# 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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

The idea is to use character count arrays.
1) Count occurrences of all characters from ‘a’ to ‘z’ in 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 Java implementation of above steps.

```// 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 occurence
// 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 occurence
// 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);
}
}
```

Output:

```eeefgkors
```

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

This article is contributed by Deepak Srivatsav. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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