Efficiently find first repeated character in a string without using any additional data structure in one traversal

Implement a space efficient algorithm to check First repeated character in a string without using any additional data structure in one traversal. Use additional data structures like count array, hash, etc is not allowed.

Examples :

Input :  abcfdeacf
Output : char = a, index= 6

The idea is to use an integer variable and uses 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.

Implementation:

C++

// Efficiently check First repeated character
// in C++ program
#include<bits/stdc++.h>
using namespace std;

// Returns -1 if all characters of str are
// unique.
// Assumptions : (1) str contains only characters
//                 from 'a' to 'z'
//             (2) integers are stored using 32
//                 bits
int FirstRepeated(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 i;

        // set bit in checker
        checker |= (1 << val);
    }

    return -1;
}

// Driver code
int main()
{
    string s = "abcfdeacf";
    int i=FirstRepeated(s);
    if (i!=-1)
        cout <<"Char = "<< s[i] << "   and Index = "<<i;
    else
        cout << "No repeated Char";
    return 0;
}

Java

// Efficiently check First repeated character
// in Java program
public class First_Repeated_char {

    static int FirstRepeated(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 i;

            // set bit in checker
            checker |= (1 << val);
        }

        return -1;
    }

    // Driver code
    public static void main(String args[])
    {
        String s = "abcfdeacf";
        int i=FirstRepeated(s);
        if (i!=-1)
           System.out.println("Char = "+ s.charAt(i) + "   and Index = "+i);
        else
            System.out.println( "No repeated Char");
    }
}
// This code is contributed by Sumit Ghosh

Python3

# Efficiently check First repeated character
# in Python

# Returns -1 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 FirstRepeated(string):

    # An integer to store presence/absence
    # of 26 characters using its 32 bits.
    checker = 0

    pos = 0
    for i in string:
        val = ord(i) - ord('a');

        # If bit corresponding to current
        # character is already set
        if ((checker & (1 << val)) > 0):
            return pos

        # set bit in checker
        checker |= (1 << val)
        pos += 1

    return -1

# Driver code
string = "abcfdeacf"
i = FirstRepeated(string)
if i != -1:
    print ("Char = ", string[i], " and Index = ", i)
else:
    print ("No repeated Char")

# This code is contributed by Sachin Bisht

C#

// C# program to Efficiently
// check First repeated character
using System;

public class First_Repeated_char {

    static int FirstRepeated(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 i;

            // set bit in checker
            checker |= (1 << val);
        }

        return -1;
    }

    // Driver code
    public static void Main()
    {
        string s = "abcfdeacf";
        int i=FirstRepeated(s);
        if (i!=-1)
           Console.WriteLine("Char = " + s[i] +
                          " and Index = " + i);
        else
            Console.WriteLine( "No repeated Char");
    }
}

// This code is contributed by vt_m.

PHP

<?php
// Efficiently check First repeated character
// in PHP program

// Returns -1 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 FirstRepeated($str)
{
    // An integer to store presence/absence
    // of 26 characters using its 32 bits.
    $checker = 0;

    for ($i = 0; $i < strlen($str); ++$i)
    {
        $val = (ord($str[$i]) - ord('a'));

        // If bit corresponding to current
        // character is already set
        if (($checker & (1 << $val)) > 0)
            return $i;

        // set bit in checker
        $checker |= (1 << $val);
    }

    return -1;
}

// Driver code
$s = "abcfdeacf";
$i=FirstRepeated($s);
if ($i!=-1)
    echo "Char = " . $s[$i] .
         " and Index = " . $i;
else
    echo "No repeated Char";

// This code is contributed by ita_c
?>

Javascript

<script>
// Efficiently check First repeated character
// in Javascript program

    function FirstRepeated(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 i;

            // set bit in checker
            checker |= (1 << val);
        }

        return -1;
    }

    // Driver code
    let s = "abcfdeacf";
    let i=FirstRepeated(s);
    if (i!=-1)
       document.write("Char = "+ s[i] + " and Index = "+i);
    else
        document.write( "No repeated Char");

    // This code is contributed by rag2127

</script>

Output

Char = a   and Index = 6

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

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.