Print all distinct characters of a string in order (3 Methods)

25 Jun, 2022
Given a string, find the all distinct (or non-repeating characters) in it. For example, if the input string is “Geeks for Geeks”, then output should be ‘for’ and if input string is “Geeks Quiz”, then output should be ‘GksQuiz’.The distinct characters should be printed in same order as they appear in input string.Examples:  
Input  : Geeks for Geeks
Output : for

Input  : Hello Geeks
Output : HoGks
 
Method 1 (Simple : O(n2)) A Simple Solution is to run two loops. Start traversing from left side. For every character, check if it repeats or not. If the character doesn’t repeat, increment count of non-repeating characters. When the count becomes 1, return each character.
C++
Java
Python3
C#
Javascript
#include <bits/stdc++.h>using namespace std; int main(){    string str = "GeeksforGeeks";     for (int i = 0; i < str.size(); i++)    {        int flag = 0;        for (int j = 0; j < str.size(); j++)        {            // checking if two characters are equal            if (str[i] == str[j] and i != j)            {                flag = 1;                break;            }        }        if (flag == 0)            cout << str[i];    }     return 0;} // This code is contributed by umadevi9616
import java.util.*; class GFG{ public static void main(String[] args){    String str = "GeeksforGeeks";     for (int i = 0; i < str.length(); i++)    {        int flag = 0;        for (int j = 0; j < str.length(); j++)        {            // checking if two characters are equal            if (str.charAt(i) == str.charAt(j) && i != j)            {                flag = 1;                break;            }        }        if (flag == 0)            System.out.print(str.charAt(i));    }}} // This code is contributed by gauravrajput1
string="GeeksforGeeks" for i in range(0,len(string)):    flag=0    for j in range(0,len(string)):        #checking if two characters are equal        if(string[i]==string[j] and i!=j):            flag=1            break    if(flag==0):        print(string[i],end="")
using System; public class GFG{ public static void Main(String[] args){    String str = "GeeksforGeeks";     for (int i = 0; i < str.Length; i++)    {        int flag = 0;        for (int j = 0; j < str.Length; j++)        {            // checking if two characters are equal            if (str[i] == str[j] && i != j)            {                flag = 1;                break;            }        }        if (flag == 0)            Console.Write(str[i]);    }}} // This code is contributed by gauravrajput1
<script>        var str = "GeeksforGeeks";         for (var i = 0; i < str.length; i++) {            var flag = 0;            for (j = 0; j < str.length; j++) {                // checking if two characters are equal                if (str.charAt(i) == str.charAt(j) && i != j) {                    flag = 1;                    break;                }            }            if (flag == 0)                document.write(str.charAt(i));        } // This code is contributed by gauravrajput1</script>
for
Time Complexity: O(n2) Auxiliary Space: O(1) 
Method 2 (Efficient but requires two traversals: O(n))  
Create an array count[] to store counts of characters.Traverse the input string str and do following for every character x = str[i]. Increment count[x].Traverse the input string again and do following for every character str[i]If count[x] is 1, then print the unique characterIf count[x] is greater than 1, then ignore the repeated character.
Create an array count[] to store counts of characters.
Traverse the input string str and do following for every character x = str[i]. Increment count[x].
Traverse the input string again and do following for every character str[i]If count[x] is 1, then print the unique characterIf count[x] is greater than 1, then ignore the repeated character.
If count[x] is 1, then print the unique characterIf count[x] is greater than 1, then ignore the repeated character.
If count[x] is 1, then print the unique character
If count[x] is greater than 1, then ignore the repeated character.
Below is the implementation of above idea.  
