Print all the duplicates in the input string

Difficulty Level : Basic
Last Updated : 10 Feb, 2021

Write an efficient program to print all the duplicates and their counts in the input string

Method 1: Using hashing
Algorithm: Let input string be “geeksforgeeks”
1: Construct character count array from the input string.
count[‘e’] = 4
count[‘g’] = 2
count[‘k’] = 2
……
2: Print all the indexes from the constructed array which have values greater than 1.
Solution

C++14

// C++ program to count all duplicates
// from string using hashing
#include <iostream>
using namespace std;
# define NO_OF_CHARS 256
 
class gfg
{
    public :
     
    /* Fills count array with
    frequency of characters */
    void fillCharCounts(char *str, int *count)
    {
        int i;
        for (i = 0; *(str + i); i++)
        count[*(str + i)]++;
    }
 
    /* Print duplicates present 
    in the passed string */
    void printDups(char *str)
    {
         
        // Create an array of size 256 and fill
        // count of every character in it
        int *count = (int *)calloc(NO_OF_CHARS, 
                                      sizeof(int));
        fillCharCounts(str, count);
 
        // Print characters having count more than 0
        int i;
        for (i = 0; i < NO_OF_CHARS; i++)
        if(count[i] > 1)
            printf("%c, count = %d \n", i, count[i]);
 
        free(count);
    }
};
 
/* Driver code*/
int main()
{
    gfg g ;
    char str[] = "test string";
    g.printDups(str);
    //getchar();
    return 0;
}
 
// This code is contributed by SoM15242

C

// C program to count all duplicates 
// from string using hashing
# include <stdio.h>
# include <stdlib.h>
# define NO_OF_CHARS 256
 
/* Fills count array with 
   frequency of characters */
void fillCharCounts(char *str, int *count)
{
   int i;
   for (i = 0; *(str+i);  i++)
      count[*(str+i)]++;
}
 
/* Print duplicates present 
   in the passed string */
void printDups(char *str)
{
   
  // Create an array of size 256 and 
  // fill count of every character in it
  int *count = (int *)calloc(NO_OF_CHARS, 
                             sizeof(int));
  fillCharCounts(str, count);
 
  // Print characters having count more than 0
  int i;
  for (i = 0; i < NO_OF_CHARS; i++)
    if(count[i] > 1)
        printf("%c,  count = %d \n", i,  count[i]);
 
  free(count);
}
 
/* Driver program to test to pront printDups*/
int main()
{
    char str[] = "test string";
    printDups(str);
    getchar();
    return 0;
}

Java

// Java program to count all 
// duplicates from string using
// hashing
 
public class GFG {
    static final int NO_OF_CHARS = 256;
 
    /* Fills count array with 
       frequency of characters */
    static void fillCharCounts(String str, 
                                   int[] count)
    {
        for (int i = 0; i < str.length(); i++)
            count[str.charAt(i)]++;
    }
 
    /* Print duplicates present 
      in the passed string */
    static void printDups(String str)
    {
        // Create an array of size 
        // 256 and fill count of
        // every character in it
        int count[] = new int[NO_OF_CHARS];
        fillCharCounts(str, count);
 
        for (int i = 0; i < NO_OF_CHARS; i++)
            if (count[i] > 1)
                System.out.println((char)(i) + 
                          ", count = " + count[i]);
    }
 
    // Driver Method
    public static void main(String[] args)
    {
        String str = "test string";
        printDups(str);
    }
}

Python

# Python program to count all 
# duplicates from string using hashing
NO_OF_CHARS = 256
 
# Fills count array with 
# frequency of characters
def fillCharCounts(string, count):
    for i in string:
        count[ord(i)] += 1
    return count
 
# Print duplicates present 
# in the passed string
def printDups(string):
     
    # Create an array of size 256 
    # and fill count of every character in it
    count = [0] * NO_OF_CHARS
    count = fillCharCounts(string,count)
 
    # Utility Variable
    k = 0
 
    # Print characters having 
    # count more than 0
    for i in count:
        if int(i) > 1:
            print chr(k) + ", count = " + str(i)
        k += 1
 
