# Efficiently check if a string has all unique characters without using any additional data structure

Implement an space efficient algorithm to determine if a string (of characters from ‘a’ to ‘z’) has all unique characters or not. Use additional data structures like count array, hash, etc is not allowed.
Expected Time Complexity : O(n)

Examples :

```Input  : str = "aaabbccdaa"
Output : No

Input  : str = "abcd"
Output : Yes```

Brute Force Approach:

The brute force approach to solve this problem is to compare each character of the string with all the other characters in the string to check if it is unique or not. We can define a function that takes the input string and returns true if all the characters in the string are unique, and false otherwise.

1. Traverse the string character by character.
2. For each character, compare it with all the other characters in the string.
3. If any other character is found to be equal to the current character, return false as the string has duplicate characters.
4. If the end of the string is reached without finding any duplicate characters, return true as the string has all unique characters.

Below is the implementation of the above approach:

## C++

 `#include` `using` `namespace` `std;`   `bool` `areChractersUnique(string str)` `{` `    ``int` `n = str.length();` `    ``for``(``int` `i = 0; i < n; i++)` `        ``for``(``int` `j = i + 1; j < n; j++)` `            ``if``(str[i] == str[j])` `                ``return` `false``;` `    ``return` `true``;` `}`   `int` `main()` `{` `    ``string s = ``"aaabbccdaa"``;` `    ``if` `(areChractersUnique(s))` `        ``cout << ``"Yes"``;` `    ``else` `        ``cout << ``"No"``;` `    ``return` `0;` `}`

## Java

 `import` `java.util.*;`   `public` `class` `Main {` `  ``public` `static` `boolean` `areCharactersUnique(String str) {` `    ``int` `n = str.length();` `    ``for` `(``int` `i = ``0``; i < n; i++)` `      ``for` `(``int` `j = i + ``1``; j < n; j++)` `        ``if` `(str.charAt(i) == str.charAt(j))` `          ``return` `false``;` `    ``return` `true``;` `  ``}`   `  ``public` `static` `void` `main(String[] args) {` `    ``String s = ``"aaabbccdaa"``;` `    ``if` `(areCharactersUnique(s))` `      ``System.out.println(``"Yes"``);` `    ``else` `      ``System.out.println(``"No"``);` `  ``}` `}`

## Python3

 `# This function checks if all characters in a string are unique.` `def` `are_characters_unique(string):`   `    ``n ``=` `len``(string)` `    ``for` `i ``in` `range``(n):` `        ``for` `j ``in` `range``(i ``+` `1``, n):` `            ``if` `string[i] ``=``=` `string[j]:` `                ``return` `False` `    ``return` `True`     `if` `__name__ ``=``=` `'__main__'``:` `    ``s ``=` `"aaabbccdaa"` `    ``if` `are_characters_unique(s):` `        ``print``(``"Yes"``)` `    ``else``:` `        ``print``(``"No"``)`

## C#

 `using` `System;`   `public` `class` `GFG` `{` `    ``public` `static` `bool` `areChractersUnique(``string` `str)` `    ``{` `        ``int` `n = str.Length;` `        ``for` `(``int` `i = 0; i < n; i++)` `            ``for` `(``int` `j = i + 1; j < n; j++)` `                ``if` `(str[i] == str[j])` `                    ``return` `false``;` `        ``return` `true``;` `    ``}`   `    ``public` `static` `void` `Main()` `    ``{` `        ``string` `s = ``"aaabbccdaa"``;` `        ``if` `(areChractersUnique(s))` `            ``Console.WriteLine(``"Yes"``);` `        ``else` `            ``Console.WriteLine(``"No"``);` `    ``}` `}`

## Javascript

 `// javascript code addition `   `// This function checks if all characters in a string are unique.` `function` `areChractersUnique(str)` `{` `    `  `    ``// Find the length of the string ` `    ``let n = str.length;` `    `  `    ``// outer loop ` `    ``for``(let i = 0; i < n; i++)` `        ``// innerloop` `        ``for``(let j = i + 1; j < n; j++)` `            ``if``(str[i] == str[j])` `                ``return` `false``;` `    ``return` `true``;` `}`   `let s = ``"aaabbccdaa"``;` `if` `(areChractersUnique(s))` `    ``// Print "Yes" if all characters are unique. ` `    ``console.log(``"Yes"``);` `else` `    ``console.log(``"No"``);`   `// The code is contributed by Arushi Goel. `

Output

`No`

Time Complexity: O(N^2)

Auxiliary Space: O(1)

The idea is to use an integer variable and use bits in its binary representation to store whether a character is present or not. Typically an integer has at-least 32 bits and we need to store presence/absence of only 26 characters.

