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Print common characters of two Strings in alphabetical order

Given two strings, print all the common characters in lexicographical order. If there are no common letters, print -1. All letters are lower case.

Examples:

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

The idea is to use character count arrays.

1. Count occurrences of all characters from ‘a’ to ‘z’ in the first and second strings. Store these counts in two arrays a1[] and a2[].
2. Traverse a1[] and a2[] (Note size of both is 26). For every index i, print character ‘a’ + i number of times equal min(a1[i], a2[i]).

Below is the implementation of the above steps.

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



Output

eeefgkors

Time Complexity: If we consider n = length(larger string), then this algorithm runs in O(n) complexity.
Auxiliary Space: O(1).

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