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++
#include <bits/stdc++.h>using namespace std;int main(){ string str = "neveropen"; 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 |
Java
import java.util.*;class GFG{public static void main(String[] args){ String str = "neveropen"; 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 |
Python3
string="neveropen"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="") |
C#
using System;public class GFG{public static void Main(String[] args){ String str = "neveropen"; 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 |
Javascript
<script> var str = "neveropen"; 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> |
Output
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 character
- If count[x] is greater than 1, then ignore the repeated character.
Below is the implementation of above idea.
C++
// 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[] = "neveropen"; printDistinct(str); return 0;} |
Java
// 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 = "neveropen"; printDistinct(str); }}// This code is contributed by Sumit Ghosh |
Python3
# Python3 program to print distinct # characters of a string.NO_OF_CHARS = 256# Print duplicates present in the # passed string def 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 = "neveropen" printDistinct(str) # This code is contributed by ita_c |
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 = "neveropen"; printDistinct(str); }}// This code is contributed by parashar. |
Javascript
<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 = "neveropen"; 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] = 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++
// 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 = "neveropen"; printDistinct(str); return 0;} |
Java
// 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 = "neveropen"; printDistinct(str); }}// This code is contributed by Sumit Ghosh |
Python
# Python3 program to find all distinct characters# in a StringMAX_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 codeStr = "neveropen"printDistinct(Str)# This code is contributed by mohit kumar 29 |
C#
// 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 = "neveropen"; printDistinct(str); }}// This code is contributed by nitin mittal. |
Javascript
<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 = "neveropen"; printDistinct(str); // This code is contributed by avanitrachhadiya2155</script> |
Output
for
Time Complexity: O(n)
Auxiliary Space: O(n)
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