# Determine if a string has all Unique Characters

Given a string, Determine if the string has all unique characters.

Examples:

```Input : abcd10jk
Output : true

Input : hutg9mnd!nk9
Output : false
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach 1 – Brute Force technique: Run 2 loops with variable i and j. Compare str[i] and str[j]. If they become equal at any point, return false.
Time Complexity: O(n2)

## C++

```// C++ program to illustrate string
// with unique characters using
// brute force technique
#include<bits/stdc++.h>
using namespace std;

bool uniqueCharacters(string str) {

// If at any time we encounter 2
// same characters, return false
for(int i = 0; i < str.length(); i++) {
for(int j = i + 1; j < str.length(); j++) {
if(str[i] == str[j]) {
return false;
}
}
}

// If no duplicate characters encountered,
// return true
return true;
}

// driver code
int main() {
string str = "GeeksforGeeks";

if(uniqueCharacters(str)) {
cout << "The String " << str
<< " has all unique characters\n";
}
else {
cout << "The String " << str
<< " has duplicate characters\n";
}
return 0;
}
// This code is contributed by Divyam Madaan
```

## Java

```// Java program to illustrate string with
// unique characters using brute force technique
import java.util.*;

class GfG
{
boolean uniqueCharacters(String str)
{
// If at any time we encounter 2 same
// characters, return false
for (int i=0; i<str.length(); i++)
for (int j=i+1; j<str.length(); j++)
if (str.charAt(i) == str.charAt(j))
return false;

// If no duplicate characters encountered,
// return true
return true;
}

public static void main(String args[])
{
GfG obj = new GfG();
String input = "GeeksforGeeks";

if (obj.uniqueCharacters(input))
System.out.println("The String " +
input + " has all unique characters");
else
System.out.println("The String " +
input + " has duplicate characters");
}
}
```

Output:

```The String GeeksforGeeks has duplicate characters
```

Note: Please note that the program is case sensitive.

Approach 2 – Sorting: Using sorting based on ASCII values of characters
Time Complexity: O(n log n)

## C++

```// C++ program to illustrate string
// with unique characters using
// brute force technique
#include<bits/stdc++.h>
using namespace std;

bool uniqueCharacters(string str) {

// Using sorting
sort(str.begin(), str.end());

for(int i = 0; i < str.length(); i++) {

// if at any time, 2 adjacent
// elements become equal,
// return false
if(str[i] == str[i+1]) {
return false;
}
}
return true;
}

// driver code
int main() {

string str = "GeeksforGeeks";

if(uniqueCharacters(str)) {
cout << "The String " << str
<< " has all unique characters\n";
}
else {

cout << "The String " << str
<< " has duplicate characters\n";
}
return 0;
}
// This code is contributed by Divyam Madaan
```

## Java

```// Java program to check string with unique
// characters using sorting technique
import java.util.*;

class GfG
{
/* Convert the string to character array
for sorting */
boolean uniqueCharacters(String str)
{
char [] chArray = str.toCharArray();

// Using sorting
Arrays.sort(chArray);

for(int i=0; i<chArray.length-1; i++)
{
// if the adjacent elements are not
// equal, move to next element
if (chArray[i] != chArray[i+1])
continue;

// if at any time, 2 adjacent elements
// become equal, return false
else
return false;
}
return true;
}

// Driver code
public static void main(String args[])
{
GfG obj = new GfG();
String input = "GeeksforGeeks";

if (obj.uniqueCharacters(input))
System.out.println("The String " + input
+ " has all unique characters");
else
System.out.println("The String " + input
+ " has duplicate characters");
}
}
```

Output:

```The String GeeksforGeeks has duplicate characters
```

Approach 3 – Use of Extra Data Structure: This approach assumes ASCII char set(8 bits). The idea is to maintain a boolean array for the characters. The 256 indices represent 256 characters. All the array elements are initially set to false. As we iterate over the string, set true at the index equal to the int value of the character. If at any time, we encounter that the array value is already true, it means the character with that int value is repeated.

Time Complexity: O(n)

## C++

