Print characters and their frequencies in order of occurrence

Given 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 frequency 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 frequency 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 frequency 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 frequency 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 frequency 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 frequency 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.

Method #2: Using Unorderd_map & Hashing

We can also use unordered_map and 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

Javascript

<script>
// Javascript implementation to
//print the characters and
// frequencies in order
// of its occurrence

function prCharWithFreq(s)
{
  // Store all characters and
  // their frequencies in dictionary
  var d = new Map();

  s.split('').forEach(element => {

        if(d.has(element))
        {
            d.set(element, d.get(element)+1);
        }
        else
            d.set(element, 1);
  });

  // Print characters and their
  // frequencies in same order
  // of their appearance

  s.split('').forEach(element => {
    // Print only if this
    // character is not printed
    // before
    if(d.has(element) && d.get(element)!=0)
    {
      document.write( element + d.get(element) + " ");
      d.set(element, 0);
    }
  });
}

// Driver Code
var s="geeksforgeeks";
prCharWithFreq(s);

</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(n), 

Method #3: Using Object-Oriented programming:

We can solve this problem without using a HashMap too. But, we then have to create our own class whose objects will have 2 properties – character and its occurrence.

1. We create a class in this case called as CharOccur and initialize 2 variables character and occurrence in its constructor.
2. In the main of our original class we will just create a list that can store these objects.

Logic – 

• Loop through the string.
• Check if string’s current character is already present in some object. If present then increment its occurrence else set its occurrence to 1.

C++

#include <bits/stdc++.h>
using namespace std;
// Creating a class CharOccur whose objects have 2
// properties - a character and its occurrence
class CharOccur {
  public:
  char character;
  int occurrence = 0;
  CharOccur(char character, int occurrence)
  {
    this->character = character;
    this->occurrence = occurrence;
  }
};
void frequency(string s)
{
  if (s.size() == 0) {
    cout << "Empty string" << endl;
    return;
  }

  vector<CharOccur> occurrences;
  // Creating vector of objects of Charoccur class

  for (int i = 0; i < s.size(); i++) {
    /* Logic
         * If a pair of character and its occurrence is
         * already present as object - increment the
         * occurrence else create a new object of
         * character with its occurrence set to 1
         */
    char c = s[i];
    int flag = 0;
    for (auto& o : occurrences) {
      if (o.character == c) {
        o.occurrence += 1;
        flag = 1;
      }
    }
    if (flag == 0) {
      CharOccur grp(c, 1);
      occurrences.push_back(grp);
    }
  }

  // Printing the character - occurrences pair
  for (auto o : occurrences) {
    cout << o.character << " " << o.occurrence << endl;
  }
}
int main()
{
  string s1 = "GFG";
  cout << "For " << s1 << endl;
  frequency(s1);

  string s2 = "aaabccccffgfghc";
  cout << "For " << s2 << endl;
  frequency(s2);
}

// This code is contributed by garg28harsh.

Java

/*package whatever //do not write package name here */

import java.io.*;
import java.util.ArrayList;

class GFG {
    public static void main(String[] args)
    {
        String s1 = "GFG";
        System.out.println("For " + s1);
        frequency(s1);

        String s2 = "aaabccccffgfghc";
        System.out.println("For " + s2);
        frequency(s2);
    }
    private static void frequency(String s)
    {
        if (s.length() == 0) {
            System.out.println("Empty string");
            return;
        }
        ArrayList<CharOccur> occurrences
            = new ArrayList<CharOccur>();
        // Creating ArrayList of objects of Charoccur class

        for (int i = 0; i < s.length(); i++) {
            /* Logic
             * If a pair of character and its occurrence is
             * already present as object - increment the
             * occurrence else create a new object of
             * character with its occurrence set to 1
             */
            char c = s.charAt(i);
            int flag = 0;
            for (CharOccur o : occurrences) {
                if (o.character == c) {
                    o.occurrence += 1;
                    flag = 1;
                }
            }
            if (flag == 0) {
                CharOccur grp = new CharOccur(c, 1);
                occurrences.add(grp);
            }
        }

