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 : forgeeksOutput :eegksExplanation: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 : gfghhmhOutput :hhh

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 implementation of above steps.

## C++

`// C++ program to print common characters ` `// of two Strings in alphabetical order ` `#include<bits/stdc++.h> ` `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; ` `} ` |

*chevron_right*

*filter_none*

## 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); ` ` ` `} ` `} ` |

*chevron_right*

*filter_none*

## 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 ` |

*chevron_right*

*filter_none*

## 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); ` ` ` `} ` `} ` |

*chevron_right*

*filter_none*

**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.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the **DSA Self Paced Course** at a student-friendly price and become industry ready.