Below is the implementation of the idea.

## C++

 `// A space efficient C++ program to check if` `// all characters of string are unique.` `#include` `using` `namespace` `std;`   `// Returns true if all characters of str are` `// unique.` `// Assumptions : (1) str contains only characters` `//                   from 'a' to 'z'` `//               (2) integers are stored using 32` `//                   bits` `bool` `areChractersUnique(string str)` `{` `    ``// An integer to store presence/absence` `    ``// of 26 characters using its 32 bits.` `    ``int` `checker = 0;`   `    ``for` `(``int` `i = 0; i < str.length(); ++i)` `    ``{` `        ``int` `val = (str[i]-``'a'``);`   `        ``// If bit corresponding to current` `        ``// character is already set` `        ``if` `((checker & (1 << val)) > 0)` `            ``return` `false``;`   `        ``// set bit in checker` `        ``checker |=  (1 << val);` `    ``}`   `    ``return` `true``;` `}`   `// Driver code` `int` `main()` `{` `    ``string s = ``"aaabbccdaa"``;` `    ``if` `(areChractersUnique(s))` `        ``cout << ``"Yes"``;` `    ``else` `        ``cout << ``"No"``;` `    ``return` `0;` `}`

## Java

 `// A space efficient Java program to check if` `// all characters of string are unique.` `class` `GFG {` `        `  `    ``// Returns true if all characters of str are` `    ``// unique.` `    ``// Assumptions : (1) str contains only characters` `    ``//                 from 'a' to 'z'` `    ``//             (2) integers are stored using 32` `    ``//                 bits` `    ``static` `boolean` `areChractersUnique(String str)` `    ``{` `        `  `        ``// An integer to store presence/absence` `        ``// of 26 characters using its 32 bits.` `        ``int` `checker = ``0``;` `    `  `        ``for` `(``int` `i = ``0``; i < str.length(); ++i)` `        ``{` `            ``int` `val = (str.charAt(i)-``'a'``);` `    `  `            ``// If bit corresponding to current` `            ``// character is already set` `            ``if` `((checker & (``1` `<< val)) > ``0``)` `                ``return` `false``;` `    `  `            ``// set bit in checker` `            ``checker |= (``1` `<< val);` `        ``}` `    `  `        ``return` `true``;` `    ``}` `    `  `    ``//driver code` `    ``public` `static` `void` `main (String[] args)` `    ``{` `        ``String s = ``"aaabbccdaa"``;` `        `  `        ``if` `(areChractersUnique(s))` `            ``System.out.print(``"Yes"``);` `        ``else` `            ``System.out.print(``"No"``);` `    ``}` `}`   `// This code is contributed by Anant Agarwal.`

## Python3

 `# A space efficient Python3 program to check if ` `# all characters of string are unique`   `# Returns true if all characters of str are ` `# unique. ` `# Assumptions : (1) str contains only characters ` `#                    from 'a' to 'z' ` `#                (2) integers are stored using 32 ` `#                    bits`   `def` `areCharactersUnique(s):` `    `  `    ``# An integer to store presence/absence ` `    ``# of 26 characters using its 32 bits` `    ``checker ``=` `0` `    `  `    ``for` `i ``in` `range``(``len``(s)):` `        `  `        ``val ``=` `ord``(s[i]) ``-` `ord``(``'a'``)` `        `  `        ``# If bit corresponding to current ` `        ``# character is already set` `        ``if` `(checker & (``1` `<< val)) > ``0``:` `            ``return` `False` `        `  `        ``# set bit in checker ` `        ``checker |``=` `(``1` `<< val)` `        `  `    ``return` `True` `    `  `# Driver code` `s ``=` `"aaabbccdaa"` `if` `areCharactersUnique(s):` `    ``print``(``"Yes"``)` `else``:` `    ``print``(``"No"``)` `    `  `# This code is contributed ` `# by Mohit Kumar`