C++
Java
Python3
C#
Javascript
// C++ program to print distinct characters of a// string.# include <iostream>using namespace std;# define NO_OF_CHARS 256 /* Print duplicates present in the passed string */void printDistinct(char *str){    // Create an array of size 256 and count of    // every character in it    int count[NO_OF_CHARS];     /* Count array with frequency of characters */    int i;    for (i = 0; *(str+i); i++)        if(*(str+i)!=' ')            count[*(str+i)]++;    int n = i;     // Print characters having count more than 0    for (i = 0; i < n; i++)        if (count[*(str+i)] == 1)            cout<< str[i];} /* Driver program*/int main(){    char str[] = "GeeksforGeeks";    printDistinct(str);    return 0;}
// Java program to print distinct characters of a// string.public class GFG {    static final int NO_OF_CHARS = 256;          /* Print duplicates present in the passed string */    static void printDistinct(String str)    {        // Create an array of size 256 and count of        // every character in it        int[] count = new int[NO_OF_CHARS];              /* Count array with frequency of characters */        int i;        for (i = 0; i < str.length(); i++)            if(str.charAt(i)!=' ')                count[(int)str.charAt(i)]++;        int n = i;              // Print characters having count more than 0        for (i = 0; i < n; i++)            if (count[(int)str.charAt(i)] == 1)                System.out.print(str.charAt(i));    }          /* Driver program*/    public static void main(String args[])    {        String str = "GeeksforGeeks";        printDistinct(str);    }}// This code is contributed by Sumit Ghosh
# Python3 program to print distinct# characters of a string.NO_OF_CHARS = 256 # Print duplicates present in the# passed stringdef printDistinct(str):     # Create an array of size 256 and    # count of every character in it    count = [0] * NO_OF_CHARS     # Count array with frequency of    # characters    for i in range (len(str)):        if(str[i] != ' '):            count[ord(str[i])] += 1    n = i     # Print characters having count    # more than 0    for i in range(n):        if (count[ord(str[i])] == 1):            print (str[i], end = "") # Driver Codeif __name__ == "__main__":     str = "GeeksforGeeks"    printDistinct(str)     # This code is contributed by ita_c
// C# program to print distinct characters// of a string.using System; public class GFG {         static int NO_OF_CHARS = 256;         /* Print duplicates present in the    passed string */    static void printDistinct(String str)    {                 // Create an array of size 256 and        // count of every character in it        int[] count = new int[NO_OF_CHARS];             /* Count array with frequency of        characters */        int i;                 for (i = 0; i < str.Length; i++)            if(str[i]!=' ')                count[(int)str[i]]++;                         int n = i;             // Print characters having count        // more than 0        for (i = 0; i < n; i++)            if (count[(int)str[i]] == 1)                Console.Write(str[i]);    }         /* Driver program*/    public static void Main()    {        String str = "GeeksforGeeks";                 printDistinct(str);    }} // This code is contributed by parashar.
<script>// Javascript program to print distinct characters of a// string.     let NO_OF_CHARS = 256;         /* Print duplicates present in the passed string */    function printDistinct(str)    {        // Create an array of size 256 and count of        // every character in it        let count = new Array(NO_OF_CHARS);        for(let i=0;i<NO_OF_CHARS;i++)        {            count[i]=0;        }                /* Count array with frequency of characters */        let i;        for (i = 0; i < str.length; i++)            if(str[i]!=' ')                count[str[i].charCodeAt(0)]++;        let n = i;                 // Print characters having count more than 0        for (i = 0; i < n; i++)            if (count[str[i].charCodeAt(0)] == 1)                document.write(str[i]);    }         /* Driver program*/    let str = "GeeksforGeeks";    printDistinct(str);         // This code is contributed by rag2127</script>
Output:  
for
Time Complexity: O(n) Auxiliary Space: O(n) 
Method 3 (O(n) and requires one traversal) The idea is to use two auxiliary arrays of size 256 (Assuming that characters are stored using 8 bits).  
Initialize all values in count[] as 0 and all values in index[] as n where n is length of string.Traverse the input string str and do following for every character c = str[i]. Increment count[x].If count[x] is 1, then store index of x in index[x], i.e., index[x] = iIf count[x] is 2, then remove x from index[], i.e., index[x] = nNow index[] has indexes of all distinct characters. Sort indexes and print characters using it. Note that this step takes O(1) time assuming number of characters are fixed (typically 256)
Initialize all values in count[] as 0 and all values in index[] as n where n is length of string.
