Print characters in decreasing order of frequency

Last Updated : 31 Aug, 2020

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:

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 <bits/stdc++.h>
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<Character,
              Integer> occ = new HashMap<Character,
                                         Integer>();
    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<Character,
                        Integer> 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

Output:

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

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