1 ; j < s . length (); j ++ ) { for ( int k = j + 1 ; k < s . length (); k ++ ) { // Making 4 string from the given string String s1 = '' s2 = '' s3 = '' s4 = '' ; try { s1 = s . substring ( 0 i ); s2 = s . substring ( i j - i ); s3 = s . substring ( j k - j ); s4 = s . substring ( k s . length () - k ); } catch ( StringIndexOutOfBoundsException e ) { } // Checking if they are distinct or not. if ( strcheck ( s1 s2 ) && strcheck ( s1 s3 ) && strcheck ( s1 s4 ) && strcheck ( s2 s3 ) && strcheck ( s2 s4 ) && strcheck ( s3 s4 )) return true ; } } } return false ; } // Driver code public static void main ( String [] args ) { String str = 'aaabb' ; if ( check ( str )) System . out . println ( 'Yes' ); else System . out . println ( 'No' ); } } // This code is contributed by // sanjeev2552
Python # Python3 program to check if we can # break a into four distinct strings. # Return if the given string can be # split or not. def check ( s ): # We can always break a of size 10 or # more into four distinct strings. if ( len ( s ) >= 10 ): return True # Brute Force for i in range ( 1 len ( s )): for j in range ( i + 1 len ( s )): for k in range ( j + 1 len ( s )): # Making 4 from the given s1 = s [ 0 : i ] s2 = s [ i : j - i ] s3 = s [ j : k - j ] s4 = s [ k : len ( s ) - k ] # Checking if they are distinct or not. if ( s1 != s2 and s1 != s3 and s1 != s4 and s2 != s3 and s2 != s4 and s3 != s4 ): return True return False # Driver Code if __name__ == '__main__' : str = 'aaabb' print ( 'Yes' ) if ( check ( str )) else print ( 'NO' ) # This code is contributed # by SHUBHAMSINGH10
C# // C# program to check if we can break a string // into four distinct strings. using System ; class GFG { // Return true if both strings are equal public static Boolean strcheck ( String s1 String s2 ) { if ( s1 . CompareTo ( s2 ) != 0 ) return false ; return true ; } // Return if the given string // can be split or not. public static Boolean check ( String s ) { if ( s . Length >= 10 ) return true ; // Brute Force for ( int i = 1 ; i < s . Length ; i ++ ) { for ( int j = i + 1 ; j < s . Length ; j ++ ) { for ( int k = j + 1 ; k < s . Length ; k ++ ) { // Making 4 string from the given string String s1 = '' s2 = '' s3 = '' s4 = '' ; try { s1 = s . Substring ( 0 i ); s2 = s . Substring ( i j - i ); s3 = s . Substring ( j k - j ); s4 = s . Substring ( k s . Length - k ); } catch ( Exception e ) { } // Checking if they are distinct or not. if ( strcheck ( s1 s2 ) && strcheck ( s1 s3 ) && strcheck ( s1 s4 ) && strcheck ( s2 s3 ) && strcheck ( s2 s4 ) && strcheck ( s3 s4 )) return true ; } } } return false ; } // Driver code public static void Main ( String [] args ) { String str = 'aaabb' ; if ( check ( str )) Console . WriteLine ( 'Yes' ); else Console . WriteLine ( 'No' ); } } // This code is contributed by Princi Singh
JavaScript < script > // JavaScript program to check if we can break a string // into four distinct strings. // Return true if both strings are equal function strcheck ( s1 s2 ) { if ( s1 . localeCompare ( s2 ) != 0 ) return false ; return true ; } // Return if the given string can be split or not. function check ( s ) { if ( s . length >= 10 ) return true ; // Brute Force for ( let i = 1 ; i < s . length ; i ++ ) { for ( let j = i + 1 ; j < s . length ; j ++ ) { for ( let k = j + 1 ; k < s . length ; k ++ ) { // Making 4 string from the given string let s1 = '' s2 = '' s3 = '' s4 = '' ; s1 = s . substring ( 0 i ); s2 = s . substring ( i i + j - i ); s3 = s . substring ( j j + k - j ); s4 = s . substring ( k k + s . length - k ); // Checking if they are distinct or not. if ( strcheck ( s1 s2 ) && strcheck ( s1 s3 ) && strcheck ( s1 s4 ) && strcheck ( s2 s3 ) && strcheck ( s2 s4 ) && strcheck ( s3 s4 )) return true ; } } } return false ; } let str = 'aaabb' ; if ( check ( str )) document . write ( 'Yes' ); else document . write ( 'No' ); < /script>
PHP // Return true if the given string can be split into four distinct strings function check ( $s ) { // We can always break a string of size 10 or more into four distinct strings if ( strlen ( $s ) >= 10 ) { return true ; } // Brute Force for ( $i = 1 ; $i < strlen ( $s ); $i ++ ) { for ( $j = $i + 1 ; $j < strlen ( $s ); $j ++ ) { for ( $k = $j + 1 ; $k < strlen ( $s ); $k ++ ) { // Making 4 from the given string $s1 = substr ( $s 0 $i ); $s2 = substr ( $s $i $j - $i ); $s3 = substr ( $s $j $k - $j ); $s4 = substr ( $s $k strlen ( $s ) - $k ); // Checking if they are distinct or not if ( $s1 != $s2 && $s1 != $s3 && $s1 != $s4 && $s2 != $s3 && $s2 != $s4 && $s3 != $s4 ) { return true ; } } } } return false ; } // Driver Code $str = 'aaabb' ; echo ( check ( $str ) ? 'Yes' : 'NO' ); ?>
Išvestis
No
Laiko sudėtingumas : O (n 3 )
Pagalbinė erdvė: O(n)
Sukurti viktoriną
