Imprime caracteres comunes de dos cadenas en orden alfabético
Pruébalo en GfG Practice 
Producción
Dadas dos cadenas, imprime todos los caracteres comunes en lexicográfico orden. Si no hay letras comunes, imprima -1. Todas las letras están en minúsculas.
Ejemplos:
Input : string1 : geeks string2 : forgeeks Output : eegks Explanation: The letters that are common between the two strings are e(2 times) k(1 time) and s(1 time). Hence the lexicographical output is 'eegks' Input : string1 : hhhhhello string2 : gfghhmh Output : hhh
La idea es utilizar matrices de recuento de caracteres.
- Cuente las apariciones de todos los caracteres desde la 'a' hasta la 'z' en la primera y segunda cadena. Almacene estos recuentos en dos matrices a1[] y a2[].
- Recorra a1[] y a2[] (tenga en cuenta que el tamaño de ambos es 26). Para cada índice imprimo el carácter 'a' + i un número de veces igual a min(a1[i] a2[i]).
A continuación se muestra la implementación de los pasos anteriores.
C++ // C++ program to print common characters // of two Strings in alphabetical order #include using namespace std ; int main () { string s1 = 'geeksforgeeks' ; string s2 = 'practiceforgeeks' ; // to store the count of // letters in the first string int a1 [ 26 ] = { 0 }; // to store the count of // letters in the second string int a2 [ 26 ] = { 0 }; int i j ; char ch ; char ch1 = 'a' ; int k = ( int ) ch1 m ; // for each letter present increment the count for ( i = 0 ; i < s1 . length () ; i ++ ) { a1 [( int ) s1 [ i ] - k ] ++ ; } for ( i = 0 ; i < s2 . length () ; i ++ ) { a2 [( int ) s2 [ i ] - k ] ++ ; } for ( i = 0 ; i < 26 ; i ++ ) { // the if condition guarantees that // the element is common that is // a1[i] and a2[i] are both non zero // means that the letter has occurred // at least once in both the strings if ( a1 [ i ] != 0 and a2 [ i ] != 0 ) { // print the letter for a number // of times that is the minimum // of its count in s1 and s2 for ( j = 0 ; j < min ( a1 [ i ] a2 [ i ]) ; j ++ ) { m = k + i ; ch = ( char )( k + i ); cout < < ch ; } } } return 0 ; }
Java // Java program to print common characters // of two Strings in alphabetical order import java.io.* ; import java.util.* ; // Function to find similar characters public class Simstrings { static final int MAX_CHAR = 26 ; static void printCommon ( String s1 String s2 ) { // two arrays of length 26 to store occurrence // of a letters alphabetically for each string int [] a1 = new int [ MAX_CHAR ] ; int [] a2 = new int [ MAX_CHAR ] ; int length1 = s1 . length (); int length2 = s2 . length (); for ( int i = 0 ; i < length1 ; i ++ ) a1 [ s1 . charAt ( i ) - 'a' ] += 1 ; for ( int i = 0 ; i < length2 ; i ++ ) a2 [ s2 . charAt ( i ) - 'a' ] += 1 ; // If a common index is non-zero it means // that the letter corresponding to that // index is common to both strings for ( int i = 0 ; i < MAX_CHAR ; i ++ ) { if ( a1 [ i ] != 0 && a2 [ i ] != 0 ) { // Find the minimum of the occurrence // of the character in both strings and print // the letter that many number of times for ( int j = 0 ; j < Math . min ( a1 [ i ] a2 [ i ] ) ; j ++ ) System . out . print ((( char )( i + 'a' ))); } } } // Driver code public static void main ( String [] args ) throws IOException { String s1 = 'geeksforgeeks' s2 = 'practiceforgeeks' ; printCommon ( s1 s2 ); } }
