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.
- 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
# 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. |
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<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; } |
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
<?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 ?> |
Javascript
<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> |
No
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" ); } } } // 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
<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" ); } // This code is contributed by umadevi9616. </script> |
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.
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