Sort a string according to the frequency of characters

Given a string str, the task is to sort the string according to the frequency of each character, in ascending order. If two elements have the same frequency, then they are sorted in lexicographical order.
Examples:

Input: str = “geeksforgeeks” 
Output: forggkksseeee 
Explanation:
Frequency of characters: g2 e4 k2 s2 f1 o1 r1 
Sorted characters according to frequency: f1 o1 r1 g2 k2 s2 e4 
f, o, r occurs one time so they are ordered lexicographically and so are g, k and s. 
Hence the final output is forggkksseeee.
Input: str = “abc” 
Output: abc 

Approach The idea is to store each character with its frequency in a vector of pairs and then sort the vector pairs according to the frequency stored. Finally, print the vector in order.
Below is the implementation of the above approach:

C++

// C++ implementation to Sort strings
// according to the frequency of
// characters in ascending order

#include <bits/stdc++.h>
using namespace std;

// Returns count of character in the string
int countFrequency(string str, char ch)
{
    int count = 0;

    for (int i = 0; i < str.length(); i++)

        // Check for vowel
        if (str[i] == ch)
            ++count;

    return count;
}

// Function to sort the string
// according to the frequency
void sortArr(string str)
{
    int n = str.length();

    // Vector to store the frequency of
    // characters with respective character
    vector<pair<int, char> > vp;

    // Inserting frequency
    // with respective character
    // in the vector pair
    for (int i = 0; i < n; i++) {

        vp.push_back(
            make_pair(
                countFrequency(str, str[i]),
                str[i]));
    }

    // Sort the vector, this will sort the pair
    // according to the number of characters
    sort(vp.begin(), vp.end());

    // Print the sorted vector content
    for (int i = 0; i < vp.size(); i++)
        cout << vp[i].second;
}

// Driver code
int main()
{
    string str = "geeksforgeeks";

    sortArr(str);

    return 0;
}

Java

import java.util.*;

class GFG {
    // Returns count of character in the string
    static int countFrequency(String str, char ch)
    {
        int count = 0;

        for (int i = 0; i < str.length(); i++) {
            // Check for character
            if (str.charAt(i) == ch) {
                ++count;
            }
        }

        return count;
    }

    // Function to sort the string according to the
    // frequency of characters in ascending order
    static void sortArr(String str)
    {
        int n = str.length();

        // Dictionary to store the frequency of characters
        Map<Character, Integer> freqDict
            = new HashMap<Character, Integer>();

        // Count the frequency of each character in the
        // input string
        for (int i = 0; i < n; i++) {
            if (freqDict.containsKey(str.charAt(i))) {
                freqDict.put(str.charAt(i),
                             freqDict.get(str.charAt(i))
                                 + 1);
            }
            else {
                freqDict.put(str.charAt(i), 1);
            }
        }

        // Sort the dictionary by value (frequency) in
        // ascending order
        List<Map.Entry<Character, Integer> > sortedDict
            = new ArrayList<Map.Entry<Character, Integer> >(
                freqDict.entrySet());
        Collections.sort(
            sortedDict,
            new Comparator<
                Map.Entry<Character, Integer> >() {
                public int compare(
                    Map.Entry<Character, Integer> o1,
                    Map.Entry<Character, Integer> o2)
                {
                    return (o1.getValue()
                            == (o2.getValue()))
                        ? o1.getKey() - o2.getKey()
                        : o1.getValue() - o2.getValue();
                }
            });

        // Print the sorted characters in the order of their
        // frequency
        for (Map.Entry<Character, Integer> entry :
             sortedDict) {
            for (int i = 0; i < entry.getValue(); i++) {
                System.out.print(entry.getKey());
            }
        }
    }

    // Driver code
    public static void main(String[] args)
    {
        String str = "geeksforgeeks";

          // Driver code
        sortArr(str);
    }
}

Python3

# Python3 implementation to Sort strings
# according to the frequency of
# characters in ascending order

# Returns count of character in the string
def countFrequency(string ,  ch) :

    count = 0;

    for i in range(len(string)) :

        # Check for vowel
        if (string[i] == ch) :
            count += 1;

    return count;

# Function to sort the string
# according to the frequency
def sortArr(string) :
    n = len(string);

    # Vector to store the frequency of
    # characters with respective character
    vp = [];

    # Inserting frequency
    # with respective character
    # in the vector pair
    for i in range(n) :

        vp.append((countFrequency(string, string[i]), string[i]));

    # Sort the vector, this will sort the pair
    # according to the number of characters
    vp.sort();

    # Print the sorted vector content
    for i in range(len(vp)) :
        print(vp[i][1],end="");

# Driver code
if __name__ == "__main__" :

    string = "geeksforgeeks";

    sortArr(string);

