Minimum number of operations on a binary string such that it gives 10^A as remainder when divided by 10^B

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: 2
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++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output:

2


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