```#include <iostream>
#include <cstring>
using namespace std;

const int MAX_CHAR = 256;

bool uniqueCharacters(string str) {

// If length is greater than 265,
// some characters must have been repeated
if (str.length() > MAX_CHAR)
return false;

bool chars[MAX_CHAR] = {0};
for (int i = 0; i < str.length(); i++) {
if (chars[int(str[i])] == true)
return false;

chars[int(str[i])] = true;
}
return true;
}

// driver code
int main() {
string str = "GeeksforGeeks";

if (uniqueCharacters(str)) {
cout << "The String " << str
<< " has all unique characters\n";
}
else {

cout << "The String " << str
<< " has duplicate characters\n";
}
return 0;
}
// This code is contributed by Divyam Madaan
```

## Java

```// Java program to illustrate String With
// Unique Characters using data structure
import java.util.*;

class GfG
{
final static MAX_CHAR = 256;

boolean uniqueCharacters(String str)
{
// If length is greater than 256,
// some characters must have been repeated
if (str.length() > MAX_CHAR)
return false;

boolean[] chars = new boolean[MAX_CHAR];
Arrays.fill(chars, false);

for (int i=0; i<str.length(); i++)
{
int index = (int) str.charAt(i);

/* If the value is already true, string
has duplicate characters, return false */
if (chars[index] == true)
return false;

chars[index] = true;
}

/* No duplicates encountered, return true */
return true;
}

// Driver code
public static void main(String args[])
{
GfG obj = new GfG();
String input = "GeeksforGeeks";

if (obj.uniqueCharacters(input))
System.out.println("The String " + input +
" has all unique characters");
else
System.out.println("The String " + input +
" has duplicate characters");
}
}
```

Output:
```The String  GeeksforGeeks has all unique characters
```

Approach 4 – Without Extra Data Structure: The approach is valid for strings having alphabet as a-z. This approach is little tricky. Instead of maintaining a boolean array, we maintain an integer value called checker(32 bits). As we iterate over the string, we find the int value of the character with respect to ‘a’ with the statement int bitAtIndex = str.charAt(i)-‘a’;
Then the bit at that int value is set to 1 with the statement 1<<bitAtIndex .
Now, if this bit is already set in the checker, the bit AND operation would make checker > 0. Return false in this case.
Else Update checker to make the bit 1 at that index with the statement checker = checker | (1<<bitAtIndex);

Time Complexity: O(n)

## C++

```// C++ program to illustrate string
// with unique characters using
// brute force technique
#include<bits/stdc++.h>
using namespace std;

bool uniqueCharacters(string str) {

// Assuming string can have characters
// a-z, this has 32 bits set to 0
int checker = 0;

for (int i = 0; i < str.length(); i++) {

int bitAtIndex = str[i] - 'a';

// if that bit is already set in
// checker, return false
if ((checker & (1 << bitAtIndex)) > 0) {
return false;
}

// otherwise update and continue by
// setting that bit in the checker
checker = checker | (1<<bitAtIndex);
}

// no duplicates encountered, return true
return true;
}

// driver code
int main() {

string str = "GeeksforGeeks";

if(uniqueCharacters(str)) {
cout << "The String " << str
<< " has all unique characters\n";
}
else {
cout << "The String " << str
<< " has duplicate characters\n";
}
return 0;
}
// This code is contributed by Divyam Madaan
```

## Java

```// Java program to illustrate String with unique
// characters without using any data structure
import java.util.*;

class GfG
{
boolean uniqueCharacters(String str)
{
// Assuming string can have characters a-z
// this has 32 bits set to 0
int checker = 0;

for (int i=0; i<str.length(); i++)
{
int bitAtIndex = str.charAt(i)-'a';

// if that bit is already set in checker,
// return false
if ((checker & (1<<bitAtIndex)) > 0)
return false;

// otherwise update and continue by
// setting that bit in the checker
checker = checker | (1<<bitAtIndex);
}

// no duplicates encountered, return true
return true;
}

// Driver Code
public static void main(String args[])
{
GfG obj = new GfG();
String input = "GeekforGeeks";

if (obj.uniqueCharacters(input))
System.out.println("The String " + input
+ " has all unique characters");
else
System.out.println("The String " + input
+ " has duplicate characters");
}
}
```

Output:

```The String GeekforGeeks has duplicate characters
```

Exercise: Above program is case sensitive, you can try making same program which is case insensitive i.e Geeks and GEeks both give similar output.

Reference:
Cracking the Coding Interview by Gayle

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