Imprimir caracteres y sus frecuencias en orden de aparición

String dada str que contiene solo caracteres en minúsculas. El problema es imprimir los caracteres junto con su frecuencia en el orden en que aparecen y en el formato dado que se explica en los ejemplos a continuación.

Ejemplos: 

Input : str = "geeksforgeeks"
Output : g2 e4 k2 s2 f1 o1 r1

Input : str = "elephant"
Output : e2 l1 p1 h1 a1 n1 t1

Fuente: Experiencia de entrevista de SAP | Conjunto 26

Enfoque: Cree una array de conteo para almacenar la frecuencia de cada carácter en la string dada str . Recorra la string str nuevamente y verifique si la frecuencia de ese carácter es 0 o no. Si no es 0, imprima el carácter junto con su frecuencia y actualice su frecuencia a 0 en la tabla hash. Esto se hace para que no se vuelva a imprimir el mismo carácter.

// C++ implementation to print the character and
// its frequency in order of its occurrence
#include <bits/stdc++.h>
using namespace std;
 
#define SIZE 26
 
// function to print the character and its frequency
// in order of its occurrence
void printCharWithFreq(string str)
{
    // size of the string 'str'
    int n = str.size();
 
    // 'freq[]' implemented as hash table
    int freq[SIZE];
 
    // initialize all elements of freq[] to 0
    memset(freq, 0, sizeof(freq));
 
    // accumulate frequency of each character in 'str'
    for (int i = 0; i < n; i++)
        freq[str[i] - 'a']++;
 
    // traverse 'str' from left to right
    for (int i = 0; i < n; i++) {
 
        // if frequency of character str[i] is not
        // equal to 0
        if (freq[str[i] - 'a'] != 0) {
 
            // print the character along with its
            // frequency
            cout << str[i] << freq[str[i] - 'a'] << " ";
 
            // update frequency of str[i] to 0 so
            // that the same character is not printed
            // again
            freq[str[i] - 'a'] = 0;
        }
    }
}
 
// Driver program to test above
int main()
{
    string str = "geeksforgeeks";
    printCharWithFreq(str);
    return 0;
}

// Java implementation to print the character and
// its frequency in order of its occurrence
public class Char_frequency {
     
    static final int SIZE = 26;
      
    // function to print the character and its
    // frequency in order of its occurrence
    static void printCharWithFreq(String str)
    {
         // size of the string 'str'
        int n = str.length();
 
        // 'freq[]' implemented as hash table
        int[] freq = new int[SIZE];
 
        // accumulate frequency of each character
        // in 'str'
        for (int i = 0; i < n; i++)
            freq[str.charAt(i) - 'a']++;
 
        // traverse 'str' from left to right
        for (int i = 0; i < n; i++) {
 
            // if frequency of character str.charAt(i)
            // is not equal to 0
            if (freq[str.charAt(i) - 'a'] != 0) {
 
                // print the character along with its
                // frequency
                System.out.print(str.charAt(i));
                System.out.print(freq[str.charAt(i) - 'a'] + " ");
 
                // update frequency of str.charAt(i) to
                // 0 so that the same character is not
                // printed again
                freq[str.charAt(i) - 'a'] = 0;
            }
        }
    }
      
    // Driver program to test above
    public static void main(String args[])
    {
        String str = "geeksforgeeks";
        printCharWithFreq(str);
    }
}
// This code is contributed by Sumit Ghosh

# Python3 implementation to pr the character and
# its frequency in order of its occurrence
 
# import library
import numpy as np
 
# Function to print the character and its
# frequency in order of its occurrence
def prCharWithFreq(str) :
     
    # Size of the 'str'
    n = len(str)
     
    # Initialize all elements of freq[] to 0
    freq = np.zeros(26, dtype = np.int)
 
    # Accumulate frequency of each
    # character in 'str'
    for i in range(0, n) :
        freq[ord(str[i]) - ord('a')] += 1
                 
    # Traverse 'str' from left to right
    for i in range(0, n) :
         
        # if frequency of character str[i] 
        # is not equal to 0
        if (freq[ord(str[i])- ord('a')] != 0) :
             
            # print the character along
            # with its frequency
            print (str[i], freq[ord(str[i]) - ord('a')],
                                                end = " ")
 
