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Find repeated character present first in a string

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  • Difficulty Level : Easy
  • Last Updated : 06 Oct, 2022
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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.  

Implementation:

C++




// C++ program to find the first
// 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 first 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 first 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 first
# 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 first 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 first
// 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
?>

Javascript




<script>
 
// Javascript program to find the first
// character that is repeated
function findRepeatFirstN2(s)
{
     
    // This is O(N^2) method
    let 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
let str = "geeksforgeeks";
let pos = findRepeatFirstN2(str);
 
if (pos == -1)
    document.write("Not found");
else
    document.write(str[pos]);
     
// This code is contributed by suresh07  
 
</script>

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.

Implementation:

C++




// C++ program to find the first 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 first 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 first 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 first
# 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 first 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.

Javascript




<script>
    // Javascript Program to find the first character that is repeated
     
    function findRepeatFirst(s)
    {
        // this is optimized method
        let p = -1, i, k;
   
        // initialized counts of occurrences of
        // elements as zero
        let MAX_CHAR = 256;
        let hash = new Array(MAX_CHAR);
        hash.fill(0);
   
        // initialized positions
        let pos = new Array(MAX_CHAR);
        pos.fill(0);
   
        for (i = 0; i < s.length; i++)
        {
            k = s[i].charCodeAt();
            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;
    }
     
    let str = "geeksforgeeks";
    let pos = findRepeatFirst(str);
    if (pos == -1)
      document.write("Not found");
    else
      document.write(str[pos]);
 
// This code is contributed by rameshtravel07.
</script>

Output

g

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

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)
Auxiliary Space: O(n)

Method #4: Solving just by single traversal of the given string.

Algorithm :

  1. Traverse the string from left to right.
  2. If current character is not present in hash map, Then push this character along with its Index.
  3. If the current character is already present in hash map, Then get the index of current character ( from hash map ) and compare it with the index of the previously found repeating character.
  4. If the current index is smaller, then update the index.

C++




#include <iostream>
#include<unordered_map>
#define INT_MAX 2147483647
using namespace std;
 
// Function to find left most repeating character.
char firstRep (string s)
    {
        unordered_map<char,int> map;
        char c='#';
        int index=INT_MAX;
         
        // single traversal of string.
        for(int i=0;i<s.size();i++)
        {
            char p=s[i];
             
            if(map.find(p)==map.end())map.insert({p,i});
            else
            {
                if(map[p]<index)
                {
                    index=map[p];
                    c=p;
                }
            }
             
        }
         
         
        return c;
    }
 
// Main function.
int main() {
 
    // Input string.
    string s="abccdbd";
    cout<<firstRep(s)<<endl;
     
    return 0;
}
 
 
// This code is contributed
// by rohan007

Java




// Java code to find the first repeating character in a
// string
import java.util.*;
 
public class GFG {
  public static int INT_MAX = 2147483647;
 
  // Function to find left most repeating character.
  public static char firstRep(String s)
  {
    HashMap<Character, Integer> map
      = new HashMap<Character, Integer>();
    char c = '#';
    int index = INT_MAX;
 
    // single traversal of string.
    for (int i = 0; i < s.length(); i++) {
      char p = s.charAt(i);
 
      if (!map.containsKey(p)) {
        map.put(p, i);
      }
      else {
        if (map.get(p) < index) {
          index = map.get(p);
          c = p;
        }
      }
    }
 
    return c;
  }
 
  // Main function.
  public static void main(String[] args)
  {
 
    // Input string.
    String s = "abccdbd";
    System.out.print(firstRep(s));
    System.out.print("\n");
  }
}
 
// This code is contributed by Aarti_Rathi

Python3




# Python3 code to find the first repeating character in a
# string
INT_MAX = 2147483647
 
# Function to find left most repeating character.
def firstRep(s):
    map = dict()
    c = '#'
    index = INT_MAX
     
    # single traversal of string.
    i = 0
    while (i < len(s)):
        p = s[i]
        if (not (p in map.keys())):
            map[p] = i
        else:
            if (map.get(p) < index):
                index = map.get(p)
                c = p
        i += 1
    return c
 
if __name__ == "__main__":
   
    # Input string.
    s = "abccdbd"
    print(firstRep(s), end="")
    print("\n", end="")
 
# This code is contributed by Aarti_Rathi

C#




// C# code to find the first repeating character in a string
using System;
using System.Collections.Generic;
 
public static class GFG {
  static int INT_MAX = 2147483647;
 
  // Function to find left most repeating character.
  public static char firstRep(string s)
  {
    Dictionary<char, int> map
      = new Dictionary<char, int>();
    char c = '#';
    int index = INT_MAX;
 
    // single traversal of string.
    for (int i = 0; i < s.Length; i++) {
      char p = s[i];
 
      if (!map.ContainsKey(p)) {
        map[p] = i;
      }
      else {
        if (map[p] < index) {
          index = map[p];
          c = p;
        }
      }
    }
 
    return c;
  }
 
  // Main function.
  public static void Main()
  {
 
    // Input string.
    string s = "abccdbd";
    Console.Write(firstRep(s));
    Console.Write("\n");
  }
}
 
// This code is contributed by Aarti_Rathi

Javascript




<script>
// JavaScript code to find the first repeating character in a string
const INT_MAX = 2147483647
 
// Function to find left most repeating character.
function firstRep(s)
{
    map = new Map();
    let c = '#';
    let index=INT_MAX;
         
    // single traversal of string.
    for(let i = 0; i < s.length; i++)
    {
        let p = s[i];
             
        if(!map.has(p))map.set(p,i);
        else
        {
            if(map.get(p) < index)
            {
                index = map.get(p);
                c = p;
            }
        }        
    }
    return c;
}
 
// Driver code
 
// Input string.
const s="abccdbd";
document.write(firstRep(s));
     
// This code is contributed by shinjanpatra
</script>

Output

b

Time complexity: O(N)
Auxiliary Space: O(1), as there will be a constant number of characters present in the string.

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 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.


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