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

• Last Updated : 22 Apr, 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 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^Bint 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 codeint 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 approachimport 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^Bdef 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 codeif __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 approachusing 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


Output:
2

Time Complexity: O(N )

Auxiliary Space: O(N)

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