Operaciones requeridas para hacer que la string esté vacía

Dada una string str , la tarea es hacer que la string esté vacía con la operación dada. En una sola operación, puede seleccionar algunos caracteres de la string (cada uno de los caracteres seleccionados debe tener la misma frecuencia) y eliminarlos de la string. Imprime el total de operaciones requeridas para dejar la string vacía.
Ejemplos: 
 

Entrada: str = “aabbccc” 
Salida: 2 
En una sola operación, los caracteres ‘a’ y ‘b’ pueden eliminarse ya que ambos tienen la misma frecuencia. 
La segunda operación puede eliminar el carácter ‘c’ que tiene una frecuencia de 3. 
Se requieren 2 operaciones en total.
Entrada: str = «geeksforgeeks» 
Salida: 3 
 

Enfoque: encuentre frecuencias únicas de los caracteres de la string. El recuento total de frecuencias únicas será el número de operaciones necesarias para que la string quede vacía. 
Para str = “aaabbbcccc” , las frecuencias únicas son 3 y 4 . El recuento total de frecuencias únicas es 2 . 
HashMap se puede usar para almacenar los caracteres y sus frecuencias, luego HashSet se puede usar para encontrar el recuento de frecuencias únicas, que es la cantidad de operaciones requeridas.
A continuación se muestra la implementación del enfoque anterior: 
 

// CPP implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the count of operations required
int totalOperations(string str, int len)
{
    // HashMap to store characters and their frequencies
    unordered_map<char, int> h;
    for (int i = 0; i < len; i++)
 
        // If already contains the character then
        // increment its frequency by 1
        h[str[i]]++;
 
    // HashSet to store unique frequency
    unordered_set<int> hs;
 
    // Insert frequencies into HashSet
    for (auto i : h)
        hs.insert(i.second);
 
    // Count of unique frequencies
    return hs.size();
}
 
// Driver Code
int main()
{
    string str = "geeksforgeeks";
    int len = str.length();
 
    cout << totalOperations(str, len) << endl;
    return 0;
}
 
// This code is contributed by
// sanjeev2552

// Java implementation of the approach
import java.util.*;
class GFG {
 
    // Function to return the count of operations required
    static int totalOperations(String str, int len)
    {
 
        // HashMap to store characters and their frequencies
        HashMap<Character, Integer> h = new HashMap<Character, Integer>();
        for (int i = 0; i < len; i++) {
 
            // If already contains the character then
            // increment its frequency by 1
            if (h.containsKey(str.charAt(i)))
                h.put(str.charAt(i), h.get(str.charAt(i)) + 1);
 
            // Else add the character to the HashMap with frequency 1
            else
                h.put(str.charAt(i), 1);
        }
 
        // Set to iterate over HashMap
        Set<Map.Entry<Character, Integer> > set = h.entrySet();
 
        // HashSet to store unique frequency
        HashSet<Integer> hs = new HashSet<Integer>();
 
        // Insert frequencies into HashSet
        for (Map.Entry<Character, Integer> me : set)
            hs.add(me.getValue());
 
        // Count of unique frequencies
        return hs.size();
    }
 
    // Driver code
    public static void main(String[] args)
    {
        String str = "geeksforgeeks";
        int len = str.length();
        System.out.println(totalOperations(str, len));
    }
}

# Python implementation of the approach
 
# Function to return the count of operations required
def totalOperations(st, length):
     
    # Dictionary to store characters and their frequencies
    d = {}
    for i in range(length):
         
        # If already contains the character then
        # increment its frequency by 1
        if st[i] in d:
            d[st[i]] += 1
         
        # Else add the character to the HashMap with frequency 1
        else:
            d[st[i]] = 1
 
    # Set to Store unique frequency
    valueSet = set()
 
    # Insert frequencies into HashSet
    for key in d.keys():
        valueSet.add(d[key])
     
    # Count of unique frequencies
    return len(valueSet)
 
# Driver Code
st = "geeksforgeeks"
l = len(st)
print(totalOperations(st, l))
 
# This code is contributed by Vivekkumar Singh

// C# implementation of the approach
using System;
using System.Collections.Generic;
 
class GFG
{
 
// Function to return
// the count of operations required
static int totalOperations(String str, int len)
{
 
    // HashMap to store characters
    // and their frequencies
    Dictionary<char,
               int> h = new Dictionary<char,
                                       int>();
    for (int i = 0; i < len; i++)
    {
 
        // If already contains the character then
        // increment its frequency by 1
        if (h.ContainsKey(str[i]))
            h[str[i]] = h[str[i]] + 1;
 
        // Else add the character
        // to the HashMap with frequency 1
        else
            h.Add(str[i], 1);
    }
 
    // Set to iterate over HashMap
    // HashSet to store unique frequency
    HashSet<int> hs = new HashSet<int>();
 
    // Insert frequencies into HashSet
    foreach(KeyValuePair<char, int> me in h)
        hs.Add(me.Value);
 
    // Count of unique frequencies
    return hs.Count;
}
 
// Driver code
public static void Main(String[] args)
{
    String str = "geeksforgeeks";
    int len = str.Length;
    Console.WriteLine(totalOperations(str, len));
}
}
 
// This code is contributed by Rajput-Ji

<script>
 
// Javascript implementation of the approach
 
// Function to return the count of operations required
function totalOperations(str, len)
{
    // HashMap to store characters and their frequencies
    var h = new Map();
    for (var i = 0; i < len; i++)
 
        // If already contains the character then
        // increment its frequency by 1
        if(h.has(str[i]))
            h.set(str[i], h.get(str[i])+1)
        else
            h.set(str[i], 1)
 
    // HashSet to store unique frequency
    var hs = new Set();
 
    // Insert frequencies into HashSet
    h.forEach((value, key) => {
         
        hs.add(value);
    });
 
    // Count of unique frequencies
    return hs.size;
}
 
// Driver Code
var str = "geeksforgeeks";
var len = str.length;
document.write( totalOperations(str, len));
 
// This code is contributed by itsok.
</script>

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