二进制字符串上的最小操作数，以使其在除以10 ^ B时得到10 ^ A的余数

给定长度为N的二进制字符串str以及两个整数A和B ，使得0≤A 。的任务是计数操作的最小数目的字符串，使得它给出10 ^A作为剩余部分在由10 ^B划分的。一个运算意味着将1更改为0或将0更改为1 。

例子：

Input: str = “1001011001”, A = 3, B = 6 
Output: 2 
The string after 2 operations is 1001001000. 
1001001000 % 10⁶ = 10³

Input: str = “11010100101”, A = 1, B = 5 
Output: 3  

方法：为了使数字除以10 ^B时得到10 ^A的余数，字符串的最后B位必须为0 ，但从^第(A + 1)位起的数字应为1 。因此，请检查字符串的最后B个数字是否满足上述条件，并为每个数字不匹配将计数增加1。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
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

// 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

Python3

# 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#

// 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

Javascript

 
// 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
 

输出：

2

时间复杂度： O(N)

辅助空间： O(N)

如果您希望与行业专家一起参加现场课程，请参阅《 Geeks现场课程》和《 Geeks现场课程美国》。