Traverse the input string str and do following for every character c = str[i]. Increment count[x].If count[x] is 1, then store index of x in index[x], i.e., index[x] = iIf count[x] is 2, then remove x from index[], i.e., index[x] = n
Increment count[x].
If count[x] is 1, then store index of x in index[x], i.e., index[x] = i
If count[x] is 2, then remove x from index[], i.e., index[x] = n
Now index[] has indexes of all distinct characters. Sort indexes and print characters using it. Note that this step takes O(1) time assuming number of characters are fixed (typically 256)
Below is the implementation of above idea.  
C++
Java
Python
C#
Javascript
// C++ program to find all distinct characters// in a string#include <bits/stdc++.h>using namespace std;const int MAX_CHAR = 256; // Function to print distinct characters in// given string str[]void printDistinct(string str){    int n = str.length();     // count[x] is going to store count of    // character 'x' in str. If x is not present,    // then it is going to store 0.    int count[MAX_CHAR];     // index[x] is going to store index of character    // 'x' in str. If x is not present or x is    // more than once, then it is going to store a value    // (for example, length of string) that cannot be    // a valid index in str[]    int index[MAX_CHAR];     // Initialize counts of all characters and indexes    // of distinct characters.    for (int i = 0; i < MAX_CHAR; i++)    {        count[i] = 0;        index[i] = n; // A value more than any index                      // in str[]    }     // Traverse the input string    for (int i = 0; i < n; i++)    {        // Find current character and increment its        // count        char x = str[i];        ++count[x];         // If this is first occurrence, then set value        // in index as index of it.        if (count[x] == 1 && x !=' ')            index[x] = i;         // If character repeats, then remove it from        // index[]        if (count[x] == 2)            index[x] = n;    }     // Since size of index is constant, below operations    // take constant time.    sort(index, index+MAX_CHAR);    for (int i=0; i<MAX_CHAR && index[i] != n; i++)       cout << str[index[i]];} // Driver codeint main(){    string str = "GeeksforGeeks";    printDistinct(str);    return 0;}
// Java program to print distinct characters of// a string.import java.util.Arrays; public class GFG {         static final int MAX_CHAR = 256;          // Function to print distinct characters in    // given string str[]    static void printDistinct(String str)    {        int n = str.length();              // count[x] is going to store count of        // character 'x' in str. If x is not present,        // then it is going to store 0.        int[] count = new int[MAX_CHAR];              // index[x] is going to store index of character        // 'x' in str. If x is not present or x is        // more than once, then it is going to store a        // value (for example, length of string) that        // cannot be a valid index in str[]        int[] index = new int[MAX_CHAR];              // Initialize counts of all characters and        // indexes of distinct characters.        for (int i = 0; i < MAX_CHAR; i++)        {            count[i] = 0;            index[i] = n; // A value more than any                          // index in str[]        }              // Traverse the input string        for (int i = 0; i < n; i++)        {            // Find current character and increment            // its count            char x = str.charAt(i);            ++count[x];                  // If this is first occurrence, then set            // value in index as index of it.            if (count[x] == 1 && x !=' ')                index[x] = i;                  // If character repeats, then remove it            // from index[]            if (count[x] == 2)                index[x] = n;        }              // Since size of index is constant, below        // operations take constant time.        Arrays.sort(index);                 for (int i = 0; i < MAX_CHAR && index[i] != n;                                                  i++)           System.out.print(str.charAt(index[i]));    }          // Driver code    public static void main(String args[])    {        String str = "GeeksforGeeks";        printDistinct(str);    }}// This code is contributed by Sumit Ghosh
