# Find uncommon characters of the two strings

• Difficulty Level : Easy
• Last Updated : 24 Jun, 2022

Find and print the uncommon characters of the two given strings in sorted order. Here uncommon character means that either the character is present in one string or it is present in other string but not in both. The strings contain only lowercase characters and can contain duplicates.
Source: Amazon Interview Experience | Set 355 (For 1 Year Experienced)
Examples:

Input: str1 = “characters”, str2 = “alphabets”
Output: b c l p r
Input: str1 = “geeksforgeeks”, str2 = “geeksquiz”
Output: f i o q r u z

Naive Approach: Using two loops, for each character of 1st string check whether it is present in the 2nd string or not. Likewise, for each character of 2nd string check whether it is present in the 1st string or not.
Time Complexity: O(n^2) and extra would be required to handle duplicates.
Efficient Approach: An efficient approach is to use hashing

• Use a hash table of size 26 for all the lowercase characters.
• Initially, mark presence of each character as ‘0’ (denoting that the character is not present in both the strings).
• Traverse the 1st string and mark presence of each character of 1st string as ‘1’ (denoting 1st string) in the hash table.
• Now, traverse the 2nd string. For each character of 2nd string, check whether its presence in the hash table is ‘1’ or not. If it is ‘1’, then mark its presence as ‘-1’ (denoting that the character is common to both the strings), else mark its presence as ‘2’ (denoting 2nd string).

Below image is a dry run of the above approach: Below is the implementation of the above approach:

## C++

 `// C++ implementation to find the uncommon``// characters of the two strings``#include ``using` `namespace` `std;` `// size of the hash table``const` `int` `MAX_CHAR = 26;` `// function to find the uncommon characters``// of the two strings``void` `findAndPrintUncommonChars(string str1, string str2)``{``    ``// mark presence of each character as 0``    ``// in the hash table 'present[]'``    ``int` `present[MAX_CHAR];``    ``for` `(``int` `i=0; i

## Java

 `// Java implementation to find the uncommon``// characters of the two strings``class` `GFG``{` `    ``// size of the hash table``    ``static` `int` `MAX_CHAR = ``26``;` `    ``// function to find the uncommon``    ``// characters of the two strings``    ``static` `void` `findAndPrintUncommonChars(String str1,``                                       ``String str2)``    ``{``        ``// mark presence of each character as 0``        ``// in the hash table 'present[]'``        ``int` `present[] = ``new` `int``[MAX_CHAR];``        ``for` `(``int` `i = ``0``; i < MAX_CHAR; i++)``        ``{``            ``present[i] = ``0``;``        ``}` `        ``int` `l1 = str1.length();``        ``int` `l2 = str2.length();` `        ``// for each character of str1, mark its``        ``// presence as 1 in 'present[]'``        ``for` `(``int` `i = ``0``; i < l1; i++)``        ``{``            ``present[str1.charAt(i) - ``'a'``] = ``1``;``        ``}` `        ``// for each character of str2``        ``for` `(``int` `i = ``0``; i < l2; i++)``        ``{``            ` `            ``// if a character of str2 is also present``            ``// in str1, then mark its presence as -1``            ``if` `(present[str2.charAt(i) - ``'a'``] == ``1``                ``|| present[str2.charAt(i) - ``'a'``] == -``1``)``            ``{``                ``present[str2.charAt(i) - ``'a'``] = -``1``;``            ``}``            ` `            ``// else mark its presence as 2``            ``else``            ``{``                ``present[str2.charAt(i) - ``'a'``] = ``2``;``            ``}``        ``}` `        ``// print all the uncommon characters``        ``for` `(``int` `i = ``0``; i < MAX_CHAR; i++)``        ``{``            ``if` `(present[i] == ``1` `|| present[i] == ``2``)``            ``{``                ``System.out.print((``char``) (i + ``'a'``) + ``" "``);``            ``}``        ``}``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``String str1 = ``"characters"``;``        ``String str2 = ``"alphabets"``;``        ``findAndPrintUncommonChars(str1, str2);``    ``}``}` `// This code is contributed by Rajput-JI`