            # Update frequency of str[i] to 0 so that
            # the same character is not printed again
            freq[ord(str[i]) - ord('a')] = 0
         
     
# Driver Code
if __name__ == "__main__" :
     
    str = "geeksforgeeks";
    prCharWithFreq(str);
     
# This code is contributed by 'Saloni1297'

// C# implementation to print the
// character and its frequency in
// order of its occurrence
using System;
 
class GFG {
    static int SIZE = 26;
 
    // function to print the character and its
    // frequency in order of its occurrence
    static void printCharWithFreq(String str)
    {
        // size of the string 'str'
        int n = str.Length;
 
        // 'freq[]' implemented as hash table
        int[] freq = new int[SIZE];
 
        // accumulate frequency of each character
        // in 'str'
        for (int i = 0; i < n; i++)
            freq[str[i] - 'a']++;
 
        // traverse 'str' from left to right
        for (int i = 0; i < n; i++) {
 
            // if frequency of character str.charAt(i)
            // is not equal to 0
            if (freq[str[i] - 'a'] != 0) {
 
                // print the character along with its
                // frequency
                Console.Write(str[i]);
                Console.Write(freq[str[i] - 'a'] + " ");
 
                // update frequency of str.charAt(i) to
                // 0 so that the same character is not
                // printed again
                freq[str[i] - 'a'] = 0;
            }
        }
    }
 
    // Driver program to test above
    public static void Main()
    {
        String str = "geeksforgeeks";
        printCharWithFreq(str);
    }
}
 
// This code is contributed by Sam007

<?php
// PHP implementation to print the
// character and its frequency in
// order of its occurrence
$SIZE = 26;
 
// function to print the character and
// its frequency in order of its occurrence
function printCharWithFreq($str)
{
    global $SIZE;
     
    // size of the string 'str'
    $n = strlen($str);
 
    // 'freq[]' implemented as hash table
    $freq = array_fill(0, $SIZE, NULL);
 
    // accumulate frequency of each
    // character in 'str'
    for ($i = 0; $i < $n; $i++)
        $freq[ord($str[$i]) - ord('a')]++;
 
    // traverse 'str' from left to right
    for ($i = 0; $i < $n; $i++)
    {
 
        // if frequency of character str[i]
        // is not equal to 0
        if ($freq[ord($str[$i]) - ord('a')] != 0)
        {
 
            // print the character along with
            // its frequency
            echo $str[$i] . $freq[ord($str[$i]) -
                                  ord('a')] . " ";
 
            // update frequency of str[i] to 0
            // so that the same character is
            // not printed again
            $freq[ord($str[$i]) - ord('a')] = 0;
        }
    }
}
 
// Driver Code
$str = "geeksforgeeks";
printCharWithFreq($str);
 
// This code is contributed by ita_c
?>

<script>
// Javascript implementation to print the character and
// its frequency in order of its occurrence
     
    let SIZE = 26;
     
    // function to print the character and its
    // frequency in order of its occurrence
    function printCharWithFreq(str)
    {
     
        // size of the string 'str'
        let n = str.length;
  
        // 'freq[]' implemented as hash table
        let freq = new Array(SIZE);
        for(let i = 0; i < freq.length; i++)
        {
            freq[i] = 0;
        }
  
        // accumulate frequency of each character
        // in 'str'
        for (let i = 0; i < n; i++)
            freq[str[i].charCodeAt(0) - 'a'.charCodeAt(0)]++;
  
        // traverse 'str' from left to right
        for (let i = 0; i < n; i++) {
  
            // if frequency of character str.charAt(i)
            // is not equal to 0
            if (freq[str[i].charCodeAt(0) - 'a'.charCodeAt(0)] != 0) {
  
                // print the character along with its
                // frequency
                document.write(str[i]);
                document.write(freq[str[i].charCodeAt(0) - 'a'.charCodeAt(0)] + " ");
  
                // update frequency of str.charAt(i) to
                // 0 so that the same character is not
                // printed again
                freq[str[i].charCodeAt(0) - 'a'.charCodeAt(0)] = 0;
            }
        }
    }
     
    // Driver program to test above
    let str = "geeksforgeeks";
    printCharWithFreq(str);
     
    // This code is contributed by rag2127.
     
</script>

g2 e4 k2 s2 f1 o1 r1 

Complejidad de tiempo: O(n), donde n es el número de caracteres de la string. 
Espacio Auxiliar: O(1), ya que solo hay letras minúsculas.