        // Printing the character - occurrences pair
        for (CharOccur o : occurrences) {
            System.out.println(o.character + " "
                               + o.occurrence);
        }
    }
}

// Creating a class CharOccur whose objects have 2
// properties - a character and its occurrence
class CharOccur {
    char character;
    int occurrence = 0;
    CharOccur(char character, int occurrence)
    {
        this.character = character;
        this.occurrence = occurrence;
    }
}
// Contributed by Soham Ratnaparkhi

Python3

class CharOccur:
    def __init__(self, character, occurrence):
        # Creating a class CharOccur whose objects have 2 properties - a character and its occurrence
        self.character = character
        self.occurrence = occurrence

def frequency(s: str):
    if not s:
        print("Empty string")
        return

    occurrences = []
    # Creating list of objects of Charoccur class
    for c in s:
        found = False
        for occur in occurrences:
            # Logic
            # If a pair of character and its occurrence is already present as object - increment the occurrence
            # else create a new object of character with its occurrence set to 1
            if occur.character == c:
                occur.occurrence += 1
                found = True
                break
        if not found:
            occurrences.append(CharOccur(c, 1))

    # Printing the character - occurrences pair
    for occur in occurrences:
        print(occur.character, occur.occurrence)

#example usage
s1 = "GFG"
print("For " + s1)
frequency(s1)

s2 = "aaabccccffgfghc"
print("For " + s2)
frequency(s2)

# This code is contributed by Shivam Tiwari

C#

using System;
using System.Collections.Generic;

class CharOccur {
  public char character;
  public int occurrence = 0;

  // Creating a class CharOccur whose objects have 2
  // properties - a character and its occurrence
  public CharOccur(char character, int occurrence)
  {
    this.character = character;
    this.occurrence = occurrence;
  }
}

class Program {
  static void frequency(string s)
  {
    if (string.IsNullOrEmpty(s)) {
      Console.WriteLine("Empty string");
      return;
    }

    List<CharOccur> occurrences = new List<CharOccur>();

    for (int i = 0; i < s.Length; i++) {

      /* Logic
             * If a pair of character and its occurrence is
             * already present as object - increment the
             * occurrence else create a new object of
             * character with its occurrence set to 1
             */
      char c = s[i];
      bool found = false;

      for (int j = 0; j < occurrences.Count; j++) {
        if (occurrences[j].character == c) {
          occurrences[j].occurrence++;
          found = true;
          break;
        }
      }

      if (!found) {
        occurrences.Add(new CharOccur(c, 1));
      }
    }

    // Printing the character - occurrences pair
    for (int i = 0; i < occurrences.Count; i++) {
      Console.WriteLine(occurrences[i].character + " "
                        + occurrences[i].occurrence);
    }
  }

  static void Main()
  {
    string s1 = "GFG";
    Console.WriteLine("For " + s1);
    frequency(s1);

    string s2 = "aaabccccffgfghc";
    Console.WriteLine("For " + s2);
    frequency(s2);
  }
}

// This code is contributed by Susobhan Akhuli

Javascript

// Creating a class CharOccur whose objects have 2
// properties - a character and its occurrence
class CharOccur{

  constructor( character, occurrence)
  {
    this.character = character;
    this.occurrence = occurrence;
  }
}
function frequency(s)
{
  if (s.length== 0) {
    document.write("Empty string");
    return;
  }

  let occurrences=[];
  // Creating vector of objects of Charoccur class

  for (let i = 0; i < s.length; i++) {
    /* Logic
         * If a pair of character and its occurrence is
         * already present as object - increment the
         * occurrence else create a new object of
         * character with its occurrence set to 1
         */
    let c = s[i];
    let flag = 0;
    for (let i=0;i< occurrences.length;i++) {

      if (occurrences[i].character == c) {
        occurrences[i].occurrence += 1;
        flag = 1;
      }
    }
    if (flag == 0) {
      let grp= new CharOccur(c, 1);
      occurrences.push(grp);
    }
  }

  // Printing the character - occurrences pair
  for (let i=0;i< occurrences.length;i++) {
      let o = occurrences[i];
    console.log(o.character +" " + o.occurrence);
  }
}

  let s1 = "GFG";
  console.log("For " + s1);
  frequency(s1);

  let s2 = "aaabccccffgfghc";
  console.log("For " + s2);
  frequency(s2);

// This code is contributed by garg28harsh.

