# First non-repeating character using one traversal of string | Set 2

• Difficulty Level : Easy
• Last Updated : 29 Aug, 2022

Given a string, find the first non-repeating character in it. For example, if the input string is “GeeksforGeeks”, then output should be ‘f’ and if input string is “GeeksQuiz”, then output should be ‘G’. We have discussed two solutions in Given a string, find its first non-repeating character . In this post, a further optimized solution (over method 2 of previous post) is discussed. The idea is to optimize space. Instead of using a pair to store count and index, we use single element that stores index if an element appears once, else stores a negative value.

Implementation:

## C++

 `// CPP program to find first non-repeating``// character using 1D array and one traversal.``#include ``using` `namespace` `std;``#define NO_OF_CHARS 256` `/* The function returns index of the first``non-repeating character in a string. If``all characters are repeating then``returns INT_MAX */``int` `firstNonRepeating(``char``* str)``{``  ` `    ``// Initialize all characters as``    ``// absent.``    ``int` `arr[NO_OF_CHARS];``    ``for` `(``int` `i = 0; i < NO_OF_CHARS; i++)``        ``arr[i] = -1;` `    ``// After below loop, the value of``    ``// arr[x] is going to be index of``    ``// of x if x appears only once. Else``    ``// the value is going to be either``    ``// -1 or -2.``    ``for` `(``int` `i = 0; str[i]; i++) {``        ``if` `(arr[str[i]] == -1)``            ``arr[str[i]] = i;``        ``else``            ``arr[str[i]] = -2;``    ``}` `    ``int` `res = INT_MAX;``    ``for` `(``int` `i = 0; i < NO_OF_CHARS; i++)` `        ``// If this character occurs only``        ``// once and appears before the``        ``// current result, then update the``        ``// result``        ``if` `(arr[i] >= 0)``            ``res = min(res, arr[i]);` `    ``return` `res;``}` `/* Driver program to test above function */``int` `main()``{``    ``char` `str[] = ``"geeksforgeeks"``;``    ``int` `index = firstNonRepeating(str);``    ``if` `(index == INT_MAX)``        ``cout <<``"Either all characters are "``               ``"repeating or string is empty"``;``    ``else``        ``cout <<``"First non-repeating character"``               ``" is "``<< str[index];``    ``return` `0;``}` `// This code is contributed by shivanisinghss210`

## C

 `// CPP program to find first non-repeating``// character using 1D array and one traversal.``#include ``#include ``#include ``#define NO_OF_CHARS 256` `/* The function returns index of the first``non-repeating character in a string. If``all characters are repeating then``returns INT_MAX */``int` `firstNonRepeating(``char``* str)``{``    ``// Initialize all characters as``    ``// absent.``    ``int` `arr[NO_OF_CHARS];``    ``for` `(``int` `i = 0; i < NO_OF_CHARS; i++)``        ``arr[i] = -1;` `    ``// After below loop, the value of``    ``// arr[x] is going to be index of``    ``// of x if x appears only once. Else``    ``// the value is going to be either``    ``// -1 or -2.``    ``for` `(``int` `i = 0; str[i]; i++) {``        ``if` `(arr[str[i]] == -1)``            ``arr[str[i]] = i;``        ``else``            ``arr[str[i]] = -2;``    ``}` `    ``int` `res = INT_MAX;``    ``for` `(``int` `i = 0; i < NO_OF_CHARS; i++)` `        ``// If this character occurs only``        ``// once and appears before the``        ``// current result, then update the``        ``// result``        ``if` `(arr[i] >= 0)``            ``res = min(res, arr[i]);` `    ``return` `res;``}` `/* Driver program to test above function */``int` `main()``{``    ``char` `str[] = ``"geeksforgeeks"``;``    ``int` `index = firstNonRepeating(str);``    ``if` `(index == INT_MAX)``        ``printf``(``"Either all characters are "``               ``"repeating or string is empty"``);``    ``else``        ``printf``(``"First non-repeating character"``               ``" is %c"``, str[index]);``    ``return` `0;``}`