Python3 # Python3 program to print common characters # of two Strings in alphabetical order # Initializing size of array MAX_CHAR = 26 # Function to find similar characters def printCommon ( s1 s2 ): # two arrays of length 26 to store occurrence # of a letters alphabetically for each string a1 = [ 0 for i in range ( MAX_CHAR )] a2 = [ 0 for i in range ( MAX_CHAR )] length1 = len ( s1 ) length2 = len ( s2 ) for i in range ( 0 length1 ): a1 [ ord ( s1 [ i ]) - ord ( 'a' )] += 1 for i in range ( 0 length2 ): a2 [ ord ( s2 [ i ]) - ord ( 'a' )] += 1 # If a common index is non-zero it means # that the letter corresponding to that # index is common to both strings for i in range ( 0 MAX_CHAR ): if ( a1 [ i ] != 0 and a2 [ i ] != 0 ): # Find the minimum of the occurrence # of the character in both strings and print # the letter that many number of times for j in range ( 0 min ( a1 [ i ] a2 [ i ])): ch = chr ( ord ( 'a' ) + i ) print ( ch end = '' ) # Driver code if __name__ == '__main__' : s1 = 'geeksforgeeks' s2 = 'practiceforgeeks' printCommon ( s1 s2 ); # This Code is contributed by Abhishek Sharma
C# // C# program to print common characters // of two Strings in alphabetical order using System ; // Function to find similar characters public class Simstrings { static int MAX_CHAR = 26 ; static void printCommon ( string s1 string s2 ) { // two arrays of length 26 to store occurrence // of a letters alphabetically for each string int [] a1 = new int [ MAX_CHAR ]; int [] a2 = new int [ MAX_CHAR ]; int length1 = s1 . Length ; int length2 = s2 . Length ; for ( int i = 0 ; i < length1 ; i ++ ) a1 [ s1 [ i ] - 'a' ] += 1 ; for ( int i = 0 ; i < length2 ; i ++ ) a2 [ s2 [ i ] - 'a' ] += 1 ; // If a common index is non-zero it means // that the letter corresponding to that // index is common to both strings for ( int i = 0 ; i < MAX_CHAR ; i ++ ) { if ( a1 [ i ] != 0 && a2 [ i ] != 0 ) { // Find the minimum of the occurrence // of the character in both strings and print // the letter that many number of times for ( int j = 0 ; j < Math . Min ( a1 [ i ] a2 [ i ]) ; j ++ ) Console . Write ((( char )( i + 'a' ))); } } } // Driver code public static void Main () { string s1 = 'geeksforgeeks' s2 = 'practiceforgeeks' ; printCommon ( s1 s2 ); } }
JavaScript < script > // Javascript program to print common characters // of two Strings in alphabetical order let MAX_CHAR = 26 ; // Function to find similar characters function printCommon ( s1 s2 ) { // two arrays of length 26 to store occurrence // of a letters alphabetically for each string let a1 = new Array ( MAX_CHAR ); let a2 = new Array ( MAX_CHAR ); for ( let i = 0 ; i < MAX_CHAR ; i ++ ) { a1 [ i ] = 0 ; a2 [ i ] = 0 ; } let length1 = s1 . length ; let length2 = s2 . length ; for ( let i = 0 ; i < length1 ; i ++ ) a1 [ s1 [ i ]. charCodeAt ( 0 ) - 'a' . charCodeAt ( 0 )] += 1 ; for ( let i = 0 ; i < length2 ; i ++ ) a2 [ s2 [ i ]. charCodeAt ( 0 ) - 'a' . charCodeAt ( 0 )] += 1 ; // If a common index is non-zero it means // that the letter corresponding to that // index is common to both strings for ( let i = 0 ; i < MAX_CHAR ; i ++ ) { if ( a1 [ i ] != 0 && a2 [ i ] != 0 ) { // Find the minimum of the occurrence // of the character in both strings and print // the letter that many number of times for ( let j = 0 ; j < Math . min ( a1 [ i ] a2 [ i ]) ; j ++ ) document . write (( String . fromCharCode ( i + 'a' . charCodeAt ( 0 )))); } } } // Driver code let s1 = 'geeksforgeeks' s2 = 'practiceforgeeks' ; printCommon ( s1 s2 ); // This code is contributed by avanitrachhadiya2155 < /script>
Producción
eeefgkors
Complejidad del tiempo: Si consideramos n = longitud (cadena más grande), entonces este algoritmo se ejecuta en En) complejidad.
Espacio Auxiliar: O(1).
