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.
- Traverse the string character by character.
- For each character, compare it with all the other characters in the string.
- If any other character is found to be equal to the current character, return false as the string has duplicate characters.
- 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<bits/stdc++.h>
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
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
function areChractersUnique(str)
{
let n = str.length;
for (let i = 0; i < n; i++)
for (let j = i + 1; j < n; j++)
if (str[i] == str[j])
return false ;
return true ;
}
let s = "aaabbccdaa" ;
if (areChractersUnique(s))
console.log( "Yes" );
else
console.log( "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++
#include<bits/stdc++.h>
using namespace std;
bool areChractersUnique(string str)
{
int checker = 0;
for ( int i = 0; i < str.length(); ++i)
{
int val = (str[i]- 'a' );
if ((checker & (1 << val)) > 0)
return false ;
checker |= (1 << val);
}
return true ;
}
int main()
{
string s = "aaabbccdaa" ;
if (areChractersUnique(s))
cout << "Yes" ;
else
cout << "No" ;
return 0;
}
|
Java
class GFG {
static boolean areChractersUnique(String str)
{
int checker = 0 ;
for ( int i = 0 ; i < str.length(); ++i)
{
int val = (str.charAt(i)- 'a' );
if ((checker & ( 1 << val)) > 0 )
return false ;
checker |= ( 1 << val);
}
return true ;
}
public static void main (String[] args)
{
String s = "aaabbccdaa" ;
if (areChractersUnique(s))
System.out.print( "Yes" );
else
System.out.print( "No" );
}
}
|
Python3
def areCharactersUnique(s):
checker = 0
for i in range ( len (s)):
val = ord (s[i]) - ord ( 'a' )
if (checker & ( 1 << val)) > 0 :
return False
checker | = ( 1 << val)
return True
s = "aaabbccdaa"
if areCharactersUnique(s):
print ( "Yes" )
else :
print ( "No" )
|
C#
using System;
class GFG {
static bool areChractersUnique( string str)
{
int checker = 0;
for ( int i = 0; i < str.Length; ++i) {
int val = (str[i] - 'a' );
if ((checker & (1 << val)) > 0)
return false ;
checker |= (1 << val);
}
return true ;
}
public static void Main()
{
string s = "aaabbccdaa" ;
if (areChractersUnique(s))
Console.Write( "Yes" );
else
Console.Write( "No" );
}
}
|
PHP
<?php
function areChractersUnique( $str )
{
$checker = 0;
for ( $i = 0; $i < $len = strlen ( $str ); ++ $i )
{
$val = ( $str [ $i ] - 'a' );
if (( $checker & (1 << $val )) > 0)
return false;
$checker |= (1 << $val );
}
return true;
}
$s = "aaabbccdaa" ;
if (areChractersUnique( $s ))
echo "Yes" ;
else
echo "No" ;
?>
|
Javascript
<script>
function areChractersUnique(str)
{
let checker = 0;
for (let i = 0; i < str.length; ++i)
{
let val = (str[i]- 'a' );
if ((checker & (1 << val)) > 0)
return false ;
checker |= (1 << val);
}
return true ;
}
var s = "aaabbccdaa" ;
if (areChractersUnique(s))
document.write( "Yes" );
else
document.write( "No" );
</script>
|
Time Complexity : O(n)
Auxiliary Space : O(1)
Another Implementation: Using STL
C++
#include <bits/stdc++.h>
using namespace std;
bool unique(string s) {
sort(s.begin(),s.end());
for ( int i=0;i<s.size()-1;i++)
{
if (s[i]==s[i+1])
{
return false ;
break ;
}
}
return true ;
}
int main() {
if (unique( "abcdd" )== true )
{
cout << "String is Unique" <<endl;
}
else
{
cout << "String is not Unique" <<endl;
}
return 0;
}
|
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" );
}
}
}
|
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" )
|
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 ;
}
public static void Main(String[] args) {
if (unique( "abcdd" ) == true ) {
Console.WriteLine( "String is Unique" );
} else {
Console.WriteLine( "String is not Unique" );
}
}
}
|
Javascript
<script>
function unique(s) {
for ( var i = 0; i < s.length-1; i++) {
if (s.charAt(i) == s.charAt(i + 1)) {
return false ;
}
}
return true ;
}
if (unique( "abcdd" ) == true ) {
document.write( "String is Unique" );
} else {
document.write( "String is not Unique" );
}
</script>
|
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.
This article is contributed by Mr. Somesh Awasthi. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.