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Find repeated character present first in a string
  • Difficulty Level : Easy
  • Last Updated : 27 Feb, 2021
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Given a string, find the repeated character present first in the string.
(Not the first repeated character, found here.) 
 

Examples: 

Input  : geeksforgeeks
Output : g
(mind that it will be g, not e.)

Asked in: Goldman Sachs internship 

Simple Solution using O(N^2) complexity 
The solution is to loop through the string for each character and search for the same in the rest of the string. This would need two loops and thus not optimal.  



C++




// C++ program to find the fist
// character that is repeated
#include <bits/stdc++.h>
#include <string.h>
 
using namespace std;
int findRepeatFirstN2(char* s)
{
    // this is O(N^2) method
    int p = -1, i, j;
    for (i = 0; i < strlen(s); i++)
    {
        for (j = i + 1; j < strlen(s); j++)
        {
            if (s[i] == s[j])
            {
                p = i;
                break;
            }
        }
        if (p != -1)
            break;
    }
 
    return p;
}
 
// Driver code
int main()
{
    char str[] = "geeksforgeeks";
    int pos = findRepeatFirstN2(str);
    if (pos == -1)
        cout << "Not found";
    else
        cout << str[pos];
    return 0;
}
 
// This code is contributed
// by Akanksha Rai

C




// C program to find the fist character that
// is repeated
#include <stdio.h>
#include <string.h>
 
int findRepeatFirstN2(char* s)
{
    // this is O(N^2) method
    int p = -1, i, j;
    for (i = 0; i < strlen(s); i++) {
        for (j = i + 1; j < strlen(s); j++) {
            if (s[i] == s[j]) {
                p = i;
                break;
            }
        }
        if (p != -1)
            break;
    }
 
    return p;
}
 
// Driver code
int main()
{
    char str[] = "geeksforgeeks";
    int pos = findRepeatFirstN2(str);
    if (pos == -1)
        printf("Not found");
    else
        printf("%c", str[pos]);
    return 0;
}

Java




// Java program to find the fist character
// that is repeated
import java.io.*;
import java.util.*;
 
class GFG {
 
    static int findRepeatFirstN2(String s)
    {
         
        // this is O(N^2) method
        int p = -1, i, j;
        for (i = 0; i < s.length(); i++)
        {
            for (j = i + 1; j < s.length(); j++)
            {
                if (s.charAt(i) == s.charAt(j))
                {
                    p = i;
                    break;
                }
            }
            if (p != -1)
                break;
        }
     
        return p;
    }
     
    // Driver code
    static public void main (String[] args)
    {
        String str = "geeksforgeeks";
        int pos = findRepeatFirstN2(str);
         
        if (pos == -1)
            System.out.println("Not found");
        else
        System.out.println( str.charAt(pos));
    }
}
 
// This code is contributed by anuj_67.

Python3




# Python3 program to find the fist
# character that is repeated
 
def findRepeatFirstN2(s):
 
    # this is O(N^2) method
    p = -1
    for i in range(len(s)):
     
        for j in range (i + 1, len(s)):
         
            if (s[i] == s[j]):
                p = i
                break
             
        if (p != -1):
            break
 
    return p
 
# Driver code
if __name__ == "__main__":
 
    str = "geeksforgeeks"
    pos = findRepeatFirstN2(str)
    if (pos == -1):
        print ("Not found")
    else:
        print (str[pos])
     
# This code is contributed
# by ChitraNayal

C#




// C# program to find the fist character
// that is repeated
using System;
 
class GFG {
 
    static int findRepeatFirstN2(string s)
    {
         
        // this is O(N^2) method
        int p = -1, i, j;
        for (i = 0; i < s.Length; i++)
        {
            for (j = i + 1; j < s.Length; j++)
            {
                if (s[i] == s[j])
                {
                    p = i;
                    break;
                }
            }
            if (p != -1)
                break;
        }
     
        return p;
    }
     
    // Driver code
    static public void Main ()
    {
        string str = "geeksforgeeks";
        int pos = findRepeatFirstN2(str);
         
        if (pos == -1)
            Console.WriteLine("Not found");
        else
        Console.WriteLine( str[pos]);
    }
}
 
// This code is contributed by anuj_67.