# Driver program to test the above function
string = "test string"
print printDups(string)
 
# This code is contributed by Bhavya Jain

C#

// C# program to count all duplicates
// from string using hashing
using System;
 
class GFG
{
     
    static int NO_OF_CHARS = 256;
     
    /* Fills count array with 
       frequency of characters */
    static void fillCharCounts(String str, 
                                 int[] count)
    {
        for (int i = 0; i < str.Length; i++)
            count[str[i]]++;
    }
     
    /* Print duplicates present in
    the passed string */
    static void printDups(String str)
    {
         
        // Create an array of size 256 and
        // fill count of every character in it
        int []count = new int[NO_OF_CHARS];
        fillCharCounts(str, count);
     
        for (int i = 0; i < NO_OF_CHARS; i++)
            if(count[i] > 1)
                Console.WriteLine((char)i + ", " +
                              "count = " + count[i]);
    }
     
    // Driver Method
    public static void Main()
    {
        String str = "test string";
        printDups(str);
    }
}
 
// This code is contributed by Sam007

PHP

<?php
// PHP program to count 
// all duplicates from 
// string using hashing
function fillCharCounts($str, $count)
{
    for ($i = 0; $i < strlen($str); $i++)
        $count[ord($str[$i])]++;
         
    for ($i = 0; $i < 256; $i++)
        if($count[$i] > 1)
            echo chr($i) . ", " ."count = " . 
                         ($count[$i]) . "\n";
}
 
// Print duplicates present 
// in the passed string
function printDups($str)
{
    // Create an array of size 
    // 256 and fill count of 
    // every character in it
    $count = array();
    for ($i = 0; $i < 256; $i++)
    $count[$i] = 0;
    fillCharCounts($str, $count);
 
     
}
 
// Driver Code
$str = "test string";
printDups($str);
     
// This code is contributed by Sam007
?>

Output :

s,  count = 2
t,  count = 3

Time Complexity: O(n), where n = length of the string passed

Space Complexity: O(NO_OF_CHARS)

Note: Hashing involves the use of an array of fixed size each time no matter whatever the string is.

For example, str = “aaaaaaaaaa”.

An array of size 256 is used for str, only 1 block out of total size (256) will be utilized to store the number of occurrences of ‘a’ in str (i.e count[‘a’] = 10).

Rest 256 – 1 = 255 blocks remain unused.

Thus, Space Complexity is potentially high for such cases. So, to avoid any discrepancies and to improve Space Complexity, maps are generally preferred over long-sized arrays.

Method 2: Using Maps

Approach: The approach is the same as discussed in Method 1, but, using a map to store the count.

Solution:

C++

// C++ program to count all duplicates
// from string using maps
#include <bits/stdc++.h>
using namespace std;
void printDups(string str)
{
    map<char, int> count;
    for (int i = 0; i < str.length(); i++) {
        count[str[i]]++;
    }
 
    for (auto it : count) {
        if (it.second > 1)
            cout << it.first << ", count = " << it.second
                 << "\n";
    }
}
/* Driver code*/
int main()
{
    string str = "test string";
    printDups(str);
    return 0;
}
// This code is contributed by yashbeersingh42

Java

// Java program to count all duplicates
// from string using maps
import java.util.*;
 
class GFG {
 
    static void printDups(String str)
    {
        HashMap<Character, Integer> count = new HashMap<>();
        /*Store duplicates present
        in the passed string */
        for (int i = 0; i < str.length(); i++) {
            if (!count.containsKey(str.charAt(i)))
                count.put(str.charAt(i), 1);
            else
                count.put(str.charAt(i),
                          count.get(str.charAt(i)) + 1);
        }
 
        /*Print duplicates in sorted order*/
        for (Map.Entry mapElement : count.entrySet()) {
            char key = (char)mapElement.getKey();
            int value = ((int)mapElement.getValue());
 
            if (value > 1)
                System.out.println(key
                                   + ", count = " + value);
        }
    }
    // Driver code
    public static void main(String[] args)
    {
        String str = "test string";
        printDups(str);
    }
}
// This code is contributed by yashbeersingh42

Output :

s,  count = 2
t,  count = 3

Time Complexity: O(N log N), where N = length of the string passed and it generally takes logN time for an element insertion in a map.

Space Complexity: O(K), where K = size of the map (0<=K<=input_string_length).

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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