Print the frequency of each character in Alphabetical order

Given a string str, the task is to print the frequency of each of the characters of str in alphabetical order.
Example:

Input: str = “aabccccddd” 
Output: a2b1c4d3 
Since it is already in alphabetical order, the frequency 
of the characters is returned for each character. 
Input: str = “geeksforgeeks” 
Output: e4f1g2k2o1r1s2 

Approach:

1. Create a Map to store the frequency of each of the characters of the given string.
2. Iterate through the string and check if the character is present in the map.
3. If the character is not present, insert it in the map with 1 as the initial value else increment its frequency by 1.
4. Finally, print the frequency of each of the character in alphabetical order.

Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

const int MAX = 26;

// Function to print the frequency
// of each of the characters of
// s in alphabetical order
void compressString(string s, int n)
{
    // To store the frequency
    // of the characters
    int freq[MAX] = { 0 };

    // Update the frequency array
    for (int i = 0; i < n; i++) {
        freq[s[i] - 'a']++;
    }

    // Print the frequency in alphatecial order
    for (int i = 0; i < MAX; i++) {

        // If the current alphabet doesn't
        // appear in the string
        if (freq[i] == 0)
            continue;

        cout << (char)(i + 'a') << freq[i];
    }
}

// Driver code
int main()
{
    string s = "geeksforgeeks";
    int n = s.length();

    compressString(s, n);

    return 0;
}

Java

// Java implementation of the approach
class GFG
{

    static int MAX = 26;

    // Function to print the frequency
    // of each of the characters of
    // s in alphabetical order
    static void compressString(String s, int n)
    {
        // To store the frequency
        // of the characters
        int freq[] = new int[MAX] ;

        // Update the frequency array
        for (int i = 0; i < n; i++)
        {
            freq[s.charAt(i) - 'a']++;
        }

        // Print the frequency in alphatecial order
        for (int i = 0; i < MAX; i++)
        {

            // If the current alphabet doesn't
            // appear in the string
            if (freq[i] == 0)
                continue;

            System.out.print((char)(i + 'a') +""+ freq[i]);
        }
    }

    // Driver code
    public static void main (String[] args)
    {
        String s = "geeksforgeeks";
        int n = s.length();

        compressString(s, n);
    }
}

// This code is contributed by AnkitRai01

Python3

# Python3 implementation of the approach
MAX = 26;

# Function to print the frequency
# of each of the characters of
# s in alphabetical order
def compressString(s, n) :

    # To store the frequency
    # of the characters
    freq = [ 0 ] * MAX;

    # Update the frequency array
    for i in range(n) :
        freq[ord(s[i]) - ord('a')] += 1;

    # Print the frequency in alphatecial order
    for i in range(MAX) :

        # If the current alphabet doesn't
        # appear in the string
        if (freq[i] == 0) :
            continue;

        print((chr)(i + ord('a')),freq[i],end = " ");

# Driver code
if __name__ == "__main__" :

    s = "geeksforgeeks";
    n = len(s);

    compressString(s, n);

# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;

class GFG
{

    static int MAX = 26;

    // Function to print the frequency
    // of each of the characters of
    // s in alphabetical order
    static void compressString(string s, int n)
    {
        // To store the frequency
        // of the characters
        int []freq = new int[MAX] ;

        // Update the frequency array
        for (int i = 0; i < n; i++)
        {
            freq[s[i] - 'a']++;
        }

        // Print the frequency in alphatecial order
        for (int i = 0; i < MAX; i++)
        {

            // If the current alphabet doesn't
            // appear in the string
            if (freq[i] == 0)
                continue;

            Console.Write((char)(i + 'a') +""+ freq[i]);
        }
    }

    // Driver code
    public static void Main()
    {
        string s = "geeksforgeeks";
        int n = s.Length;

        compressString(s, n);
    }
}

// This code is contributed by AnkitRai01

Javascript

<script>

    // Javascript implementation of the approach

    let MAX = 26;

    // Function to print the frequency
    // of each of the characters of
    // s in alphabetical order
    function compressString(s, n)
    {
        // To store the frequency
        // of the characters
        let freq = new Array(MAX);
        freq.fill(0);

        // Update the frequency array
        for (let i = 0; i < n; i++)
        {
            freq[s[i].charCodeAt() - 'a'.charCodeAt()]++;
        }