# Python3 program to find all distinct characters# in a String MAX_CHAR = 256 # Function to print distinct characters in# given Str[]def printDistinct(Str):     n = len(Str)     # count[x] is going to store count of    # character 'x' in Str. If x is not present,    # then it is going to store 0.    count = [0 for i in range(MAX_CHAR)]     # index[x] is going to store index of character    # 'x' in Str. If x is not present or x is    # more than once, then it is going to store a value    # (for example, length of String) that cannot be    # a valid index in Str[]    index = [n for i in range(MAX_CHAR)]      # Traverse the input String    for i in range(n):                 # Find current character and increment its        # count        x = ord(Str[i])        count[x] += 1         # If this is first occurrence, then set value        # in index as index of it.        if (count[x] == 1 and x !=' '):            index[x] = i         # If character repeats, then remove it from        # index[]        if (count[x] == 2):            index[x] = n      # Since size of index is constant, below operations    # take constant time.    index=sorted(index)     for i in range(MAX_CHAR):        if index[i] == n:            break        print(Str[index[i]],end="") # Driver code Str = "GeeksforGeeks"printDistinct(Str) # This code is contributed by mohit kumar 29
// C# program to print distinct characters of// a string.using System; public class GFG {         static int MAX_CHAR = 256;         // Function to print distinct characters in    // given string str[]    static void printDistinct(string str)    {        int n = str.Length;             // count[x] is going to store count of        // character 'x' in str. If x is not        // present, then it is going to store 0.        int []count = new int[MAX_CHAR];             // index[x] is going to store index of        // character 'x' in str. If x is not        // present or x is more than once, then        // it is going to store a value (for        // example, length of string) that        // cannot be a valid index in str[]        int []index = new int[MAX_CHAR];             // Initialize counts of all characters        // and indexes of distinct characters.        for (int i = 0; i < MAX_CHAR; i++)        {            count[i] = 0;                         // A value more than any index            // in str[]            index[i] = n;                                 }             // Traverse the input string        for (int i = 0; i < n; i++)        {            // Find current character and            // increment its count            char x = str[i];            ++count[x];                 // If this is first occurrence, then            // set value in index as index of it.            if (count[x] == 1 && x !=' ')                index[x] = i;                 // If character repeats, then remove            // it from index[]            if (count[x] == 2)                index[x] = n;        }             // Since size of index is constant, below        // operations take constant time.        Array.Sort(index);                 for (int i = 0; i < MAX_CHAR &&                              index[i] != n; i++)        Console.Write(str[index[i]]);    }         // Driver code    public static void Main()    {        string str = "GeeksforGeeks";        printDistinct(str);    }} // This code is contributed by nitin mittal.
<script>// Javascript program to print distinct characters of// a string.         let MAX_CHAR = 256;         // Function to print distinct characters in    // given string str[]    function printDistinct(str)    {        let n = str.length;               // count[x] is going to store count of        // character 'x' in str. If x is not present,        // then it is going to store 0.        let count = new Array(MAX_CHAR);               // index[x] is going to store index of character        // 'x' in str. If x is not present or x is        // more than once, then it is going to store a        // value (for example, length of string) that        // cannot be a valid index in str[]        let index = new Array(MAX_CHAR);               // Initialize counts of all characters and        // indexes of distinct characters.        for (let i = 0; i < MAX_CHAR; i++)        {            count[i] = 0;            index[i] = n; // A value more than any                          // index in str[]        }                      // Traverse the input string        for (let i = 0; i < n; i++)        {            // Find current character and increment            // its count            let x = str[i].charCodeAt(0);            ++count[x];                   // If this is first occurrence, then set            // value in index as index of it.            if (count[x] == 1 && x !=' ')                index[x] = i;                   // If character repeats, then remove it            // from index[]            if (count[x] == 2)                index[x] = n;        }               // Since size of index is constant, below        // operations take constant time.        index.sort(function(a,b){return a-b});                  for (let i = 0; i < MAX_CHAR && index[i] != n;                                                  i++)           document.write(str[index[i]]);    }         // Driver code    let str = "GeeksforGeeks";    printDistinct(str);              // This code is contributed by avanitrachhadiya2155</script>
for
Time Complexity: O(n) Auxiliary Space: O(n) 
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parashar
nitin mittal
ukasp
mohit kumar 29
rag2127
avanitrachhadiya2155
geeky_bhaskar
pulamolusaimohan
umadevi9616
GauravRajput1
kapoorsagar226
simranarora5sos
sachinvinod1904
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