# Print characters in decreasing order of frequency

Given a string str, the task is to print the characters in decreasing order of their frequency. If the frequency of two characters is 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

• 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:

 `// 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; ` `} `

Output:

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

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