    # This code is contributed by Yash_R

C#

using System;
using System.Collections.Generic;
using System.Linq;

public class Program
{
  // Returns count of character in the string
  static int countFrequency(string str, char ch)
  {
    int count = 0;

    for (int i = 0; i < str.Length; i++)
    {
      // Check for character
      if (str[i] == ch)
      {
        ++count;
      }
    }

    return count;
  }

  // Function to sort the string according to the
  // frequency of characters in ascending order
  static void sortArr(string str)
  {
    int n = str.Length;

    // Dictionary to store the frequency of characters
    Dictionary<char, int> freqDict = new Dictionary<char, int>();

    // Count the frequency of each character in the input string
    for (int i = 0; i < n; i++)
    {
      if (freqDict.ContainsKey(str[i]))
      {
        freqDict[str[i]]++;
      }
      else
      {
        freqDict[str[i]] = 1;
      }
    }

    // Sort the dictionary by value (frequency) in ascending order
    var sortedDict = freqDict.OrderBy(x => x.Value);

    // Print the sorted characters in the order of their frequency
    foreach (var kvp in sortedDict)
    {
      for (int i = 0; i < kvp.Value; i++)
      {
        Console.Write(kvp.Key);
      }
    }
  }

  // Driver code
  static void Main(string[] args)
  {
    string str = "geeksforgeeks";

    sortArr(str);
  }
}

Javascript

<script>

// JavaScript implementation of the above approach

// Returns count of character in the string
function countFrequency(str, ch)
{
    var count = 0;

    for (var i = 0; i < str.length; i++)

        // Check for vowel
        if (str[i] == ch)
            ++count;

    return count;
}

// Function to sort the string
// according to the frequency
function sortArr(str)
{
    var n = str.length;

    // Vector to store the frequency of
    // characters with respective character
    vp = new Array(n);

    // Inserting frequency
    // with respective character
    // in the vector pair
    for (var i = 0; i < n; i++) {

        vp[i] = [countFrequency(str, str[i]), str[i]];
    }

    // Sort the vector, this will sort the pair
    // according to the number of characters
    vp.sort();

    // Print the sorted vector content
    for (var i = 0; i < n; i++)
        document.write(vp[i][1]);
}

// Driver Code

// Array of points
let str = "geeksforgeeks";
sortArr(str);

</script>

Output

forggkksseeee

Time Complexity: O(n²)

Auxiliary Space: O(n)

Method 2: (Optimized Approach – Min Heap Based)

Algorithm:

1. Take the frequency of each character into a map.

2 .Take a MIN Heap, store in FREQUENCY, CHAR

3. After all insertions, Topmost element is the less frequent character

4. We keep a CUSTOM COMPARATOR for LESS FREQ, WHEN SAME FREQ – Ascending Order Characters.

5. Then Pop one by one and append in ANS String for FREQ no. of times.

CODE:

C++

#include <bits/stdc++.h>
using namespace std;

//O(N*LogN) Time, O(Distinct(N)) Space
//MIN HEAP Based - as we need less frequent element first
#define ppi pair<int,char>

//CUSTOM COMPARATOR for Heap
class Compare{
  public:
  //Override
  bool operator()(pair<int,char>below, pair<int,char> above){
    if(below.first == above.first){
      //freq same
      return below.second > above.second; //lexicographically smaller is TOP
    }
    return below.first > above.first; //less freq at TOP
  }
};

string frequencySort(string s) {

  unordered_map<char,int> mpp;
  priority_queue<ppi,vector<ppi>,Compare> minH; // freq , character

  for(char ch : s){
    mpp[ch]++;
  }

  for(auto m : mpp){
    minH.push({m.second, m.first}); // as freq is 1st , char is 2nd
  }

  string ans="";
  //Now we have in the TOP - Less Freq chars

  while(minH.size()>0){

    int freq = minH.top().first;
    char ch = minH.top().second;
    for(int i=0; i<freq; i++){
      ans+=ch; // append as many times of freq
    }
    minH.pop(); //Heapify happens
  }

  return ans;

}

// Driver code
int main()
{
    string str = "geeksforgeeks";

    cout<<frequencySort(str)<<"\n";

    return 0;
}

//Code is Contributed by Balakrishnan R (rbkraj000)

Java

import java.util.*;
class Pair implements Comparable<Pair>
{
    int first;
    char second;
    Pair(int first,char second)
    {
        this.first  = first;
        this.second  = second;
    }
    // Custom comparator useful for heap

    public int compareTo(Pair a)
    {
       // If frequencies are same for two characters
       // sort according to their order
        if(this.first==a.first)
            return this.second-a.second;
         return this.first-a.first;
    }