## Python 3

 `# Python 3 implementation to find the``# uncommon characters of the two strings` `# size of the hash table``MAX_CHAR ``=` `26` `# function to find the uncommon characters``# of the two strings``def` `findAndPrintUncommonChars(str1, str2):``    ` `    ``# mark presence of each character as 0``    ``# in the hash table 'present[]'``    ``present ``=` `[``0``] ``*` `MAX_CHAR``    ``for` `i ``in` `range``(``0``, MAX_CHAR):``        ``present[i] ``=` `0` `    ``l1 ``=` `len``(str1)``    ``l2 ``=` `len``(str2)``    ` `    ``# for each character of str1, mark its``    ``# presence as 1 in 'present[]'``    ``for` `i ``in` `range``(``0``, l1):``        ``present[``ord``(str1[i]) ``-` `ord``(``'a'``)] ``=` `1``        ` `    ``# for each character of str2``    ``for` `i ``in` `range``(``0``, l2):``        ` `        ``# if a character of str2 is also present``        ``# in str1, then mark its presence as -1``        ``if``(present[``ord``(str2[i]) ``-` `ord``(``'a'``)] ``=``=` `1` `or``           ``present[``ord``(str2[i]) ``-` `ord``(``'a'``)] ``=``=` `-``1``):``            ``present[``ord``(str2[i]) ``-` `ord``(``'a'``)] ``=` `-``1` `        ``# else mark its presence as 2``        ``else``:``            ``present[``ord``(str2[i]) ``-` `ord``(``'a'``)] ``=` `2` `    ``# print all the uncommon characters``    ``for` `i ``in` `range``(``0``, MAX_CHAR):``        ``if``(present[i] ``=``=` `1` `or` `present[i] ``=``=` `2``):``            ``print``(``chr``(i ``+` `ord``(``'a'``)), end ``=` `" "``)` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:``    ``str1 ``=` `"characters"``    ``str2 ``=` `"alphabets"``    ``findAndPrintUncommonChars(str1, str2)` `# This code is contributed``# by Sairahul099`

## C#

 `// C# implementation to find the uncommon``// characters of the two strings``using` `System;` `class` `GFG``{` `    ``// size of the hash table``    ``static` `int` `MAX_CHAR = 26;` `    ``// function to find the uncommon``    ``// characters of the two strings``    ``static` `void` `findAndPrintUncommonChars(String str1,``                                    ``String str2)``    ``{``        ``// mark presence of each character as 0``        ``// in the hash table 'present[]'``        ``int` `[]present = ``new` `int``[MAX_CHAR];``        ``for` `(``int` `i = 0; i < MAX_CHAR; i++)``        ``{``            ``present[i] = 0;``        ``}` `        ``int` `l1 = str1.Length;``        ``int` `l2 = str2.Length;` `        ``// for each character of str1, mark its``        ``// presence as 1 in 'present[]'``        ``for` `(``int` `i = 0; i < l1; i++)``        ``{``            ``present[str1[i] - ``'a'``] = 1;``        ``}` `        ``// for each character of str2``        ``for` `(``int` `i = 0; i < l2; i++)``        ``{``            ` `            ``// if a character of str2 is also present``            ``// in str1, then mark its presence as -1``            ``if` `(present[str2[i] - ``'a'``] == 1``                ``|| present[str2[i] - ``'a'``] == -1)``            ``{``                ``present[str2[i] - ``'a'``] = -1;``            ``}``            ` `            ``// else mark its presence as 2``            ``else``            ``{``                ``present[str2[i] - ``'a'``] = 2;``            ``}``        ``}` `        ``// print all the uncommon characters``        ``for` `(``int` `i = 0; i < MAX_CHAR; i++)``        ``{``            ``if` `(present[i] == 1 || present[i] == 2)``            ``{``                ``Console.Write((``char``) (i + ``'a'``) + ``" "``);``            ``}``        ``}``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``String str1 = ``"characters"``;``        ``String str2 = ``"alphabets"``;``        ``findAndPrintUncommonChars(str1, str2);``    ``}``}` `// This code is contributed by PrinciRaj1992`

## Javascript

 ``

Output

`b c l p r `

Time Complexity: O(m + n), where m and n are the sizes of the two strings respectively.

Auxiliary Space: O(26), no any other extra space is required, so it is a constant.