Solución alternativa (usar hashing) 
También podemos usar hash para resolver el problema. 

// C++ implementation to
//print the characters and
// frequencies in order
// of its occurrence
#include <bits/stdc++.h>
using namespace std;
 
void prCharWithFreq(string s)
{
  // Store all characters and
  // their frequencies in dictionary
  unordered_map<char, int> d;
 
  for(char i : s)
  {
    d[i]++;
  }
 
  // Print characters and their
  // frequencies in same order
  // of their appearance
  for(char i : s)
  {
    // Print only if this
    // character is not printed
    // before
    if(d[i] != 0)
    {
      cout << i << d[i] << " ";
      d[i] = 0;
    }
  }
}
  
// Driver Code
int main()
{
  string s="geeksforgeeks";
  prCharWithFreq(s);
}
 
// This code is contributed by rutvik_56

// Java implementation to
// print the characters and
// frequencies in order
// of its occurrence
import java.util.*;
class Gfg{
    public static void prCharWithFreq(String s)
    {
       
        // Store all characters and
        // their frequencies in dictionary
        Map<Character, Integer> d = new HashMap<Character, Integer>();
         
         
        for(int i = 0; i < s.length(); i++)
        {
            if(d.containsKey(s.charAt(i)))
            {
                d.put(s.charAt(i), d.get(s.charAt(i)) + 1);
            }
            else
            {
                d.put(s.charAt(i), 1);
            }
        }
         
        // Print characters and their
        // frequencies in same order
        // of their appearance
        for(int i = 0; i < s.length(); i++)
        {
           
            // Print only if this
            // character is not printed
            // before
            if(d.get(s.charAt(i)) != 0)
            {
                System.out.print(s.charAt(i));
                System.out.print(d.get(s.charAt(i)) + " ");
                d.put(s.charAt(i), 0);
            }           
        }
    }
   
    // Driver code
     public static void main(String []args)
     {
       String S = "geeksforgeeks";
       prCharWithFreq(S);
     }
}
 
// This code is contributed by avanitrachhadiya2155

# Python3 implementation to print the characters and
# frequencies in order of its occurrence
 
def prCharWithFreq(str):
 
    # Store all characters and their frequencies
    # in dictionary
    d = {}
    for i in str:
        if i in d:
            d[i] += 1
        else:
            d[i] = 1
     
    # Print characters and their frequencies in
    # same order of their appearance
    for i in str:
 
        # Print only if this character is not printed
        # before. 
        if d[i] != 0:
            print("{}{}".format(i,d[i]), end =" ")
            d[i] = 0
      
     
# Driver Code
if __name__ == "__main__" :
     
    str = "geeksforgeeks";
    prCharWithFreq(str);
     
# This code is contributed by 'Ankur Tripathi'

// C# implementation to
// print the characters and
// frequencies in order
// of its occurrence
 
using System;
using System.Collections;
using System.Collections.Generic;
 
class GFG
{
     
public static void prCharWithFreq(string s)
{
   
  // Store all characters and
  // their frequencies in dictionary
  Dictionary<char,int> d = new Dictionary<char,int>();
  
  foreach(char i in s)
  {
      if(d.ContainsKey(i))
      {
        d[i]++;
      }
      else
      {
        d[i]=1; 
      }
  }
  
  // Print characters and their
  // frequencies in same order
  // of their appearance
  foreach(char i in s)
  {
    // Print only if this
    // character is not printed
    // before
    if(d[i] != 0)
    {
        Console.Write(i+d[i].ToString() + " ");
        d[i] = 0;
    }
  }
}
   
// Driver Code
public static void Main(string []args)
{
  string s="geeksforgeeks";
  prCharWithFreq(s);
}
}
 
// This code is contributed by pratham76

<script>
// Javascript implementation to
//print the characters and
// frequencies in order
// of its occurrence
 
function prCharWithFreq(s)
{
  // Store all characters and
  // their frequencies in dictionary
  var d = new Map();
 
  s.split('').forEach(element => {
     
        if(d.has(element))
        {
            d.set(element, d.get(element)+1);
        }
        else
            d.set(element, 1);
  });
 
