GFG App
Open App
Browser
Continue

# Find uncommon characters of the two strings

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 another 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

Recommended Practice

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.

Note: In the practice area of gfg the string has to be sorted in order to match with the output.

Algorithm:

1. Take two strings str1 and str2 as input.
2. Initialize an empty string ans to store the uncommon characters.
3. Initialize a boolean vector used of size 26 to keep track of characters already visited.
4. Traverse str1 and for each character check if it is present in str2.
5. If the character is not present in str2 and not already added to ans, then add it to ans and mark it as used.
6. Traverse str2 and for each character check if it is present in str1.
7. If the character is not present in str1 and not already added to ans, then add it to ans and mark it as used.
8. Sort the ans string in lexicographical order.
9. If ans is empty, print -1. Otherwise, print the contents of ans.

Below is the implementation of the above approach:

## C++

 // C++ implementation to find the uncommon // characters of the two strings #include using namespace std;   // function to find the uncommon characters // of the two strings void findAndPrintUncommonChars(string str1, string str2) {     // to store the answer     string ans = "";             // to handle the case of duplicates     vector used(26, false);       // check first for str1     for (int i = 0; i < str1.size(); i++) {         // keeping a flag variable         bool found = false;           for (int j = 0; j < str2.size(); j++) {             // if found change the flag             // and break from loop             if (str1[i] == str2[j]) {                 found = true;                 break;             }         }           // if duplicate character not found         // then add it to ans         if (!found and !used[str1[i] - 'a']) {             used[str1[i] - 'a'] = true;             ans += str1[i];         }     }       // now check for str2     for (int i = 0; i < str2.size(); i++) {         // keeping a flag variable         bool found = false;           for (int j = 0; j < str1.size(); j++) {             // if found change the flag             // and break from loop             if (str2[i] == str1[j]) {                 found = true;                 break;             }         }           // if duplicate character not found         // then add it to ans         if (!found and !used[str2[i] - 'a']) {             used[str2[i] - 'a'] = true;             ans += str2[i];         }     }       // to match with output     sort(ans.begin(), ans.end());         // if not found any character     if (ans.size() == 0)         cout << "-1";             // else print the answer       else         cout << ans << " "; }   // Driver program to test above int main() {     string str1 = "characters";     string str2 = "alphabets";     findAndPrintUncommonChars(str1, str2);     return 0; }

## Java

 // Java implementation to find the uncommon // characters of the two strings import java.util.*;   public class Solution {     // function to find the uncommon characters   // of the two strings   static void findAndPrintUncommonChars(String str1,                                         String str2)   {     // to store the answer     String ans = "";       // to handle the case of duplicates     boolean[] used = new boolean[26];       // check first for str1     for (int i = 0; i < str1.length(); i++)     {         // keeping a flag variable       boolean found = false;         for (int j = 0; j < str2.length(); j++)       {           // if found change the flag         // and break from loop         if (str1.charAt(i) == str2.charAt(j)) {           found = true;           break;         }       }         // if duplicate character not found       // then add it to ans       if (!found && !used[str1.charAt(i) - 'a']) {         used[str1.charAt(i) - 'a'] = true;         ans += str1.charAt(i);       }     }       // now check for str2     for (int i = 0; i < str2.length(); i++)     {         // keeping a flag variable       boolean found = false;         for (int j = 0; j < str1.length(); j++)       {           // if found change the flag         // and break from loop         if (str2.charAt(i) == str1.charAt(j)) {           found = true;           break;         }       }         // if duplicate character not found       // then add it to ans       if (!found && !used[str2.charAt(i) - 'a']) {         used[str2.charAt(i) - 'a'] = true;         ans += str2.charAt(i);       }     }       // to match with output     char tempArray[] = ans.toCharArray();       // Sorting temp array using     Arrays.sort(tempArray);     ans = new String(tempArray);       // if not found any character     if (ans.length() == 0)       System.out.println("-1");       // else print the answer     else       System.out.println(ans + " ");   }     // Driver program to test above   public static void main(String[] args)   {     String str1 = "characters";     String str2 = "alphabets";     findAndPrintUncommonChars(str1, str2);   } }   // This code is contributed by karandeep1234

