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Print characters and their frequencies in order of occurrence
  • Difficulty Level : Easy
  • Last Updated : 30 Apr, 2021

Given a string str containing only lowercase characters. The problem is to print the characters along with their frequency in the order of their occurrence and in the given format explained in the examples below.
Examples: 

Input : str = "geeksforgeeks"
Output : g2 e4 k2 s2 f1 o1 r1

Input : str = "elephant"
Output : e2 l1 p1 h1 a1 n1 t1

Source: SAP Interview Experience | Set 26

Approach: Create a count array to store the frequency of each character in the given string str. Traverse the string str again and check whether the frequency of that character is 0 or not. If not 0, then print the character along with its frequency and update its frequency to 0 in the hash table. This is done so that the same character is not printed again.

C++




// C++ implementation to print the character and
// its frequency in order of its occurrence
#include <bits/stdc++.h>
using namespace std;
 
#define SIZE 26
 
// function to print the character and its frequency
// in order of its occurrence
void printCharWithFreq(string str)
{
    // size of the string 'str'
    int n = str.size();
 
    // 'freq[]' implemented as hash table
    int freq[SIZE];
 
    // initialize all elements of freq[] to 0
    memset(freq, 0, sizeof(freq));
 
    // accumulate freqeuncy of each character in 'str'
    for (int i = 0; i < n; i++)
        freq[str[i] - 'a']++;
 
    // traverse 'str' from left to right
    for (int i = 0; i < n; i++) {
 
        // if frequency of character str[i] is not
        // equal to 0
        if (freq[str[i] - 'a'] != 0) {
 
            // print the character along with its
            // frequency
            cout << str[i] << freq[str[i] - 'a'] << " ";
 
            // update frequency of str[i] to 0 so
            // that the same character is not printed
            // again
            freq[str[i] - 'a'] = 0;
        }
    }
}
 
// Driver program to test above
int main()
{
    string str = "geeksforgeeks";
    printCharWithFreq(str);
    return 0;
}
Java



// Java implementation to print the character and
// its frequency in order of its occurrence
public class Char_frequency {
     
    static final int SIZE = 26;
      
    // function to print the character and its
    // frequency in order of its occurrence
    static void printCharWithFreq(String str)
    {
         // size of the string 'str'
        int n = str.length();
 
        // 'freq[]' implemented as hash table
        int[] freq = new int[SIZE];
 
        // accumulate freqeuncy of each character
        // in 'str'
        for (int i = 0; i < n; i++)
            freq[str.charAt(i) - 'a']++;
 
        // traverse 'str' from left to right
        for (int i = 0; i < n; i++) {
 
            // if frequency of character str.charAt(i)
            // is not equal to 0
            if (freq[str.charAt(i) - 'a'] != 0) {
 
                // print the character along with its
                // frequency
                System.out.print(str.charAt(i));
                System.out.print(freq[str.charAt(i) - 'a'] + " ");
 
                // update frequency of str.charAt(i) to
                // 0 so that the same character is not
                // printed again
                freq[str.charAt(i) - 'a'] = 0;
            }
        }
    }
      
    // Driver program to test above
    public static void main(String args[])
    {
        String str = "geeksforgeeks";
        printCharWithFreq(str);
    }
}
// This code is contributed by Sumit Ghosh
Python3



# Python3 implementation to pr the character and
# its frequency in order of its occurrence
 
# import library
import numpy as np
 
# Function to print the character and its
# frequency in order of its occurrence
def prCharWithFreq(str) :
     
    # Size of the 'str'
    n = len(str)
     
    # Initialize all elements of freq[] to 0
    freq = np.zeros(26, dtype = np.int)
 
    # Accumulate freqeuncy of each
    # character in 'str'
    for i in range(0, n) :
        freq[ord(str[i]) - ord('a')] += 1
                 
    # Traverse 'str' from left to right
    for i in range(0, n) :
         
        # if frequency of character str[i] 
        # is not equal to 0
        if (freq[ord(str[i])- ord('a')] != 0) :
             
            # print the character along
            # with its frequency
            print (str[i], freq[ord(str[i]) - ord('a')],
                                                end = " ")
 
            # Update frequency of str[i] to 0 so that
            # the same character is not printed again
            freq[ord(str[i]) - ord('a')] = 0
         
     
# Driver Code
if __name__ == "__main__" :
     
    str = "geeksforgeeks";
    prCharWithFreq(str);
     
