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Find the duplicate characters in a string in O(1) space

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  • Difficulty Level : Medium
  • Last Updated : 29 Jan, 2022
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Given a string str, the task is to find all the duplicate characters present in a given string in lexicographical order without using any additional data structure.

Examples:

Input: str = “geeksforgeeks” 
Output: e g k s 
Explanation: 
Frequency of character ‘g’ = 2 
Frequency of character ‘e’ = 4 
Frequency of character ‘k’ = 2 
Frequency of character ‘s’ = 2 
Therefore, the required output is e g k s.

Input: str = “apple” 
Output:
Explanation: 
Frequency of character ‘p’ = 2. 
Therefore, the required output is p.

Approach: Follow the steps below to solve the problem:

Below is the implementation of the above approach:

C++




// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find duplicate characters
// in string without using any additional
// data structure
void findDuplicate(string str, int N)
{
 
    // Check if (i + 'a') is present
    // in str at least once or not.
    int first = 0;
 
    // Check if (i + 'a') is present
    // in str at least twice or not.
    int second = 0;
 
    // Iterate over the characters
    // of the string str
    for (int i = 0; i < N; i++) {
 
        // If str[i] has already occurred in str
        if (first & (1 << (str[i] - 'a'))) {
 
            // Set (str[i] - 'a')-th bit of second
            second
                = second | (1 << (str[i] - 'a'));
        }
        else {
 
            // Set (str[i] - 'a')-th bit of second
            first
                = first | (1 << (str[i] - 'a'));
        }
    }
 
    // Iterate over the range [0, 25]
    for (int i = 0; i < 26; i++) {
 
        // If i-th bit of both first
        // and second is Set
        if ((first & (1 << i))
            && (second & (1 << i))) {
 
            cout << char(i + 'a') << " ";
        }
    }
}
 
// Driver Code
int main()
{
    string str = "geeksforgeeks";
    int N = str.length();
 
    findDuplicate(str, N);
}

Java




// Java program for the above approach
public class GFG
{
 
  // Function to find duplicate characters
  // in string without using any additional
  // data structure
  static void findDuplicate(String str, int N)
  {
 
    // Check if (i + 'a') is present
    // in str at least once or not.
    int first = 0;
 
    // Check if (i + 'a') is present
    // in str at least twice or not.
    int second = 0;
 
    // Iterate over the characters
    // of the string str
    for (int i = 0; i < N; i++)
    {
 
      // If str[i] has already occurred in str
      if ((first & (1 << (str.charAt(i) - 'a'))) != 0)
      {
 
        // Set (str[i] - 'a')-th bit of second
        second
          = second | (1 << (str.charAt(i) - 'a'));
      }
      else
      {
 
        // Set (str[i] - 'a')-th bit of second
        first
          = first | (1 << (str.charAt(i) - 'a'));
      }
    }
 
    // Iterate over the range [0, 25]
    for (int i = 0; i < 26; i++)
    {
 
      // If i-th bit of both first
      // and second is Set
      if (((first & (1 << i))
           & (second & (1 << i))) != 0) {
 
        System.out.print((char)(i + 'a') + " ");
      }
    }
  }
 
  // Driver Code
  static public void main(String args[])
  {
    String str = "geeksforgeeks";
    int N = str.length();
 
    findDuplicate(str, N);
  }
}
 
// This code is contributed by AnkThon.

Python3




# Python 3 code added. program to implement
# the above approach
 
# Function to find duplicate characters
# in string without using any additional
# data structure
def findDuplicate(str1, N):
   
    # Check if (i + 'a') is present
    # in str1 at least once or not.
    first = 0
 
    # Check if (i + 'a') is present
    # in str1 at least twice or not.
    second = 0
 
    # Iterate over the characters
    # of the string str1
    for i in range(N):
       
        # If str1[i] has already occurred in str1
        if (first & (1 << (ord(str1[i]) - 97))):
           
            # Set (str1[i] - 'a')-th bit of second
            second = second | (1 << (ord(str1[i]) - 97))
        else:
            # Set (str1[i] - 'a')-th bit of second
            first = first | (1 << (ord(str1[i]) - 97))
 
    # Iterate over the range [0, 25]
    for i in range(26):
       
        # If i-th bit of both first
        # and second is Set
        if ((first & (1 << i)) and (second & (1 << i))):
            print(chr(i + 97), end = " ")
 
# Driver Code
if __name__ == '__main__':
    str1 = "geeksforgeeks"
    N = len(str1)
    findDuplicate(str1, N)
 
    # This code is contributed by SURENDRA_GANGWAR.

C#




// C# program for the above approach
using System;
class GFG
{
 
  // Function to find duplicate characters
  // in string without using any additional
  // data structure
  static void findDuplicate(string str, int N)
  {
 
    // Check if (i + 'a') is present
    // in str at least once or not.
    int first = 0;
 
    // Check if (i + 'a') is present
    // in str at least twice or not.
    int second = 0;
 
    // Iterate over the characters
    // of the string str
    for (int i = 0; i < N; i++) {
 
      // If str[i] has already occurred in str
      if ((first & (1 << (str[i] - 'a'))) != 0)
      {
 
        // Set (str[i] - 'a')-th bit of second
        second
          = second | (1 << (str[i] - 'a'));
      }
      else
      {
 
        // Set (str[i] - 'a')-th bit of second
        first
          = first | (1 << (str[i] - 'a'));
      }
    }
 
    // Iterate over the range [0, 25]
    for (int i = 0; i < 26; i++)
    {
 
      // If i-th bit of both first
      // and second is Set
      if (((first & (1 << i))
           & (second & (1 << i))) != 0) {
 
        Console.Write((char)(i + 'a') + " ");
      }
    }
  }
 
  // Driver Code
  static public void Main()
  {
    string str = "geeksforgeeks";
    int N = str.Length;
 
    findDuplicate(str, N);
  }
}
 
// This code is contributed by susmitakundugoaldanga.

Javascript




<script>
 
// Javascript program for the above approach
 
// Function to find duplicate characters
// in string without using any additional
// data structure
function findDuplicate(str, N)
{
 
    // Check if (i + 'a') is present
    // in str at least once or not.
    let first = 0;
     
    // Check if (i + 'a') is present
    // in str at least twice or not.
    let second = 0;
     
    // Iterate over the characters
    // of the string str
    for(let i = 0; i < N; i++)
    {
         
        // If str[i] has already occurred in str
        if ((first & (1 << (str[i].charCodeAt() -
                               'a'.charCodeAt()))) != 0)
        {
         
            // Set (str[i] - 'a')-th bit of second
            second = second | (1 << (str[i].charCodeAt() -
                                        'a'.charCodeAt()));
        }
        else
        {
             
            // Set (str[i] - 'a')-th bit of second
            first = first | (1 << (str[i].charCodeAt() -
                                      'a'.charCodeAt()));
        }
    }
     
    // Iterate over the range [0, 25]
    for(let i = 0; i < 26; i++)
    {
         
        // If i-th bit of both first
        // and second is Set
        if (((first & (1 << i)) &
            (second & (1 << i))) != 0)
        {
            document.write(String.fromCharCode(
                i + 'a'.charCodeAt()) + " ");
        }
    }
}
 
// Driver code
let str = "geeksforgeeks";
let N = str.length;
 
findDuplicate(str, N);
 
// This code is contributed by divyesh072019
 
</script>

Output: 

e g k s

 

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


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