## C#

 // C# implementation to find the uncommon // characters of the two strings using System;   public class GFG {       // function to find the uncommon characters     // of the two strings     static void findAndPrintUncommonChars(string str1,                                           string str2)     {         // to store the answer         string ans = "";           // to handle the case of duplicates         bool[] used = new bool[26];           // check first for str1         for (int i = 0; i < str1.Length; i++) {               // keeping a flag variable             bool found = false;               for (int j = 0; j < str2.Length; j++) {                   // if found change the flag                 // and break from loop                 if (str1[i] == str2[j]) {                     found = true;                     break;                 }             }               // if duplicate character not found             // then add it to ans             if (!found && !used[str1[i] - 'a']) {                 used[str1[i] - 'a'] = true;                 ans += str1[i];             }         }           // now check for str2         for (int i = 0; i < str2.Length; i++) {               // keeping a flag variable             bool found = false;               for (int j = 0; j < str1.Length; j++) {                   // if found change the flag                 // and break from loop                 if (str2[i] == str1[j]) {                     found = true;                     break;                 }             }               // if duplicate character not found             // then add it to ans             if (!found && !used[str2[i] - 'a']) {                 used[str2[i] - 'a'] = true;                 ans += str2[i];             }         }           // to match with output         char[] tempArray = ans.ToCharArray();           // Sorting temp array using         Array.Sort(tempArray);         ans = new String(tempArray);           // if not found any character         if (ans.Length == 0)             Console.WriteLine("-1");           // else print the answer         else             Console.WriteLine(ans + " ");     }       // Driver program to test above     public static void Main(string[] args)     {         string str1 = "characters";         string str2 = "alphabets";         findAndPrintUncommonChars(str1, str2);     } }   // This code is contributed by karandeep1234

## Javascript

 // JavaScript implementation to find the uncommon // characters of the two strings   function findAndPrintUncommonChars(str1, str2) {     // to store the answer     let ans = "";       // to handle the case of duplicates     let used = new Array(26).fill(false);       // check first for str1     for (let i = 0; i < str1.length; i++) {         // keeping a flag variable         let found = false;           for (let j = 0; j < str2.length; j++) {             // if found change the flag             // and break from loop             if (str1[i] === str2[j]) {                 found = true;                 break;             }         }           // if duplicate character not found         // then add it to ans         if (!found && !used[str1.charCodeAt(i) - 97]) {             used[str1.charCodeAt(i) - 97] = true;             ans += str1[i];         }     }       // now check for str2     for (let i = 0; i < str2.length; i++) {         // keeping a flag variable         let found = false;           for (let j = 0; j < str1.length; j++) {             // if found change the flag             // and break from loop             if (str2[i] === str1[j]) {                 found = true;                 break;             }         }           // if duplicate character not found         // then add it to ans         if (!found && !used[str2.charCodeAt(i) - 97]) {             used[str2.charCodeAt(i) - 97] = true;             ans += str2[i];         }     }       // to match with output     ans = ans.split('').sort().join('');       // if not found any character     if (ans.length === 0) {         console.log("-1");     }     // else print the answer     else {         console.log(ans);     } }   // Driver program to test above let str1 = "characters"; let str2 = "alphabets"; findAndPrintUncommonChars(str1, str2);

Output

bclpr

Time Complexity: O(n1*n2)
Auxiliary Space:  O(1), as a constant-size array is used 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 the presence of each character as ‘0’ (denoting that the character is not present in both strings).
• Traverse the 1st string and mark the 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 the 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).

The 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 import java.io.*; 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(1), no any other extra space is required, so it is a constant.

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.

This approach is contributed by  Bibhash Ghosh.

