Minimice el número cambiando como máximo K dígitos

Dado un número N , la tarea es minimizar el número cambiando como máximo K dígitos. Tenga en cuenta que el número no debe contener ceros a la izquierda.
Ejemplos: 
 

Entrada: N = 91945, K = 3 
Salida: 10045
Entrada: N = 1, K = 0 
Salida: 1 
 

Acercarse: 
 

• Reemplace el primer dígito con 1 si aún no es 1 y actualice K en consecuencia.
• Ahora, para el resto de los dígitos, reemplace los siguientes K – 1 dígitos distintos de cero con un 0 .

A continuación se muestra la implementación del enfoque anterior: 
 

// C++ implementation of the approach
#include <iostream>
using namespace std;
 
// Function to return the minimized number
string minNum(string num, int k)
{
 
    // Total digits in the number
    int len = num.length();
 
    // If the string is empty or there
    // are no operations to perform
    if (len == 0 || k == 0)
        return num;
 
    // "0" is a valid number
    if (len == 1)
        return "0";
 
    // If the first digit is not already 1 then
    // update it to 1 and decrement k
    if (num[0] != '1') {
        num[0] = '1';
        k--;
    }
 
    int i = 1;
    // While there are operations left
    // and the number can still be updated
    while (k > 0 && i < len) {
 
        // If the current digit is not already 0
        // then update it to 0 and decrement k
        if (num[i] != '0') {
            num[i] = '0';
            k--;
        }
 
        i++;
    }
 
    // Return the minimized number
    return num;
}
 
// Driver code
int main()
{
    string num = "91945";
    int k = 3;
 
    cout << minNum(num, k);
 
    return 0;
}

// Java implementation of the approach
class GFG
{
     
    // Function to return the minimized number
    static String minNum(char num[], int k)
    {
     
        // Total digits in the number
        int len = num.length;
     
        // If the string is empty or there
        // are no operations to perform
        if (len == 0 || k == 0)
        {
            String num_str = new String(num);
            return num_str;
        }
     
        // "0" is a valid number
        if (len == 1)
            return "0";
     
        // If the first digit is not already 1 then
        // update it to 1 and decrement k
        if (num[0] != '1')
        {
            num[0] = '1';
            k--;
        }
     
        int i = 1;
         
        // While there are operations left
        // and the number can still be updated
        while (k > 0 && i < len)
        {
     
            // If the current digit is not already 0
            // then update it to 0 and decrement k
            if (num[i] != '0')
            {
                num[i] = '0';
                k--;
            }
            i++;
        }
         
        String num_str = new String(num);
         
        // Return the minimised number
        return num_str;
    }
     
    // Driver code
    public static void main(String args[])
    {
        String num = "91945";
        int k = 3;
     
        System.out.println(minNum(num.toCharArray(), k));
    }
}
 
// This code is contributed by AnkitRai01

# Python 3 implementation of the approach
 
# Function to return the minimized number
def minNum(num, k) :
 
    # Total digits in the number
    len_ = len(num)
 
    # If the string is empty or there
    # are no operations to perform
    if len_ == 0 or k == 0 :
        return num
 
    # "0" is a valid number
    if len_ == 1:
        return "0"
 
    # If the first digit is not already 1 then
    # update it to 1 and decrement k
    if num[0] != '1' :
        num = '1' + num[1:]
        k -= 1
 
    i = 1
     
    # While there are operations left
    # and the number can still be updated
    while k > 0 and i < len_ :
 
        # If the current digit is not already 0
        # then update it to 0 and decrement k
        if num[i] != '0' :
            num = num[:i] + '0' + num[i + 1:]
            k -= 1
 
        i += 1
  
    # Return the minimised number
    return num
 
# Driver code
num = "91945"
k = 3
 
print(minNum(num, k))
 
# This code is contributed by divyamohan123

// C# implementation of the approach
using System;
     
class GFG
{
     
    // Function to return the minimized number
    static String minNum(char []num, int k)
    {
     
        // Total digits in the number
        int len = num.Length;
     
        // If the string is empty or there
        // are no operations to perform
        if (len == 0 || k == 0)
        {
            return String.Join("", num);
        }
     
        // "0" is a valid number
        if (len == 1)
            return "0";
     
        // If the first digit is not already 1 then
        // update it to 1 and decrement k
        if (num[0] != '1')
        {
            num[0] = '1';
            k--;
        }
     
        int i = 1;
         
        // While there are operations left
        // and the number can still be updated
        while (k > 0 && i < len)
        {
     
            // If the current digit is not already 0
            // then update it to 0 and decrement k
            if (num[i] != '0')
            {
                num[i] = '0';
                k--;
            }
            i++;
        }
         
        // Return the minimised number
        return String.Join("", num);
    }
     
    // Driver code
    public static void Main(String []args)
    {
        String num = "91945";
        int k = 3;
     
        Console.WriteLine(minNum(num.ToCharArray(), k));
    }
}
 
// This code is contributed by 29AjayKumar

<script>
// Javascript implementation of the approach
 
// Function to return the minimized number
function minNum(num, k)
{
 
    // Total digits in the number
    let len = num.length;
 
    // If the string is empty or there
    // are no operations to perform
    if (len == 0 || k == 0) {
        let num_str = num.join("");
        return num_str;
    }
 
    // "0" is a valid number
    if (len == 1)
        return "0";
 
    // If the first digit is not already 1 then
    // update it to 1 and decrement k
    if (num[0] != '1') {
        num[0] = '1';
        k--;
    }
 
    let i = 1;
 
    // While there are operations left
    // and the number can still be updated
    while (k > 0 && i < len) {
 
        // If the current digit is not already 0
        // then update it to 0 and decrement k
        if (num[i] != '0') {
            num[i] = '0';
            k--;
        }
        i++;
    }
 
    let num_str = num.join("");
 
    // Return the minimised number
    return num_str;
}
 
// Driver code
let num = "91945";
let k = 3;
 
document.write(minNum(num.split(""), k));
 
// This code is contributed by _saurabh_jaiswal.
</script>

10045

Complejidad de tiempo: O(min(k, |num|))

Espacio Auxiliar: O(1)