}
class Main {
    // O(N*LogN) Time, O(Distinct(N)) Space
    public static String frequencySort(String s) {

        // Creating a HashMap to store the frequency of characters
        HashMap<Character, Integer> mpp = new HashMap<Character, Integer>();

        // Creating a min heap to store the frequency and corresponding character
        PriorityQueue<Pair> min_heap = new PriorityQueue<Pair>();

        // Looping through the string to calculate the frequency of each character
        for (char ch : s.toCharArray()) {
            mpp.put(ch, mpp.getOrDefault(ch, 0) + 1);
        }

        // Adding the frequency and character to the min heap
        for (char m : mpp.keySet()) {
            min_heap.offer(new Pair(mpp.get(m), m));
        }

        String ans = "";
        // Now we have in the TOP - Less Freq chars
        while (!min_heap.isEmpty()) {
            Pair pair = min_heap.poll();
            int freq = pair.first;
            char ch = pair.second;
            // Append as many times of frequency
            for (int i = 0; i < freq; i++) {
                ans += ch;
            }
        }
        return ans;
    }

    // Driver code
    public static void main(String[] args) {
        String str = "geeksforgeeks";
        System.out.println(frequencySort(str));
    }

}

Python3

import heapq

# O(N*LogN) Time, O(Distinct(N)) Space
def frequencySort(s):
    mpp = {}
    min_heap = []

    for ch in s:
        if ch in mpp:
            mpp[ch] += 1
        else:
            mpp[ch] = 1

    for m in mpp:
        heapq.heappush(min_heap, (mpp[m], m)) # as freq is 1st , char is 2nd

    ans = ""
    #Now we have in the TOP - Less Freq chars
    while min_heap:
        freq, ch = heapq.heappop(min_heap)
        ans += ch * freq # append as many times of freq
    return ans

# Driver code
if __name__ == '__main__':
    str = "geeksforgeeks"
    print(frequencySort(str))

# This code is contributed by Prince Kumar

C#

// C# approach
using System;
using System.Collections.Generic;

class Pair : IComparable<Pair> {
    public int first;
    public char second;
    public Pair(int first, char second)
    {
        this.first = first;
        this.second = second;
    }

    // Custom comparator useful for heap
    public int CompareTo(Pair a)
    {
        // If frequencies are same for two characters
        // sort according to their order
        if (this.first == a.first)
            return this.second - a.second;
        return this.first - a.first;
    }
}
class Program {
    // O(N*LogN) Time, O(Distinct(N)) Space
    public static string frequencySort(string s)
    {
        // Creating a HashMap to store the frequency of
        // characters
        Dictionary<char, int> mpp
            = new Dictionary<char, int>();

        // Creating a min heap to store the frequency and
        // corresponding character
        SortedSet<Pair> min_heap = new SortedSet<Pair>();

        // Looping through the string to calculate the
        // frequency of each character
        foreach(char ch in s.ToCharArray())
        {
            if (mpp.ContainsKey(ch))
                mpp[ch] = mpp[ch] + 1;
            else
                mpp[ch] = 1;
        }

        // Adding the frequency and character to the min
        // heap
        foreach(char m in mpp.Keys)
        {
            min_heap.Add(new Pair(mpp[m], m));
        }

        string ans = "";
        // Now we have in the TOP - Less Freq chars
        while (min_heap.Count > 0) {
            Pair pair = min_heap.Min;
            int freq = pair.first;
            char ch = pair.second;
            // Append as many times of frequency
            for (int i = 0; i < freq; i++) {
                ans += ch;
            }
            min_heap.Remove(pair);
        }
        return ans;
    }

    // Driver code
    public static void Main(string[] args)
    {
        string str = "geeksforgeeks";
        Console.WriteLine(frequencySort(str));
    }
}

// This code is contributed by Susobhan Akhuli

Output

forggkksseeee

Time Complexity:

O(N+ N* Log N + N* Log N)  =  O(N* Log N) 

Reason:

• 1 insertion in heap takes O(Log N), For N insertions O(N*LogN). (HEAP = Priority Queue)
• 1 deletion in heap takes O(Log N), For N insertions O(N*LogN).
• Unordered Map takes O(1) for 1 Insertion. 
• N is the length of the String S (input)

Extra Space Complexity:

O(N)

Reason:

• Map takes O(Distinct(N)) Space.
• Heap also takes O(Distinct(N)) Space.
• N is the length of the String S (input)

The Code, Approach, and Idea are proposed by Balakrishnan R (rbkraj000 GFG ID)