Output

For GFG
G 2
F 1
For aaabccccffgfghc
a 3
b 1
c 5
f 3
g 2
h 1

Complexity analysis:

• Time Complexity: O(n), where n is the number of characters in the string. 
• Auxiliary Space: O(n)

Method #4: 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

C++

// CPP implementation to print the characters and
// frequencies in order of its occurrence

#include <bits/stdc++.h>
using namespace std;

void prCharWithFreq(string str)
{
    // Store all characters and their
    // frequencies using Counter function
    map<char, int> mp;
    for (int i = 0; i < str.length(); i++)
        mp[str[i]]++;

    // Print characters and their frequencies in
    // same order of their appearance
    int count[256] = { 0 };
    for (int i = 0; i < str.length(); i++) {
        if (count[str[i]] == 0) {
            cout << str[i] << mp[str[i]] << " ";
            count[str[i]]++;
        }
    }
    cout << endl;
}

int main()
{
    string str = "geeksforgeeks";
    prCharWithFreq(str);
    return 0;
}

// This code is contributed by Susobhan Akhuli

Java

// Java implementation to print the characters and
// frequencies in order of its occurrence

import java.util.HashMap;
import java.util.Map;

public class Main {

    public static void prCharWithFreq(String string)
    {
        Map<Character, Integer> d = new HashMap<>();
        // Store all characters and their frequencies
        for (int i = 0; i < string.length(); i++) {
            char c = string.charAt(i);
            if (d.containsKey(c)) {
                d.put(c, d.get(c) + 1);
            }
            else {
                d.put(c, 1);
            }
        }

        // Print characters and their frequencies in same
        // order of their appearance
        for (int i = 0; i < string.length(); i++) {
            char c = string.charAt(i);
            if (d.containsKey(c)) {
                System.out.print(
                    c + Integer.toString(d.get(c)) + " ");
                d.remove(c);
            }
        }
    }

    public static void main(String[] args)
    {
        String string = "geeksforgeeks";
        prCharWithFreq(string);
    }
}

// This code is contributed by shivamsharma215

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 using 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

C#

// C# implementation to print the characters and
// frequencies in order of its occurrence

using System;
using System.Collections.Generic;
using System.Linq;

public class GFG
{
    public static void PrCharWithFreq(string str)
    {
        // Store all characters and their frequencies using Dictionary
        Dictionary<char, int> freqDict = new Dictionary<char, int>();
        foreach (char c in str)
        {
            if (freqDict.ContainsKey(c))
            {
                freqDict++;
            }
            else
            {
                freqDict = 1;
            }
        }

        // Print characters and their frequencies in order of their appearance
        foreach (char c in str)
        {
            if (freqDict.ContainsKey(c))
            {
                Console.Write(c + freqDict.ToString() + " ");
                freqDict.Remove(c);
            }
        }
    }

    public static void Main()
    {
        string str = "geeksforgeeks";
        PrCharWithFreq(str);
    }
}

// This code is contributed by Susobhan Akhuli

Javascript

function prCharWithFreq(str) {
  // Store all characters and their
  // frequencies using an object
  let mp = {};
  for (let i = 0; i < str.length; i++) {
    mp[str[i]] = mp[str[i]] ? mp[str[i]] + 1 : 1;
  }

  // Print characters and their frequencies in
  // same order of their appearance
  let count = {};
  for (let i = 0; i < str.length; i++) {
    if (!count[str[i]]) {
      console.log(str[i] + mp[str[i]] + " ");
      count[str[i]] = true;
    }
  }
  console.log("\n");
}

let str = "geeksforgeeks";
prCharWithFreq(str);
// This code is contributed By Shivam Tiwari

Output

g2 e4 k2 s2 f1 o1 r1 

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

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