PHP




<?php
// PHP program to find the fist
// character that is repeated
 
function findRepeatFirstN2($s)
{
    // this is O(N^2) method
    $p = -1;
    for ($i = 0; $i < strlen($s); $i++)
    {
        for ($j = ($i + 1);
             $j < strlen($s); $j++)
        {
            if ($s[$i] == $s[$j])
            {
                $p = $i;
                break;
            }
        }
        if ($p != -1)
            break;
    }
 
    return $p;
}
 
// Driver code
$str = "geeksforgeeks";
$pos = findRepeatFirstN2($str);
 
if ($pos == -1)
    echo ("Not found");
else
    echo ($str[$pos]);
 
// This code is contributed by jit_t
?>
Output
g

Optimization by counting occurrences
This solution is optimized by using the following techniques: 
1. We loop through the string and hash the characters using ASCII codes. Store 1 if found and store 2 if found again. Also, store the position of the letter first found in.
2. We run a loop on the hash array and now we find the minimum position of any character repeated.
This will have a runtime of O(N).

C++




// C++ program to find the fist character that
// is repeated
#include<bits/stdc++.h>
 
using namespace std;
// 256 is taken just to ensure nothing is left,
// actual max ASCII limit is 128
#define MAX_CHAR 256
 
int findRepeatFirst(char* s)
{
    // this is optimized method
    int p = -1, i, k;
 
    // initialized counts of occurrences of
    // elements as zero
    int hash[MAX_CHAR] = { 0 };
 
    // initialized positions
    int pos[MAX_CHAR];
 
    for (i = 0; i < strlen(s); i++) {
        k = (int)s[i];
        if (hash[k] == 0) {
            hash[k]++;
            pos[k] = i;
        } else if (hash[k] == 1)
            hash[k]++;
    }
 
    for (i = 0; i < MAX_CHAR; i++) {
        if (hash[i] == 2) {
            if (p == -1) // base case
                p = pos[i];
            else if (p > pos[i])
                p = pos[i];
        }
    }
 
    return p;
}
 
// Driver code
int main()
{
    char str[] = "geeksforgeeks";
    int pos = findRepeatFirst(str);
    if (pos == -1)
        cout << "Not found";
    else
        cout << str[pos];
    return 0;
}
 
// This code is contributed
// by Akanksha Rai

C




// C program to find the fist character that
// is repeated
#include <stdio.h>
#include <string.h>
 
// 256 is taken just to ensure nothing is left,
// actual max ASCII limit is 128
#define MAX_CHAR 256
 
int findRepeatFirst(char* s)
{
    // this is optimized method
    int p = -1, i, k;
 
    // initialized counts of occurrences of
    // elements as zero
    int hash[MAX_CHAR] = { 0 };
 
    // initialized positions
    int pos[MAX_CHAR];
 
    for (i = 0; i < strlen(s); i++) {
        k = (int)s[i];
        if (hash[k] == 0) {
            hash[k]++;
            pos[k] = i;
        } else if (hash[k] == 1)
            hash[k]++;
    }
 
    for (i = 0; i < MAX_CHAR; i++) {
        if (hash[i] == 2) {
            if (p == -1) // base case
                p = pos[i];
            else if (p > pos[i])
                p = pos[i];
        }
    }
 
    return p;
}
 
// Driver code
int main()
{
    char str[] = "geeksforgeeks";
    int pos = findRepeatFirst(str);
    if (pos == -1)
        printf("Not found");
    else
        printf("%c", str[pos]);
    return 0;
}

Java




// Java Program to find the fist character 
// that is repeated
 
import java.util.*;
import java.lang.*;
 
public class GFG
{
    public static int findRepeatFirst(String s)
    {
        // this is optimized method
        int p = -1, i, k;
 
        // initialized counts of occurrences of
        // elements as zero
        int MAX_CHAR = 256;
        int hash[] = new int[MAX_CHAR];
 
