# Print characters and their frequencies in order of occurrence

• Difficulty Level : Easy
• Last Updated : 19 Jul, 2022

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```

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 ``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

 ``

## Javascript

 ``

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 ``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 d = ``new` `HashMap();``        ` `        ` `        ``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

 ``

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.

## 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 "` `+ s1);``        ``frequency(s2);``    ``}``    ``private` `static` `void` `frequency(String s)``    ``{``        ``if` `(s.length() == ``0``) {``            ``System.out.println(``"Empty string"``);``            ``return``;``        ``}``        ``ArrayList occurrences``            ``= ``new` `ArrayList();``        ``// 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.occurrenece += ``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.occurrenece);``        ``}``    ``}``}` `// Creating a class CharOccur whose objects have 2``// properties - a character and its occurrence``class` `CharOccur {``    ``char` `character;``    ``int` `occurrenece = ``0``;``    ``CharOccur(``char` `character, ``int` `occurrenece)``    ``{``        ``this``.character = character;``        ``this``.occurrenece = occurrenece;``    ``}``}``// Contributed by Soham Ratnaparkhi`

Output

```For GFG
G 2
F 1
For GFG
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

## 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`

Output

`g2 e4 k2 s2 f1 o1 r1 `

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