        // Print the frequency in alphatecial order
        for (let i = 0; i < MAX; i++)
        {

            // If the current alphabet doesn't
            // appear in the string
            if (freq[i] == 0)
                continue;

            document.write(String.fromCharCode(i +
            'a'.charCodeAt()) +""+ freq[i]);
        }
    }

    let s = "geeksforgeeks";
    let n = s.length;

    compressString(s, n);

</script>

Output:

e4f1g2k2o1r1s2

Time Complexity: O(n) 
Auxiliary Space: O(1)

Using Map(stl):-

 The Approach:

The approach is very simple we are go on traversing and print the all the char with there frequency.

C++

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

int main() {
     //given string.
     string s = "geeksforgeeks";
     //map char-int pair.
     map<char,int>mp;
     //map over string.
     for(auto it:s)mp[it]++;
     //printing the char with there frequency.
     cout<<"All character in string with there frequency :"<<endl;
     for(auto it:mp)cout<<it.first<<it.second;
    return 0;
}

Python3

# Python program for the above approach
s = "geeksforgeeks"
# map of char-int pair
mp = {}

# map over string
for i in range(len(s)):
    if(mp.get(s[i]) != None):
        mp[s[i]] = mp.get(s[i])+1
    else:
        mp[s[i]] = 1

# printing the char with frequency
myKeys = list(mp.keys())
myKeys.sort()
mp = {i : mp[i] for i in myKeys}
for i in mp:
    print(i, end="")
    print(mp[i], end="")

C#

using System;
using System.Collections.Generic;

class GFG {
    static void Main(string[] args) {
        //given string.
        string s = "geeksforgeeks";
        //map char-int pair.
        Dictionary<char, int> mp = new Dictionary<char, int>();
        //map over string.
        foreach (char c in s) {
            if (mp.ContainsKey(c)) {
                mp++;
            } else {
                mp = 1;
            }
        }
        //printing the char with there frequency.
        Console.WriteLine("All character in string with their frequency:");
        foreach (KeyValuePair<char, int> pair in mp) {
            Console.Write(pair.Key + "" + pair.Value);
        }
    }
}

// code by ksam24000

Javascript

// JavaScript program for above approach
// given string
let s = "geeksforgeeks";

// map char-int pair
let mp = new Map();

// map over string
for(let i = 0; i<s.length; i++){
    if(mp.has(s[i])){
        mp.set(s[i], mp.get(s[i])+1);
    }else{
        mp.set(s[i], 1);
    }
}

// printing the cha with there frequency.
console.log("All characters in string with there frequency : ");
var unsortedArray = [ ...mp];
var sortedArray = unsortedArray.sort(([key1, value1], [key2, value2]) => key1.localeCompare(key2))
mp = new Map(sortedArray)
mp.forEach(function(value, key){
    console.log(key+value);
})

// This code is contributed by Yash Agarwal(yashagarwal2852002)

Java

import java.util.*;

public class Main {
    public static void main(String[] args) {
        // given string
        String s = "geeksforgeeks";
        // map char-int pair
        Map<Character, Integer> mp = new HashMap<Character, Integer>();
        // map over string
        for (char c : s.toCharArray()) {
            if (mp.containsKey(c)) {
                mp.put(c, mp.get(c) + 1);
            } else {
                mp.put(c, 1);
            }
        }
        // printing the char with their frequency
        System.out.println("All character in string with their frequency :");
        for (Map.Entry<Character, Integer> entry : mp.entrySet()) {
            System.out.print(entry.getKey() + "" + entry.getValue());
        }
    }
}
// ksam24000

Output

All character in string with there frequency :
e4f1g2k2o1r1s2

Complexity Analysis:

Time Complexity: O(n),for tarversing string.
Auxiliary Space: O(n),for map.

Approach 3: Recursive approach

Algorithm:

1. Take the string “s” as input and calculate size in n.
2. Run a loop from ‘a’ to ‘z’ and call the  “compressString” function for each time  with count = 0 to store the frequency of the letter.
3. The  “compressString” function check if counter i>size of the string then return the function else increment the count if character equal to s[i] and call the  “compressString” again with i+1;
4. Now check if count is not equal to 0 then print the letter with count.
   