  // Print characters and their
  // frequencies in same order
  // of their appearance
 
  s.split('').forEach(element => {
    // Print only if this
    // character is not printed
    // before
    if(d.has(element) && d.get(element)!=0)
    {
      document.write( element + d.get(element) + " ");
      d.set(element, 0);
    }
  });
}
  
// Driver Code
var s="geeksforgeeks";
prCharWithFreq(s);
 
</script>

g2 e4 k2 s2 f1 o1 r1 

Complejidad de tiempo: O(n), donde n es el número de caracteres de la string. 
Espacio Auxiliar: O(n), 

Método #3: Usando la programación Orientada a Objetos:

Podemos resolver este problema sin usar un HashMap también. Pero luego tenemos que crear nuestra propia clase cuyos objetos tendrán 2 propiedades: el carácter y su ocurrencia.

1. Creamos una clase en este caso llamada CharOccur e inicializamos 2 variables carácter y ocurrencia en su constructor.
2. En la parte principal de nuestra clase original, simplemente crearemos una lista que pueda almacenar estos objetos.

lógica – 

• Bucle a través de la cuerda.
• Compruebe si el carácter actual de la string ya está presente en algún objeto. Si está presente, incremente su ocurrencia; de lo contrario, establezca su ocurrencia en 1.

/*package whatever //do not write package name here */
 
import java.io.*;
import java.util.ArrayList;
 
class GFG {
    public static void main(String[] args)
    {
        String s1 = "GFG";
        System.out.println("For " + s1);
        frequency(s1);
 
        String s2 = "aaabccccffgfghc";
        System.out.println("For " + s1);
        frequency(s2);
    }
    private static void frequency(String s)
    {
        if (s.length() == 0) {
            System.out.println("Empty string");
            return;
        }
        ArrayList<CharOccur> occurrences
            = new ArrayList<CharOccur>();
        // Creating ArrayList of objects of Charoccur class
 
        for (int i = 0; i < s.length(); i++) {
            /* Logic
             * If a pair of character and its occurrence is
             * already present as object - increment the
             * occurrence else create a new object of
             * character with its occurrence set to 1
             */
            char c = s.charAt(i);
            int flag = 0;
            for (CharOccur o : occurrences) {
                if (o.character == c) {
                    o.occurrenece += 1;
                    flag = 1;
                }
            }
            if (flag == 0) {
                CharOccur grp = new CharOccur(c, 1);
                occurrences.add(grp);
            }
        }
 
        // Printing the character - occurrences pair
        for (CharOccur o : occurrences) {
            System.out.println(o.character + " "
                               + o.occurrenece);
        }
    }
}
 
// Creating a class CharOccur whose objects have 2
// properties - a character and its occurrence
class CharOccur {
    char character;
    int occurrenece = 0;
    CharOccur(char character, int occurrenece)
    {
        this.character = character;
        this.occurrenece = occurrenece;
    }
}
// Contributed by Soham Ratnaparkhi

For GFG
G 2
F 1
For GFG
a 3
b 1
c 5
f 3
g 2
h 1

Análisis de complejidad:

• Complejidad de tiempo: O(n), donde n es el número de caracteres de la string. 
• Espacio Auxiliar: O(n)
   

Método n.º 4: uso de funciones integradas de Python:

Podemos resolver este problema rápidamente usando el método python Counter() . El enfoque es muy simple.

1) Primero cree un diccionario usando el método Counter que tenga strings como claves y sus frecuencias como valores.

2) Recorra en este diccionario las claves de impresión junto con sus valores

# Python3 implementation to print the characters and
# frequencies in order of its occurrence
from collections import Counter
 
def prCharWithFreq(string):
 
    # Store all characters and their
    # frequencies using Counter function
    d = Counter(string)
 
    # Print characters and their frequencies in
    # same order of their appearance
    for i in d:
        print(i+str(d[i]), end=" ")
 
 
string = "geeksforgeeks"
prCharWithFreq(string)
 
# This code is contributed by vikkycirus

g2 e4 k2 s2 f1 o1 r1 

Este artículo es una contribución de Ayush Jauhari . Si te gusta GeeksforGeeks y te gustaría contribuir, también puedes escribir un artículo usando write.geeksforgeeks.org o enviar tu artículo por correo a review-team@geeksforgeeks.org. Vea su artículo que aparece en la página principal de GeeksforGeeks y ayude a otros Geeks.
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