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

Another map based approach:

• Take two maps and initialize their value as 0.
• traverse the first string, for each character present in first string, set 1 in the 1st map.
• Do the same for second string also.
• Iterate through all 26 characters, if the xor of map 1 and map 2 is 1 then it is present in one of the string only. i.e those characters are uncommon characters. Add them in the result string.
• return the result string, if the string is empty, return -1.

Below is the implementation of the above approach:

## C++

 `#include``using` `namespace` `std;` `string UncommonChars(string a, string b)``{``    ``int` `mp1 = {0}, mp2 = {0};``    ``int` `n = a.size(), m = b.size();` `    ``for``(``auto` `&x: a){``      ``mp1[x-``'a'``] = 1;``    ``}` `    ``for``(``auto` `&x: b){``      ``mp2[x-``'a'``] = 1;``    ``}` `    ``string chars = ``""``;` `    ``for``(``int` `i = 0; i < 26; ++i){``      ``if``(mp1[i]^mp2[i])``        ``chars+=``char``(i+``'a'``);``    ``}``    ``if``(chars == ``""``)``      ``return` `"-1"``;``    ``else``      ``return` `chars;``}` `int` `main(){``    ``string a = ``"geeksforgeeks"``;``    ``string b = ``"geeksquiz"``;``    ``string result = UncommonChars(a,b);``    ``cout << result << endl;``    ``return` `0;``}`

## Java

 `// Java implementation to find the uncommon``// characters of the two strings``class` `GFG {` `    ``// size of the hash table``    ``static` `int` `MAX_CHAR = ``26``;` `    ``// function to find the uncommon``    ``// characters of the two strings``    ``static` `String UncommonChars(String a, String b)``    ``{``        ``int` `mp1[] = ``new` `int``[MAX_CHAR];``        ``int` `mp2[] = ``new` `int``[MAX_CHAR];``        ``int` `n = a.length();``        ``int` `m = b.length();``        ``for` `(``int` `i = ``0``; i < n; i++) {``            ``mp1[a.charAt(i) - ``'a'``] = ``1``;``        ``}``        ``for` `(``int` `i = ``0``; i < m; i++) {``            ``mp2[b.charAt(i) - ``'a'``] = ``1``;``        ``}` `        ``String chars = ``""``;``        ``for` `(``int` `i = ``0``; i < ``26``; i++) {``            ``if` `((mp1[i] ^ mp2[i]) != ``0``) {``                ``chars += (``char``)(i + ``'a'``);``            ``}``        ``}``        ``if` `(chars == ``""``)``            ``return` `"-1"``;``        ``else``            ``return` `chars;``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``String a = ``"geeksforgeeks"``;``        ``String b = ``"geeksquiz"``;``        ``String result = UncommonChars(a, b);``        ``System.out.print(result);``    ``}``}` `// This code is contributed by Aarti_Rathi`

## C#

 `using` `System;` `public` `static` `class` `GFG {``  ``public` `static` `string` `UncommonChars(``string` `a, ``string` `b)``  ``{``    ``int``[] mp1 = ``new` `int``;``    ``int``[] mp2 = ``new` `int``;``    ``int` `n = a.Length;``    ``int` `m = b.Length;` `    ``foreach``(``var` `x ``in` `a) { mp1[x - ``'a'``] = 1; }` `    ``foreach``(``var` `x ``in` `b) { mp2[x - ``'a'``] = 1; }` `    ``string` `chars = ``""``;` `    ``for` `(``int` `i = 0; i < 26; ++i) {``      ``if` `((mp1[i] ^ mp2[i]) != 0) {``        ``chars += (``char``)(i + ``'a'``);``      ``}``    ``}``    ``if` `(chars == ``""``) {``      ``return` `"-1"``;``    ``}``    ``else` `{``      ``return` `chars;``    ``}``  ``}` `  ``public` `static` `void` `Main()``  ``{``    ``string` `a = ``"geeksforgeeks"``;``    ``string` `b = ``"geeksquiz"``;``    ``string` `result = UncommonChars(a, b);``    ``Console.Write(result);``    ``Console.Write(``"\n"``);``  ``}``}` `// This code is contributed by Aarti_Rathi`

Output

`fioqruz`

Time Complexity: O(m+n) Where m is the length of the first string and n is the length of second string.

Auxiliary Space: O(26), no any other extra space is required, so it is a constant.

This approach is contributed by Aarti_Rathi and Bibhash Ghosh.

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