Número mínimo de operaciones en una string binaria tal que da 10^A como resto cuando se divide por 10^B

Dada una string binaria str de longitud N y dos enteros A y B tales que 0 ≤ A < B < n . La tarea es contar el número mínimo de operaciones en la string tal que dé 10 ^A como resto cuando se divide por 10 ^B . Una operación significa cambiar 1 a 0 o 0 a 1 .

Ejemplos: 

Entrada: str = “1001011001”, A = 3, B = 6 
Salida: 2 
La string después de 2 operaciones es 1001001000. 
1001001000 % 10 ⁶ = 10 ³

Entrada: str = “11010100101”, A = 1, B = 5 
Salida: 3  

Enfoque: para que el número dé 10 ^A como resto cuando se divide por 10 ^B , los últimos dígitos B de la string deben ser 0 , excepto el dígito en (A + 1) ^enésima posición desde el último, que debe ser 1 . Por lo tanto, verifique los últimos dígitos B de la string para la condición anterior y aumente el conteo en 1 por cada discrepancia de dígitos.

A continuación se muestra la implementación del enfoque anterior:  

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the minimum number
// of operations on a binary string such that
// it gives 10^A as remainder when divided by 10^B
int findCount(string s, int n, int a, int b)
{
    // Initialize result
    int res = 0;
 
    // Loop through last b digits
    for (int i = 0; i < b; i++) {
        if (i == a)
            res += (s[n - i - 1] != '1');
        else
            res += (s[n - i - 1] != '0');
    }
 
    return res;
}
 
// Driver code
int main()
{
    string str = "1001011001";
    int N = str.size();
    int A = 3, B = 6;
 
    cout << findCount(str, N, A, B);
 
    return 0;
}

// Java implementation of the approach
import java.util.*;
 
class GFG
{
     
    // Function to return the minimum number
    // of operations on a binary string such that
    // it gives 10^A as remainder when divided by 10^B
    static int findCount(String s, int n, int a, int b)
    {
        // Initialize result
        int res = 0;
        char []s1 = s.toCharArray();
         
        // Loop through last b digits
        for (int i = 0; i < b; i++)
        {
             
            if (i == a)
            {
                if (s1[n - i - 1] != '1')
                    res += 1;
            }
            else
            {
                if (s1[n - i - 1] != '0')
                        res += 1 ;
            }
                 
        }
     
        return res;
    }
     
    // Driver code
    static public void main (String []args)
    {
         
        String str = "1001011001";
        int N = str.length() ;
        int A = 3, B = 6;
     
        System.out.println(findCount(str, N, A, B));
     
    }
}
 
// This code is contributed by ChitraNayal

# Python 3 implementation of the approach
 
# Function to return the minimum number
# of operations on a binary string such that
# it gives 10^A as remainder when divided by 10^B
def findCount(s, n, a, b):
    # Initialize result
    res = 0
 
    # Loop through last b digits
    for i in range(b):
        if (i == a):
            res += (s[n - i - 1] != '1')
        else:
            res += (s[n - i - 1] != '0')
 
    return res
 
# Driver code
if __name__ == '__main__':
    str = "1001011001"
    N = len(str)
    A = 3
    B = 6
 
    print(findCount(str, N, A, B))
 
# This code is contributed by
# Surendra_Gangwar

// C# implementation of the approach
using System;
 
class GFG
{
     
    // Function to return the minimum number
    // of operations on a binary string such that
    // it gives 10^A as remainder when divided by 10^B
    static int findCount(string s, int n, int a, int b)
    {
        // Initialize result
        int res = 0;
     
        // Loop through last b digits
        for (int i = 0; i < b; i++)
        {
             
            if (i == a)
            {
                if (s[n - i - 1] != '1')
                    res += 1;
            }
            else
            {
                if (s[n - i - 1] != '0')
                        res += 1 ;
            }
                 
        }
     
        return res;
    }
     
    // Driver code
    static public void Main ()
    {
         
        string str = "1001011001";
        int N = str.Length ;
        int A = 3, B = 6;
     
        Console.WriteLine(findCount(str, N, A, B));
     
    }
}
 
// This code is contributed by AnkitRai01

<script>
 
// Javascript implementation of the approach
 
// Function to return the minimum number
// of operations on a binary string such that
// it gives 10^A as remainder when divided by 10^B
function findCount(s, n, a, b)
{
     
    // Initialize result
    var res = 0;
 
    // Loop through last b digits
    for(var i = 0; i < b; i++)
    {
        if (i == a)
            res += (s[n - i - 1] != '1');
        else
            res += (s[n - i - 1] != '0');
    }
    return res;
}
 
// Driver code
var str = "1001011001";
var N = str.length;
var A = 3, B = 6;
 
document.write(findCount(str, N, A, B));
 
// This code is contributed by itsok
 
</script>

2

Complejidad de tiempo: O(N)

Espacio Auxiliar: O(1)