String lexicográficamente más grande posible después de la eliminación de K caracteres

Dada una string S que consta solo de letras minúsculas, la tarea es encontrar la string lexicográficamente más grande que se puede obtener eliminando K caracteres de la string dada.

Ejemplos: 

Entrada: s = “zyxedcba”, K=1 
Salida: zyxedcb 
Explicación: El carácter con el valor ASCII más pequeño de la string dada es ‘a’. 
La eliminación de ‘a’ genera la string lexicográficamente más grande posible.

Entrada: s = “abcde”, K=2 
Salida: cde

Enfoque:
La idea es usar Stack Data Structure para resolver el problema. Siga los pasos a continuación para resolver el problema:  

• Atraviesa la cuerda.
• Para cada carácter, compruebe si es mayor que el carácter en la parte superior de la pila . Si se determina que es cierto, extrae el elemento superior de la pila si K > 0 .
• Inserta el personaje en la pila.
• Después de completar el recorrido de la string, si K > 0 , elimine los elementos K superiores de la pila.
• Finalmente, almacene los caracteres en la pila como la respuesta. Imprime la respuesta .

A continuación se muestra la implementación del enfoque anterior:

// C++ Program to implement the
// above approach
#include <bits/stdc++.h>
using namespace std;
 
string largestString(string num, int k)
{
    // final result string
    string ans = "";
 
    for (auto i : num) {
 
        // If the current char exceeds the
        // character at the top of the stack
        while (ans.length() && ans.back() < i
               && k > 0) {
 
            // Remove from the end of the string
            ans.pop_back();
 
            // Decrease k for the removal
            k--;
        }
 
        // Insert current character
        ans.push_back(i);
    }
 
    // Perform remaining K deletions
    // from the end of the string
    while (ans.length() and k--) {
        ans.pop_back();
    }
 
    // Return the string
    return ans;
}
 
// Driver Code
int main()
{
    string str = "zyxedcba";
    int k = 1;
 
    cout << largestString(str, k) << endl;
}

// Java program to implement the
// above approach
class GFG{
 
static String largestString(String num, int k)
{
     
    // Final result String
    String ans = "";
 
    for(char i : num.toCharArray())
    {
         
        // If the current char exceeds the
        // character at the top of the stack
        while (ans.length() > 0 &&
               ans.charAt(ans.length() - 1) < i &&
                                          k > 0)
        {
             
            // Remove from the end of the String
            ans = ans.substring(0, ans.length() - 1);
 
            // Decrease k for the removal
            k--;
        }
 
        // Insert current character
        ans += i;
    }
 
    // Perform remaining K deletions
    // from the end of the String
    while (ans.length() > 0 && k-- > 0)
    {
        ans = ans.substring(0, ans.length() - 1);
    }
     
    // Return the String
    return ans;
}
 
// Driver Code
public static void main(String[] args)
{
    String str = "zyxedcba";
    int k = 1;
 
    System.out.print(largestString(str, k) + "\n");
}
}
 
// This code is contributed by 29AjayKumar

# Python3 program to implement the
# above approach
def largestString(num, k):
     
    # Final result string
    ans = []
     
    for i in range(len(num)):
         
        # If the current char exceeds the
        # character at the top of the stack
        while(len(ans) and ans[-1] < num[i] and
                                 k > 0):
             
            # Remove from the end of the string
            ans.pop()
             
            # Decrease k for the removal
            k -= 1
         
        # Insert current character
        ans.append(num[i])
     
    # Perform remaining K deletions
    # from the end of the string
    while(len(ans) and k):
        k -= 1
        ans.pop()
     
    # Return the string
    return ans
     
# Driver code
str = "zyxedcba"
k = 1
 
print(*largestString(str, k), sep = "")
 
# This code is contributed by divyeshrabadiya07

// C# program to implement the
// above approach
using System;
 
class GFG{
 
static String largestString(String num, int k)
{
     
    // Final result String
    String ans = "";
 
    foreach(char i in num.ToCharArray())
    {
         
        // If the current char exceeds the
        // character at the top of the stack
        while (ans.Length > 0 &&
           ans[ans.Length - 1] < i && k > 0)
        {
             
            // Remove from the end of the String
            ans = ans.Substring(0, ans.Length - 1);
 
            // Decrease k for the removal
            k--;
        }
 
        // Insert current character
        ans += i;
    }
 
    // Perform remaining K deletions
    // from the end of the String
    while (ans.Length > 0 && k-- > 0)
    {
        ans = ans.Substring(0, ans.Length - 1);
    }
     
    // Return the String
    return ans;
}
 
// Driver Code
public static void Main(String[] args)
{
    String str = "zyxedcba";
    int k = 1;
 
    Console.Write(largestString(str, k) + "\n");
}
}
 
// This code is contributed by 29AjayKumar

<script>
    // JavaScript program to implement the
    // above approach
    function largestString(num, k) {
        // Final result String
        var ans = "";
 
        var str = num.split("");
 
        for (const i of str) {
          // If the current char exceeds the
          // character at the top of the stack
          while (ans.length > 0 && ans[ans.length - 1] < i && k > 0) {
            // Remove from the end of the String
            ans = ans.substring(0, ans.length - 1);
 
            // Decrease k for the removal
            k--;
          }
 
          // Insert current character
          ans += i;
        }
 
        // Perform remaining K deletions
        // from the end of the String
        while (ans.length > 0 && k-- > 0) {
          ans = ans.substring(0, ans.length - 1);
        }
 
        // Return the String
        return ans;
    }
 
     // Driver Code
     var str = "zyxedcba";
     var k = 1;
 
     document.write(largestString(str, k) + "<br>");
</script>

zyxedcb

Complejidad temporal: O(N) 
Espacio auxiliar: O(N)