Efficiently check if a string has all unique characters without using any additional data structure
Last Updated :
20 Jul, 2024
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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:
#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;
}
public class Main {
public static boolean areCharactersUnique(String str) {
boolean[] seen = new boolean[256]; // Assuming ASCII characters
for (char c : str.toCharArray()) {
if (seen[c]) {
return false; // Character already seen
}
seen[c] = true;
}
return true; // No duplicate characters found
}
public static void main(String[] args) {
String s = "abca";
if (areCharactersUnique(s)) {
System.out.println("Yes");
} else {
System.out.println("No");
}
}
}
# 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")
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 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.
// A space efficient C++ program to check if
// all characters of string are unique.
#include<bits/stdc++.h>
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;
}
// 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.
# 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
// 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.
<script>
// Javascript program for the above approach
// 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
function areChractersUnique(str)
{
// An integer to store presence/absence
// of 26 characters using its 32 bits.
let checker = 0;
for (let i = 0; i < str.length; ++i)
{
let 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
var s = "aaabbccdaa";
if (areChractersUnique(s))
document.write("Yes");
else
document.write("No");
</script>
<?php
// A space efficient PHP 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
function areChractersUnique($str)
{
// An integer to store presence/absence
// of 26 characters using its 32 bits.
$checker = 0;
for ($i = 0; $i < $len = strlen($str); ++$i)
{
$val = ($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
$s = "aaabbccdaa";
if (areChractersUnique($s))
echo "Yes";
else
echo "No";
// This code is contributed by aj_36
?>
Output
No
Time Complexity : O(n)
Auxiliary Space : O(1)
Another Implementation: Using STL
#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("abca")==true)
{
cout <<"String is Unique"<<endl;
}
else
{
cout <<"String is not Unique"<<endl;
}
return 0;
}
import java.util.HashSet;
class GFG {
static boolean unique(String s)
{
HashSet<Character> set = new HashSet<>();
for (int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
if (set.contains(c)) {
return false;
}
set.add(c);
}
return true;
}
public static void main(String[] args)
{
if (unique("abca")) {
System.out.println("String is Unique");
}
else {
System.out.println("String is not Unique");
}
}
}
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("abca") == True):
print("String is Unique")
else:
print("String is not Unique")
# This code is contributed by shivanisinghss2110
using System;
public class GFG {
static bool Unique(string s) {
// Convert string to char array and sort it
char[] charArray = s.ToCharArray();
Array.Sort(charArray);
// Check for adjacent duplicates
for (int i = 0; i < charArray.Length - 1; i++) {
if (charArray[i] == charArray[i + 1]) {
return false;
}
}
return true;
}
// Driver code
public static void Main(string[] args) {
if (Unique("ab")) {
Console.WriteLine("String is Unique");
} else {
Console.WriteLine("String is not Unique");
}
}
}
function unique(s)
{
let charSet = new Set();
for (let char of s) {
if (charSet.has(char)) {
return false;
}
charSet.add(char);
}
return true;
}
// Driver code
if (unique("abcdd")) {
console.log("String is Unique");
}
else {
console.log("String is not Unique");
}
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.