        // initialized positions
        int pos[] = new int[MAX_CHAR];
 
        for (i = 0; i < s.length(); i++)
        {
            k = (int)s.charAt(i);
            if (hash[k] == 0)
            {
                hash[k]++;
                pos[k] = i;
            }
            else if (hash[k] == 1)
                hash[k]++;
        }
 
        for (i = 0; i < MAX_CHAR; i++)
        {
            if (hash[i] == 2)
            {
                if (p == -1) // base case
                    p = pos[i];
                else if (p > pos[i])
                    p = pos[i];
            }
        }
 
        return p;
    }
 
// Driver code
    public static void main(String[] args)
    {
        String str = "geeksforgeeks";
        int pos = findRepeatFirst(str);
        if (pos == -1)
            System.out.println("Not found");
        else
            System.out.println(str.charAt(pos));
    }
}
 
// Code Contributed by Mohit Gupta_OMG

Python3




# Python 3 program to find the fist
# character that is repeated
 
# 256 is taken just to ensure nothing
# is left, actual max ASCII limit is 128
 
MAX_CHAR = 256
 
def findRepeatFirst(s):
     
    # this is optimized method
    p = -1
 
    # initialized counts of occurrences
    # of elements as zero
    hash = [0 for i in range(MAX_CHAR)]
 
    # initialized positions
    pos = [0 for i in range(MAX_CHAR)]
 
    for i in range(len(s)):
        k = ord(s[i])
        if (hash[k] == 0):
            hash[k] += 1
            pos[k] = i
        elif(hash[k] == 1):
            hash[k] += 1
 
    for i in range(MAX_CHAR):
        if (hash[i] == 2):
            if (p == -1): # base case
                p = pos[i]
            elif(p > pos[i]):
                p = pos[i]
    return p
 
# Driver code
if __name__ == '__main__':
    str = "geeksforgeeks"
    pos = findRepeatFirst(str);
    if (pos == -1):
        print("Not found")
    else:
        print(str[pos])
         
# This code is contributed by
# Shashank_Sharma

C#




// C# Program to find the fist character 
// that is repeated
using System;
public class GFG
{
    public static int findRepeatFirst(string s)
    {
        // this is optimized method
        int p = -1, i, k;
  
        // initialized counts of occurrences of
        // elements as zero
        int MAX_CHAR = 256;
        int []hash = new int[MAX_CHAR];
  
        // initialized positions
        int []pos = new int[MAX_CHAR];
  
        for (i = 0; i < s.Length; i++)
        {
            k = (int)s[i];
            if (hash[k] == 0)
            {
                hash[k]++;
                pos[k] = i;
            }
            else if (hash[k] == 1)
                hash[k]++;
        }
  
        for (i = 0; i < MAX_CHAR; i++)
        {
            if (hash[i] == 2)
            {
                if (p == -1) // base case
                    p = pos[i];
                else if (p > pos[i])
                    p = pos[i];
            }
        }
  
        return p;
    }
  
    // Driver code
    public static void Main()
    {
        string str = "geeksforgeeks";
        int pos = findRepeatFirst(str);
        if (pos == -1)
            Console.Write("Not found");
        else
            Console.Write(str[pos]);
    }
}
  
// This code is contributed by nitin mittal.
Output
g

Method #3:Using Built in Python Functions:

Approach:

  • Calculate all frequencies of all characters using Counter() function.
  • Traverse the string and check if any element has frequency greater than 1.
  • Print the character and break the  loop

Below is the implementation:

Python3




# Python implementation
from collections import Counter
 
# Function which repeats
# first repeating character
def printrepeated(string):
   
    # Calculating frequencies
    # using Counter function
    freq = Counter(string)
     
    # Traverse the string
    for i in string:
        if(freq[i] > 1):
            print(i)
            break
 
 
# Driver code
string = "geeksforgeeks"
 
# passing string to printrepeated function
printrepeated(string)
 
# this code is contributed by vikkycirus
Output
g

Time Complexity:O(n)

Space Complexity:O(n)
More optimized Solution Repeated Character Whose First Appearance is Leftmost
This article is contributed by Suprotik Dey. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 

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