# This code is contributed by 'Saloni1297'
C#



// C# implementation to print the
// character and its frequency in
// order of its occurrence
using System;
 
class GFG {
    static int SIZE = 26;
 
    // function to print the character and its
    // frequency in order of its occurrence
    static void printCharWithFreq(String str)
    {
        // size of the string 'str'
        int n = str.Length;
 
        // 'freq[]' implemented as hash table
        int[] freq = new int[SIZE];
 
        // accumulate freqeuncy of each character
        // in 'str'
        for (int i = 0; i < n; i++)
            freq[str[i] - 'a']++;
 
        // traverse 'str' from left to right
        for (int i = 0; i < n; i++) {
 
            // if frequency of character str.charAt(i)
            // is not equal to 0
            if (freq[str[i] - 'a'] != 0) {
 
                // print the character along with its
                // frequency
                Console.Write(str[i]);
                Console.Write(freq[str[i] - 'a'] + " ");
 
                // update frequency of str.charAt(i) to
                // 0 so that the same character is not
                // printed again
                freq[str[i] - 'a'] = 0;
            }
        }
    }
 
    // Driver program to test above
    public static void Main()
    {
        String str = "geeksforgeeks";
        printCharWithFreq(str);
    }
}
 
// This code is contributed by Sam007
PHP



<?php
// PHP implementation to print the
// character and its frequency in
// order of its occurrence
$SIZE = 26;
 
// function to print the character and
// its frequency in order of its occurrence
function printCharWithFreq($str)
{
    global $SIZE;
     
    // size of the string 'str'
    $n = strlen($str);
 
    // 'freq[]' implemented as hash table
    $freq = array_fill(0, $SIZE, NULL);
 
    // accumulate freqeuncy of each
    // character in 'str'
    for ($i = 0; $i < $n; $i++)
        $freq[ord($str[$i]) - ord('a')]++;
 
    // traverse 'str' from left to right
    for ($i = 0; $i < $n; $i++)
    {
 
        // if frequency of character str[i]
        // is not equal to 0
        if ($freq[ord($str[$i]) - ord('a')] != 0)
        {
 
            // print the character along with
            // its frequency
            echo $str[$i] . $freq[ord($str[$i]) -
                                  ord('a')] . " ";
 
            // update frequency of str[i] to 0
            // so that the same character is
            // not printed again
            $freq[ord($str[$i]) - ord('a')] = 0;
        }
    }
}
 
// Driver Code
$str = "geeksforgeeks";
printCharWithFreq($str);
 
// This code is contributed by ita_c
?>
Javascript



<script>
// Javascript implementation to print the character and
// its frequency in order of its occurrence
     
    let SIZE = 26;
     
    // function to print the character and its
    // frequency in order of its occurrence
    function printCharWithFreq(str)
    {
     
        // size of the string 'str'
        let n = str.length;
  
        // 'freq[]' implemented as hash table
        let freq = new Array(SIZE);
        for(let i = 0; i < freq.length; i++)
        {
            freq[i] = 0;
        }
  
        // accumulate freqeuncy of each character
        // in 'str'
        for (let i = 0; i < n; i++)
            freq[str[i].charCodeAt(0) - 'a'.charCodeAt(0)]++;
  
        // traverse 'str' from left to right
        for (let i = 0; i < n; i++) {
  
            // if frequency of character str.charAt(i)
            // is not equal to 0
            if (freq[str[i].charCodeAt(0) - 'a'.charCodeAt(0)] != 0) {
  
                // print the character along with its
                // frequency
                document.write(str[i]);
                document.write(freq[str[i].charCodeAt(0) - 'a'.charCodeAt(0)] + " ");
  
                // update frequency of str.charAt(i) to
                // 0 so that the same character is not
                // printed again
                freq[str[i].charCodeAt(0) - 'a'.charCodeAt(0)] = 0;
            }
        }
    }
     
    // Driver program to test above
    let str = "geeksforgeeks";
    printCharWithFreq(str);
     
    // This code is contributed by rag2127.
     
</script>
Output
g2 e4 k2 s2 f1 o1 r1 
Time Complexity: O(n), where n is the number of characters in the string. 
Auxiliary Space: O(1), as there are only lowercase letters.
Alternate Solution (Use Hashing) 
We can also use hashing to solve the problem. 