## Java

 `// Java program to find first``// non-repeating character``// using 1D array and one``// traversal.``import` `java.io.*;``import` `java.util.*;``import` `java.lang.*;` `class` `GFG``{``/* The function returns index``of the first non-repeating``character in a string. If``all characters are repeating``then returns INT_MAX */``static` `int` `firstNonRepeating(String str)``{``    ``int` `NO_OF_CHARS = ``256``;``    ` `    ``// Initialize all characters``    ``// as absent.``    ``int` `arr[] = ``new` `int``[NO_OF_CHARS];``    ``for` `(``int` `i = ``0``;``             ``i < NO_OF_CHARS; i++)``        ``arr[i] = -``1``;` `    ``// After below loop, the``    ``// value of arr[x] is going``    ``// to be index of x if x``    ``// appears only once. Else``    ``// the value is going to be``    ``// either -1 or -2.``    ``for` `(``int` `i = ``0``;``             ``i < str.length(); i++)``    ``{``        ``if` `(arr[str.charAt(i)] == -``1``)``            ``arr[str.charAt(i)] = i;``        ``else``            ``arr[str.charAt(i)] = -``2``;``    ``}` `    ``int` `res = Integer.MAX_VALUE;``    ``for` `(``int` `i = ``0``; i < NO_OF_CHARS; i++)` `        ``// If this character occurs``        ``// only once and appears before``        ``// the current result, then``        ``// update the result``        ``if` `(arr[i] >= ``0``)``            ``res = Math.min(res, arr[i]);` `    ``return` `res;``}` `// Driver Code``public` `static` `void` `main(String args[])``{``    ``String str = ``"geeksforgeeks"``;``    ` `    ``int` `index = firstNonRepeating(str);``    ``if` `(index == Integer.MAX_VALUE)``        ``System.out.print(``"Either all characters are "` `+``                       ``"repeating or string is empty"``);``    ``else``        ``System.out.print(``"First non-repeating character"``+``                             ``" is "` `+ str.charAt(index));``}``}`

## Python3

 `'''``Python3 implementation to find non repeating character using``1D array and one traversal'''``import` `math as mt` `NO_OF_CHARS ``=` `256` `'''``The function returns index of the first``non-repeating character in a string. If``all characters are repeating then``returns INT_MAX '''` `def` `firstNonRepeating(string):``    ``#initialize all character as absent``    ` `    ``arr``=``[``-``1` `for` `i ``in` `range``(NO_OF_CHARS)]``    ` `    ``'''``    ``After below loop, the value of``    ``arr[x] is going to be index of``    ``of x if x appears only once. Else``    ``the value is going to be either``    ``-1 or -2.'''``    ` `    ``for` `i ``in` `range``(``len``(string)):``        ``if` `arr[``ord``(string[i])]``=``=``-``1``:``            ``arr[``ord``(string[i])]``=``i``        ``else``:``            ``arr[``ord``(string[i])]``=``-``2``    ``res``=``10``*``*``18``    ` `    ``for` `i ``in` `range``(NO_OF_CHARS):``        ``'''``        ``If this character occurs only``        ``once and appears before the``        ``current result, then update the``        ``result'''``        ``if` `arr[i]>``=``0``:``            ``res``=``min``(res,arr[i])``    ``return` `res` `#Driver program to test above function` `string``=``"geeksforgeeks"` `index``=``firstNonRepeating(string)` `if` `index``=``=``10``*``*``18``:``    ``print``(``"Either all characters are repeating or string is empty"``)``else``:``    ``print``(``"First non-repeating character is"``,string[index])` `#this code is contributed by Mohit Kumar 29``           `

## C#

 `// C# program to find first``// non-repeating character``// using 1D array and one``// traversal.``using` `System;` `class` `GFG``{``    ``/* The function returns index``    ``of the first non-repeating``    ``character in a string. If``    ``all characters are repeating``    ``then returns INT_MAX */``    ``static` `int` `firstNonRepeating(String str)``    ``{``        ``int` `NO_OF_CHARS = 256;` `        ``// Initialize all characters``        ``// as absent.``        ``int` `[]arr = ``new` `int``[NO_OF_CHARS];``        ``for` `(``int` `i = 0; i < NO_OF_CHARS; i++)``            ``arr[i] = -1;` `        ``// After below loop, the``        ``// value of arr[x] is going``        ``// to be index of x if x``        ``// appears only once. Else``        ``// the value is going to be``        ``// either -1 or -2.``        ``for` `(``int` `i = 0; i < str.Length; i++)``        ``{``            ``if` `(arr[str[i]] == -1)``                ``arr[str[i]] = i;``            ``else``                ``arr[str[i]] = -2;``        ``}` `        ``int` `res = ``int``.MaxValue;``        ``for` `(``int` `i = 0; i < NO_OF_CHARS; i++)` `            ``// If this character occurs``            ``// only once and appears before``            ``// the current result, then``            ``// update the result``            ``if` `(arr[i] >= 0)``                ``res = Math.Min(res, arr[i]);` `        ``return` `res;``    ``}` `    ``// Driver Code``    ``public` `static` `void` `Main()``    ``{``        ``String str = ``"geeksforgeeks"``;``        ``int` `index = firstNonRepeating(str);``        ``if` `(index == ``int``.MaxValue)``            ``Console.Write(``"Either all characters are "` `+``                        ``"repeating or string is empty"``);``        ``else``            ``Console.Write(``"First non-repeating character"``+``                                ``" is "` `+ str[index]);``    ``}``}` `// This code is contributed by Rajput-Ji`

## Javascript

 ``

Output

`First non-repeating character is f`

Time Complexity: O(n)

Alternate Implementation:

This is coded using a HashMap or Hashing Technique.

