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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 <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

Javascript




<script>
// Javascript implementation of the approach
 
// Function to print the characters
// of the given String in decreasing
// order of their frequencies
function printChar(arr, len)
{
 
    // To store the
    let  occ = new Map();
    for (let i = 0; i < len; i++)
        if(occ.has(arr[i]))
        {
            occ.set(arr[i], occ.get(arr[i]) + 1);
        }
          else
        {
            occ.set(arr[i], 1);
        }
  
    // Map's size
    let size = occ.size;
  
    // While there are elements in the map
    while (size-- > 0)
    {
  
        // Finding the maximum value
        // from the map
        let currentMax = 0;
        let arg_max = 0;
        for (let [key, value] of occ.entries())
        {
            if (value > currentMax ||
               (value == currentMax &&
                key > arg_max))
            {
                arg_max = key;
                currentMax = value;
            }
        }
  
        // Print the character
        // alongwith its frequency
        document.write(arg_max + " - " +
                         currentMax + "<br>");
  
        // Delete the maximum value
        occ.delete(arg_max);
    }
}
 
// Driver code
let str = "geeksforgeeks";
let len = str.length;
 
printChar(str.split(""), len);
 
// This code is contributed by patel2127
</script>
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<Character>[] 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)

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