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Minimum number of operations on a binary string such that it gives 10^A as remainder when divided by 10^B
  • Last Updated : 02 Feb, 2021

Given a binary string str of length N and two integers A and B such that 0 ≤ A < B < n. The task is to count the minimum number of operations on the string such that it gives 10A as remainder when divided by 10B. An operation means changing 1 to 0 or 0 to 1.
Examples: 

Input: str = “1001011001”, A = 3, B = 6 
Output:
The string after 2 operations is 1001001000. 
1001001000 % 106 = 103
Input: str = “11010100101”, A = 1, B = 5 
Output: 3  

Approach: In order for the number to give 10A as remainder when divided by 10B, the last B digits of the string has to be 0 except the digit at (A + 1)th position from the last which should be 1. Therefore, check the last B digits of the string for the above condition and increase the count by 1 for each mismatch of digit.
Below is the implementation of the above approach:  

C++




// 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




// 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
Output: 
2

 

Time Complexity: O(N )

Auxiliary Space: O(N)

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