C++



// C++ implementation to
//print the characters and
// frequencies in order
// of its occurrence
#include <bits/stdc++.h>
using namespace std;
 
void prCharWithFreq(string s)
{
  // Store all characters and
  // their frequencies in dictionary
  unordered_map<char, int> d;
 
  for(char i : s)
  {
    d[i]++;
  }
 
  // Print characters and their
  // frequencies in same order
  // of their appearance
  for(char i : s)
  {
    // Print only if this
    // character is not printed
    // before
    if(d[i] != 0)
    {
      cout << i << d[i] << " ";
      d[i] = 0;
    }
  }
}
  
// Driver Code
int main()
{
  string s="geeksforgeeks";
  prCharWithFreq(s);
}
 
// This code is contributed by rutvik_56
Java



// Java implementation to
// print the characters and
// frequencies in order
// of its occurrence
import java.util.*;
class Gfg{
    public static void prCharWithFreq(String s)
    {
       
        // Store all characters and
        // their frequencies in dictionary
        Map<Character, Integer> d = new HashMap<Character, Integer>();
         
         
        for(int i = 0; i < s.length(); i++)
        {
            if(d.containsKey(s.charAt(i)))
            {
                d.put(s.charAt(i), d.get(s.charAt(i)) + 1);
            }
            else
            {
                d.put(s.charAt(i), 1);
            }
        }
         
        // Print characters and their
        // frequencies in same order
        // of their appearance
        for(int i = 0; i < s.length(); i++)
        {
           
            // Print only if this
            // character is not printed
            // before
            if(d.get(s.charAt(i)) != 0)
            {
                System.out.print(s.charAt(i));
                System.out.print(d.get(s.charAt(i)) + " ");
                d.put(s.charAt(i), 0);
            }           
        }
    }
   
    // Driver code
     public static void main(String []args)
     {
       String S = "geeksforgeeks";
       prCharWithFreq(S);
     }
}
 
// This code is contributed by avanitrachhadiya2155
Python3



# Python3 implementation to print the characters and
# frequencies in order of its occurrence
 
def prCharWithFreq(str):
 
    # Store all characters and their frequencies
    # in dictionary
    d = {}
    for i in str:
        if i in d:
            d[i] += 1
        else:
            d[i] = 1
     
    # Print characters and their frequencies in
    # same order of their appearance
    for i in str:
 
        # Print only if this character is not printed
        # before. 
        if d[i] != 0:
            print("{}{}".format(i,d[i]), end =" ")
            d[i] = 0
      
     
# Driver Code
if __name__ == "__main__" :
     
    str = "geeksforgeeks";
    prCharWithFreq(str);
     
# This code is contributed by 'Ankur Tripathi'
C#



// C# implementation to
// print the characters and
// frequencies in order
// of its occurrence
 
using System;
using System.Collections;
using System.Collections.Generic;
 
class GFG
{
     
public static void prCharWithFreq(string s)
{
   
  // Store all characters and
  // their frequencies in dictionary
  Dictionary<char,int> d = new Dictionary<char,int>();
  
  foreach(char i in s)
  {
      if(d.ContainsKey(i))
      {
        d[i]++;
      }
      else
      {
        d[i]=1; 
      }
  }
  
  // Print characters and their
  // frequencies in same order
  // of their appearance
  foreach(char i in s)
  {
    // Print only if this
    // character is not printed
    // before
    if(d[i] != 0)
    {
        Console.Write(i+d[i].ToString() + " ");
        d[i] = 0;
    }
  }
}
   
// Driver Code
public static void Main(string []args)
{
  string s="geeksforgeeks";
  prCharWithFreq(s);
}
}
 
// This code is contributed by pratham76
Output
g2 e4 k2 s2 f1 o1 r1 
Method #3: Using built-in Python functions:
We can solve this problem quickly using the python Counter() method. The approach is very simple.
1) First create a dictionary using the Counter method having strings as keys and their frequencies as values.
2)Traverse in this dictionary print keys along with their values
Python3



# Python3 implementation to print the characters and
# frequencies in order of its occurrence
from collections import Counter
 
def prCharWithFreq(string):
 
    # Store all characters and their
    # frequencies suing Counter function
    d = Counter(string)
 
    # Print characters and their frequencies in
    # same order of their appearance
    for i in d:
        print(i+str(d[i]), end=" ")
 
 
string = "geeksforgeeks"
prCharWithFreq(string)
 
# This code is contributed by vikkycirus
Output
g2 e4 k2 s2 f1 o1 r1 
This article is contributed by Ayush Jauhari. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.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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