## C#

 `// A space efficient program` `// to check if all characters` `// of string are unique.` `using` `System;`   `class` `GFG {`   `    ``// Returns true if all characters` `    ``// of str are unique. Assumptions:` `    ``// (1)str contains only characters` `    ``// from 'a' to 'z'.(2)integers are` `    ``// stored using 32 bits` `    ``static` `bool` `areChractersUnique(``string` `str)` `    ``{` `        ``// An integer to store presence` `        ``// or absence of 26 characters` `        ``// using its 32 bits.` `        ``int` `checker = 0;`   `        ``for` `(``int` `i = 0; i < str.Length; ++i) {` `            ``int` `val = (str[i] - ``'a'``);`   `            ``// If bit corresponding to current` `            ``// character is already set` `            ``if` `((checker & (1 << val)) > 0)` `                ``return` `false``;`   `            ``// set bit in checker` `            ``checker |= (1 << val);` `        ``}`   `        ``return` `true``;` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `Main()` `    ``{` `        ``string` `s = ``"aaabbccdaa"``;`   `        ``if` `(areChractersUnique(s))` `            ``Console.Write(``"Yes"``);` `        ``else` `            ``Console.Write(``"No"``);` `    ``}` `}`   `// This code is contributed by Anant Agarwal.`

## PHP

 ` 0)` `            ``return` `false;`   `        ``// set bit in checker` `        ``\$checker` `|= (1 << ``\$val``);` `    ``}`   `    ``return` `true;` `}`   `// Driver code` `\$s` `= ``"aaabbccdaa"``;` `if` `(areChractersUnique(``\$s``))` `    ``echo` `"Yes"``;` `else` `    ``echo` `"No"``;`   `// This code is contributed by aj_36` `?>`

## Javascript

 ``

Output

`No`

Time Complexity : O(n)
Auxiliary Space : O(1)

Another Implementation: Using STL

## C++

 `#include ` `using` `namespace` `std;` `bool` `unique(string s) {` `    ``sort(s.begin(),s.end());` `    ``for``(``int` `i=0;i

## Java

 `import` `java.util.Arrays;`   `class` `GFG {`   `    ``static` `boolean` `unique(String s)` `    ``{` `        ``Arrays.sort(s.toCharArray());` `        ``for` `(``int` `i = ``0``; i < s.length()-``1``; i++) {` `            ``if` `(s.charAt(i) == s.charAt(i + ``1``)) {` `                ``return` `false``;` `            ``}` `        ``}` `        ``return` `true``;` `    ``}`   `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``if` `(unique(``"abcdd"``) == ``true``) {` `            ``System.out.println(``"String is Unique"``);` `        ``}` `        ``else` `{` `            ``System.out.println(``"String is not Unique"``);` `        ``}` `    ``}` `}`   `// This code is contributed by rajsanghavi9.`

## Python3

 `def` `unique(s):` `    ``s ``=` `list``(s)` `    ``s.sort()` `    ``for` `i ``in` `range` `(``len``(s) ``-` `1``):` `    `  `        ``if``(s[i] ``=``=` `s[i ``+` `1``]):` `        `  `            ``return` `False` `            ``break` `     `  `    ``return` `True`   `if``(unique(``"abcdd"``) ``=``=` `True``):` `  ``print``(``"String is Unique"``)` `    `  `else``:` `  ``print``(``"String is not Unique"``)`   `# This code is contributed by shivanisinghss2110`

## C#

 `using` `System;`   `public` `class` `GFG {`   `    ``static` `bool` `unique(String s) {` `        ``Array.Sort(s.ToCharArray());` `        ``for` `(``int` `i = 0; i < s.Length-1; i++) {` `            ``if` `(s[i] == s[i + 1]) {` `                ``return` `false``;` `            ``}` `        ``}` `        ``return` `true``;` `    ``}`   `  ``// Driver code` `    ``public` `static` `void` `Main(String[] args) {` `        ``if` `(unique(``"abcdd"``) == ``true``) {` `            ``Console.WriteLine(``"String is Unique"``);` `        ``} ``else` `{` `            ``Console.WriteLine(``"String is not Unique"``);` `        ``}` `    ``}` `}`   `// This code is contributed by umadevi9616`

## Javascript

 ``

Output

`String is not Unique`

Time Complexity : O(nlogn), where n is the length of the given string.
Auxiliary Space : O(1), no extra space is required so it is a constant.

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