Below is the implementation of the above approach:

## C++

 #include using namespace std;   string UncommonChars(string a, string b) {     int mp1[26] = {0}, mp2[26] = {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 import java.io.*; 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[26];     int[] mp2 = new int[26];     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

## Javascript

 function UncommonChars(a, b) {           let mp1 = [], mp2 = [];     for(let i = 0; i < 26; i++)     {         mp1.push(0);         mp2.push(0);     }     let n = a.length, m = b.length;       for(let i = 0; i < n; i++) {         let index = a.charCodeAt(i) - 97;       mp1[index] = 1;     }       for(let i = 0; i < m; i++)     {         let index = b.charCodeAt(i) - 97;       mp2[index] = 1;     }       let chars = "";      for(let i = 0; i < 26; ++i){       if(mp1[i]^mp2[i])           {               let char = String.fromCharCode(97+i);               chars += char;           }     }     if(chars == "")       return "-1";     else       return chars; }       let a = "geeksforgeeks";     let b = "geeksquiz";     let result = UncommonChars(a,b);     console.log(result);           // This code is contributed by garg28harsh.

## Python3

 def uncommon_chars(a: str, b: str) -> str:     mp1 = [0] * 26     mp2 = [0] * 26     n = len(a)     m = len(b)       for x in a:         mp1[ord(x) - ord('a')] = 1       for x in b:         mp2[ord(x) - ord('a')] = 1       chars = ""       for i in range(26):         if mp1[i] ^ mp2[i]:             chars += chr(i + ord('a'))       if chars == "":         return "-1"     else:         return chars     a = "geeksforgeeks" b = "geeksquiz" result = uncommon_chars(a, b) print(result)

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(1), no any other extra space is required, so it is a constant.

### Another Python-specific approach using set() and symmetric_difference().

In this approach we will convert both the strings into sets and use symmetric_diffference() to find out the uncommon characters between them.

## C++

 #include #include #include #include #include using namespace std;   int main() {     string str1 = "characters";     string str2 = "alphabets";       // Converting both the string into sets     set set1(str1.begin(), str1.end());     set set2(str2.begin(), str2.end());       // Using symmetric difference operator     string result1;     set_symmetric_difference(set1.begin(), set1.end(), set2.begin(), set2.end(),         inserter(result1, result1.begin()));     cout << result1 << endl;       // Using symmetric_difference() method     string result2;     set_symmetric_difference(set1.begin(), set1.end(), set2.begin(), set2.end(),         inserter(result2, result2.begin()));     cout << result2 << endl;       return 0; }

## Java

 import java.util.HashSet; import java.util.Set;   public class Main { public static void main(String[] args) { String str1 = "characters"; String str2 = "alphabets";       Set set1 = new HashSet<>();     Set set2 = new HashSet<>();       // Adding characters of str1 to set1     for (int i = 0; i < str1.length(); i++) {         set1.add(str1.charAt(i));     }       // Adding characters of str2 to set2     for (int i = 0; i < str2.length(); i++) {         set2.add(str2.charAt(i));     }       // Using symmetric difference operator     Set diff = new HashSet<>(set1);     diff.addAll(set2);     Set temp = new HashSet<>(set1);     temp.retainAll(set2);     diff.removeAll(temp);       StringBuilder sb = new StringBuilder();     for (char ch : diff) {         sb.append(ch);     }     String result = sb.toString();     System.out.println(result); } }

## Python3

 str1 = "characters" str2 = "alphabets"   # Converting both the string # into sets str1 = set(str1) str2 = set(str2)     # Using symmetric difference operator print("".join(sorted(str1 ^ str2)))   # using symmetric_difference() method print("".join(sorted(str1.symmetric_difference(str2))))

## Javascript

 let str1 = "characters"; let str2 = "alphabets"; let set1 = new Set(); let set2 = new Set();   // Adding characters of str1 to set1 for (let i = 0; i < str1.length; i++) {   set1.add(str1.charAt(i)); }   // Adding characters of str2 to set2 for (let i = 0; i < str2.length; i++) {   set2.add(str2.charAt(i)); }   // Using symmetric difference operator let diff = new Set([...set1, ...set2]); for (let char of set1) {   if (set2.has(char)) {     diff.delete(char);   } }   let result = [...diff].sort().join(""); console.log(result);

Output

bclpr
bclpr

Time Complexity – O(nlogn) # Just for the sorted function.
Space Complexity – O(1) # No extra space has been used

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.

My Personal Notes arrow_drop_up