# Print characters in decreasing order of frequency

• Last Updated : 09 Sep, 2021

Given string str, the task is to print the characters in decreasing order of their frequency. If the frequency of two characters is the same then sort them in descending order alphabetically.
Examples:

Input: str = “geeksforgeeks”
Output:
e – 4
s – 2
k – 2
g – 2
r – 1
o – 1
f – 1
Input: str = “bbcc”
Output:
c – 2
b – 2

Approach 1:

• Use an unordered_map to store the frequencies of all the elements of the given string.
• Find the maximum frequency element from the map, print it with its frequency, and remove it from the map.
• Repeat the previous step while the map is not empty.

Below is the implementation of the above approach:

## CPP

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Function to print the characters``// of the given string in decreasing``// order of their frequencies``void` `printChar(string str, ``int` `len)``{` `    ``// To store the``    ``unordered_map<``char``, ``int``> occ;``    ``for` `(``int` `i = 0; i < len; i++)``        ``occ[str[i]]++;` `    ``// Map's size``    ``int` `size = occ.size();``    ``unordered_map<``char``, ``int``>::iterator it;` `    ``// While there are elements in the map``    ``while` `(size--) {` `        ``// Finding the maximum value``        ``// from the map``        ``unsigned currentMax = 0;``        ``char` `arg_max;``        ``for` `(it = occ.begin(); it != occ.end(); ++it) {``            ``if` `(it->second > currentMax``                ``|| (it->second == currentMax``                    ``&& it->first > arg_max)) {``                ``arg_max = it->first;``                ``currentMax = it->second;``            ``}``        ``}` `        ``// Print the character``        ``// alongwith its frequency``        ``cout << arg_max << ``" - "` `<< currentMax << endl;` `        ``// Delete the maximum value``        ``occ.erase(arg_max);``    ``}``}` `// Driver code``int` `main()``{` `    ``string str = ``"geeksforgeeks"``;``    ``int` `len = str.length();` `    ``printChar(str, len);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``import` `java.util.*;``class` `GFG{` `// Function to print the characters``// of the given String in decreasing``// order of their frequencies``static` `void` `printChar(``char` `[]arr, ``int` `len)``{` `    ``// To store the``    ``HashMap occ = ``new` `HashMap();``    ``for` `(``int` `i = ``0``; i < len; i++)``        ``if``(occ.containsKey(arr[i]))``        ``{``            ``occ.put(arr[i], occ.get(arr[i]) + ``1``);``        ``}``          ``else``        ``{``            ``occ.put(arr[i], ``1``);``        ``}` `    ``// Map's size``    ``int` `size = occ.size();` `    ``// While there are elements in the map``    ``while` `(size-- > ``0``)``    ``{` `        ``// Finding the maximum value``        ``// from the map``        ``int` `currentMax = ``0``;``        ``char` `arg_max = ``0``;``        ``for` `(Map.Entry it : occ.entrySet())``        ``{``            ``if` `(it.getValue() > currentMax ||``               ``(it.getValue() == currentMax &&``                ``it.getKey() > arg_max))``            ``{``                ``arg_max = it.getKey();``                ``currentMax = it.getValue();``            ``}``        ``}` `        ``// Print the character``        ``// alongwith its frequency``        ``System.out.print(arg_max + ``" - "` `+``                         ``currentMax + ``"\n"``);` `        ``// Delete the maximum value``        ``occ.remove(arg_max);``    ``}``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``String str = ``"geeksforgeeks"``;``    ``int` `len = str.length();` `    ``printChar(str.toCharArray(), len);``}``}` `// This code is contributed by gauravrajput1`