C++

// C++ code for above approach
#include <bits/stdc++.h>
using namespace std;

// Recursive function
void compressString(string s, int n, int i, char letter, int &count)
{
  // Base case: if i>size of string the return
  if(i>n-1)
    return;

  // If the letter equal to the s[i] then increment count by 1
  if(s[i]==letter)
    count++;

  // Call the recursive function again with i+1
  compressString(s,n,i+1,letter,count);
}

// Driver code
int main()
{
    string s = "geeksforgeeks"; // input string
    int n = s.length(); // size of string

    for(char letter = 'a'; letter<='z'; letter++)
    {
        int count  = 0; // frequency counter

        // Call the recursive function
        compressString(s,n,0,letter,count);

        if(count==0)  // if count equal to 0 continue.
            continue;

        cout<<letter<<count; // Print the letter with count.
    }

    return 0;
}

Java

import java.util.*;

class Main {

    // Recursive function
    static void compressString(String s, int n, int i, char letter, int[] count) {
        // Base case: if i > size of string then return
        if(i > n - 1) {
            return;
        }

        // If the letter is equal to s[i], then increment count by 1
        if(s.charAt(i) == letter) {
            count[0]++;
        }

        // Call the recursive function again with i+1
        compressString(s, n, i+1, letter, count);
    }

    // Driver code
    public static void main(String[] args) {
        String s = "geeksforgeeks"; // input string
        int n = s.length(); // size of string

        for(char letter = 'a'; letter <= 'z'; letter++) {
            int[] count = {0}; // frequency counter

            // Call the recursive function
            compressString(s, n, 0, letter, count);

            if(count[0] == 0) { // if count is equal to 0, continue
                continue;
            }

            System.out.print(letter + "" + count[0]); // Print the letter with count.
        }
    }
}

C#

using System;

class GFG {
    // Recursive function
    static void compressString(string s, int n, int i,
                               char letter, ref int count)
    {
        // Base case: if i>size of string the return
        if (i > n - 1)
            return;
        // If the letter equal to the s[i] then increment
        // count by 1
        if (s[i] == letter)
            count++;
        // Call the recursive function again with i+1
        compressString(s, n, i + 1, letter, ref count);
    }

    static void Main()
    {
        string s = "geeksforgeeks"; // input string
        int n = s.Length; // size of string

        for (char letter = 'a'; letter <= 'z'; letter++) {
            int count = 0;
            // Call the recursive function
            compressString(s, n, 0, letter, ref count);
            // if count equal to 0 continue.
            if (count == 0)
                continue;

            // Print the letter with count.
            Console.Write(letter);
            Console.Write(count);
        }
    }
}

Python3

# Recursive function to compress a string
def compressString(s, n, i, letter, count):
    # Base case: if i>size of string then return
    if i > n - 1:
        return

    # If the letter is equal to the s[i], then increment count by 1
    if s[i] == letter:
        count += 1

    # Call the recursive function again with i+1
    compressString(s, n, i + 1, letter, count)

    return count

# Driver code
if __name__ == '__main__':
    s = "geeksforgeeks" # input string
    n = len(s) # size of string

    for letter in range(ord('a'), ord('z') + 1):
        letter = chr(letter)
        count = compressString(s, n, 0, letter, 0)

        if count == 0: # if count equal to 0 continue
            continue

        print(letter, count, end = '') # Print the letter with count

Javascript

// Recursive function
function compressString(s, n, i, letter, count) {
  // Base case: if i > size of string then return
  if (i > n - 1) {
    return;
  }

  // If the letter is equal to s[i], then increment count by 1
  if (s.charAt(i) === letter) {
    count[0]++;
  }

  // Call the recursive function again with i+1
  compressString(s, n, i + 1, letter, count);
}

// Driver code
const s = "geeksforgeeks"; // input string
const n = s.length; // size of string
ans = "";
for (let letter = "a"; letter <= "z"; letter = String.fromCharCode(letter.charCodeAt(0) + 1)) {
  const count = [0]; // frequency counter

  // Call the recursive function
  compressString(s, n, 0, letter, count);

  if (count[0] === 0) { // if count is equal to 0, continue
    continue;
  }

  ans  = ans +letter + "" + count[0]; // Print the letter with count.
} console.log(ans);

Output

e4f1g2k2o1r1s2

Time Complexity:  O(n)
Auxiliary Space: O(1)