Tutorialspoint.dev

Print all the duplicates in the input string

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

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 value greater than 1.

Solution

C++



// 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 ", 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

/div>

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 ", 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.printf("%c,  count = %d ", i,  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]) . " ";
}
  
// 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)

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



This article is attributed to GeeksforGeeks.org

You Might Also Like

leave a comment

code

0 Comments

load comments

Subscribe to Our Newsletter