## C#

 `// C# implementation of the approach``using` `System;``using` `System.Collections.Generic;` `class` `GFG{` `// Function to print the characters``// of the given String in decreasing``// order of their frequencies``static` `void` `printChar(``char` `[]arr, ``int` `len)``{` `    ``// To store the``    ``Dictionary<``char``,``               ``int``> occ = ``new` `Dictionary<``char``,``                                         ``int``>();``    ``for` `(``int` `i = 0; i < len; i++)``        ``if``(occ.ContainsKey(arr[i]))``        ``{``            ``occ[arr[i]]  = occ[arr[i]] + 1;``        ``}``          ``else``        ``{``            ``occ.Add(arr[i], 1);``        ``}` `    ``// Map's size``    ``int` `size = occ.Count;` `    ``// While there are elements in the map``    ``while` `(size-- > 0)``    ``{` `        ``// Finding the maximum value``        ``// from the map``        ``int` `currentMax = 0;``        ``char` `arg_max = (``char``)0;``        ``foreach` `(KeyValuePair<``char``, ``int``> it ``in` `occ)``        ``{``            ``if` `(it.Value > currentMax ||``               ``(it.Value == currentMax &&``                ``it.Key > arg_max))``            ``{``                ``arg_max = it.Key;``                ``currentMax = it.Value;``            ``}``        ``}` `        ``// Print the character``        ``// alongwith its frequency``        ``Console.Write(arg_max + ``" - "` `+``                      ``currentMax + ``"\n"``);` `        ``// Delete the maximum value``        ``occ.Remove(arg_max);``    ``}``}` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ``String str = ``"geeksforgeeks"``;``    ``int` `len = str.Length;` `    ``printChar(str.ToCharArray(), len);``}``}` `// This code is contributed by Princi Singh`

## Javascript

 ``
Output
```e - 4
s - 2
k - 2
g - 2
r - 1
o - 1
f - 1
```

Approach 2 : We will make an array arr of size one more than the size of given string length in which we will store List of characters whose frequency is equal to the index of arr and follow the below steps :

• Make a frequency map using array of characters present in the given string.
• Traverse frequency array, if its value is greater than zero let say k.
• On kth index of arr store it’s character value in List at index 0(As we need descending order of alphabets if frequency is same).
• Traverse arr from backwards as we need greater frequency first if List at that index is not empty than print its frequency and character.

Implementation of above approach :

## Java

 `// Java implementation of above approach``import` `java.util.*;` `class` `GFG {``    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``String str = ``"geeksforgeeks"``;``        ``printChar(str);``    ``}``    ``@SuppressWarnings``(``"unchecked"``)``    ``// Function to print the characters``    ``// of the given string in decreasing``    ``// order of their frequencies``    ``public` `static` `void` `printChar(String str)``    ``{``        ``// Initializing array of List type.``        ``List[] arr = ``new` `List[str.length() + ``1``];``        ``for` `(``int` `i = ``0``; i <= str.length(); i++) {``            ``// Initializing List of type Character.``            ``arr[i] = ``new` `ArrayList<>();``        ``}``        ``int``[] freq = ``new` `int``[``256``];``        ``// Mapking frequency map``        ``for` `(``int` `i = ``0``; i < str.length(); i++) {``            ``freq[(``char``)str.charAt(i)]++;``        ``}``        ``// Traversing frequency array``        ``for` `(``int` `i = ``0``; i < ``256``; i++) {``            ``if` `(freq[i] > ``0``) {``                ``// If frequency array is greater than zero``                ``// then storing its character on``                ``// i-th(frequency of that character) index``                ``// of arr``                ``arr[freq[i]].add(``0``, (``char``)(i));``            ``}``        ``}``        ``// Traversing arr from backwards as we need greater``        ``// frequency character first``        ``for` `(``int` `i = arr.length - ``1``; i >= ``0``; i--) {``            ``if` `(!arr[i].isEmpty()) {``                ``for` `(``char` `ch : arr[i]) {``                    ``System.out.println(ch + ``"-"` `+ i);``                ``}``            ``}``        ``}``    ``}``}`
Output
```e-4
s-2
k-2
g-2
r-1
o-1
f-1
```

Time Complexity : O(n), n is the length of given string

Auxiliary Space : O(n)

My Personal Notes arrow_drop_up