If the element(or key) repeats in the string the HashMap (or Dictionary) will change the value of that key to None.

This way we will later on only be finding keys whose value is “not None”.

Implementation:

## Python3

 `# Python implementation of``# above approach``from` `collections ``import` `Counter`  `def` `makeHashMap(string):` `    ``# Use Counter to make a frequency``    ``# dictionary of characters in a string.` `    ``d1 ``=` `Counter(string)` `    ``# Update value of each key such that``    ``# if frequency of  frequency of  a key``    ``# is equal to 1, then it is set to 0,``    ``# else set the value equal to None``    ``d1 ``=` `{(key): (``0` `if` `d1[key] ``=``=` `1` `else` `None``)``          ``for` `key, value ``in` `d1.items()}` `    ``return` `d1`  `def` `firstNotRepeatingCharacter(s):` `    ``d ``=` `makeHashMap(s)` `    ``# variable to store the first``    ``# non repeating character.``    ``nonRep ``=` `None` `    ``# Traversing through the string only once.``    ``for` `i ``in` `s:``        ``if` `d[i] ``is` `not` `None``:` `            ``'''``            ``As soon as the first character in the string is``            ``found whose value in the HashMap is "not None",``            ``that is our first non-repeating character.``            ``So we store it in nonRep and break out of the``            ``loop. Thus saving on time.``            ``'''``            ``nonRep ``=` `i``            ``break` `    ``if` `nonRep ``is` `not` `None``:``        ``return` `nonRep``    ``else``:` `        ``# If there are no non-repeating characters``        ``# then "_" is returned``        ``return` `str``(``"_"``)`  `# Driver Code``print``(firstNotRepeatingCharacter(``'bbcdcca'``))` `# This code is contributed by Vivek Nigam (@viveknigam300)`

Output

`d`

Time Complexity: O(N)

#### Method 3:  Another simple approach to this problem without using any hashmap or array is mentioned below. We can find the first non-repeating character by just using single for loop.

Another Approach:

To count the frequency of character we can do the following step:

frequency of a character = length_of_string – length_of_string_without_that _character

for example: Give String is “helloh” and we want to count frequency of character “h” so by using the above formula we can say

frequency of character “h” = 6(length of string) – 4(length of string without “h”)  = 2

So by this way, we can count the frequency of every character in a string and then if we found count == 1 that means that character is the first non-repeating character in the string.

Implementation: Implementation of the above method in java is shown below:

## Java

 `// Java program for the above approach``public` `class` `first_non_repeating {` `    ` `    ``static` `int` `firstNonRepeating(String str)``    ``{``        ``boolean` `flag = ``false``;` `        ``int` `index = -``1``;` `        ``for` `(``int` `i = ``0``; i < s.length(); i++) {` `            ``// Here I am replacing a character with "" and``            ``// then finding the length after replacing and``            ``// then subtracting  length of that replaced``            ``// string from the total length of the original``            ``// string``            ``int` `count_occurrence``                ``= s.length()``                  ``- s.replace(``                         ``Character.toString(s.charAt(i)),``                         ``""``)``                        ``.length();` `            ``if` `(count_occurrence == ``1``) {` `                ``flag = ``true``;` `                ``index = i;` `                ``break``;``            ``}``        ``}` `        ``if` `(flag)``            ``System.out.println(``                ``"First non repeating character is "``                ``+ s.charAt(index));` `        ``else``            ``System.out.println(``                ``"There is no non-repeating character``                         ``is present in the string");``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String arg[])``    ``{` `        ``String s = ``"GeeksforGeeks"``;` `        ``firstNonRepeating(s);``    ``}``}``// This Solution is contributed by Sourabh Sharma.`

## Python3

 `# Python implementation of above approach``def` `firstNonRepeating(s):` `    ``flag ``=` `False``    ``index ``=` `-``1``    ``for` `i ``in` `range``(``len``(s)):` `        ``# Here I am replacing a character with "" and``        ``# then finding the length after replacing and``        ``# then subtracting  length of that replaced``        ``# string from the total length of the original``        ``# string``        ``count_occurrence ``=` `len``(s) ``-` `len``(s.replace(s[i],''))` `        ``if` `(count_occurrence ``=``=` `1``):``            ``flag ``=` `True``            ``index ``=` `i``            ``break` `    ``if` `(flag):``        ``print``(f``"First non repeating character is {s[index]}"``)` `    ``else``:``        ``print``(``"There is no non-repeating character is present in the string"``)` `# Driver Code``s ``=` `"GeeksforGeeks"``firstNonRepeating(s)` `# This code is contributed by shinjanpatra`

## Javascript

 ``

Output

`First non repeating character is f`

Complexity Analysis:

• Time Complexity: O(N)
• Auxiliary Space: O(1)

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