# Modulo of a large Binary String

Given a large binary string str and an integer K, the task is to find the value of str % K.

Examples:

Input: str = “1101”, K = 45
Output: 13
decimal(1101) % 45 = 13 % 45 = 13

Input: str = “11010101”, K = 112
Output: 101
decimal(11010101) % 112 = 213 % 112 = 101

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: It is known that (str % K) where str is a binary string can be written as ((str[n – 1] * 20) + (str[n – 2] * 21) + … + (str[0] * 2n – 1)) % K which in turn can be written as (((str[n – 1] * 20) % K) + ((str[n – 2] * 21) % K) + … + ((str[0] * 2n – 1)) % K) % K. This can be used to find the required answer without actually converting the given binary string to its decimal equivalent.

Below is the implementation of the above approach:

## C++

 // C++ implementation of the approach  #include   using namespace std;     // Function to return the value of (str % k)  int getMod(string str, int n, int k)  {         // pwrTwo[i] will store ((2^i) % k)      int pwrTwo[n];      pwrTwo[0] = 1 % k;      for (int i = 1; i < n; i++)      {          pwrTwo[i] = pwrTwo[i - 1] * (2 % k);          pwrTwo[i] %= k;      }         // To store the result      int res = 0;      int i = 0, j = n - 1;      while (i < n)      {             // If current bit is 1          if (str[j] == '1')          {                 // Add the current power of 2              res += (pwrTwo[i]);              res %= k;          }          i++;          j--;      }      return res;  }     // Driver code  int main()  {      string str = "1101";      int n = str.length();      int k = 45;         cout << getMod(str, n, k) << endl;  }     // This code is contributed by ashutosh450

## Java

 // Java implementation of the approach import java.util.*;    class GFG {        // Function to return the value of (str % k)     static int getMod(String str, int n, int k)     {            // pwrTwo[i] will store ((2^i) % k)         int pwrTwo[] = new int[n];         pwrTwo[0] = 1 % k;         for (int i = 1; i < n; i++) {             pwrTwo[i] = pwrTwo[i - 1] * (2 % k);             pwrTwo[i] %= k;         }            // To store the result         int res = 0;         int i = 0, j = n - 1;         while (i < n) {                // If current bit is 1             if (str.charAt(j) == '1') {                    // Add the current power of 2                 res += (pwrTwo[i]);                 res %= k;             }             i++;             j--;         }            return res;     }        // Driver code     public static void main(String[] args)     {         String str = "1101";         int n = str.length();         int k = 45;            System.out.print(getMod(str, n, k));     } }

## Python3

 # Python3 implementation of the approach     # Function to return the value of (str % k)  def getMod(_str, n, k) :         # pwrTwo[i] will store ((2^i) % k)      pwrTwo = [0] * n      pwrTwo[0] = 1 % k      for i in range(1, n):          pwrTwo[i] = pwrTwo[i - 1] * (2 % k)          pwrTwo[i] %= k         # To store the result      res = 0     i = 0     j = n - 1     while (i < n) :             # If current bit is 1          if (_str[j] == '1') :                 # Add the current power of 2              res += (pwrTwo[i])              res %= k                         i += 1         j -= 1        return res     # Driver code  _str = "1101" n = len(_str)  k = 45    print(getMod(_str, n, k))     # This code is contributed by # divyamohan123

## C#

 // C# implementation of the above approach  using System;    class GFG {            // Function to return the value of (str % k)      static int getMod(string str, int n, int k)      {          int i;                    // pwrTwo[i] will store ((2^i) % k)          int []pwrTwo = new int[n];                     pwrTwo[0] = 1 % k;                     for (i = 1; i < n; i++)          {              pwrTwo[i] = pwrTwo[i - 1] * (2 % k);              pwrTwo[i] %= k;          }             // To store the result          int res = 0;          i = 0;         int j = n - 1;          while (i < n)         {                 // If current bit is 1              if (str[j] == '1')              {                     // Add the current power of 2                  res += (pwrTwo[i]);                  res %= k;              }              i++;              j--;          }          return res;      }         // Driver code      public static void Main()      {          string str = "1101";          int n = str.Length;          int k = 45;             Console.Write(getMod(str, n, k));      }  }    // This code is contributed by AnkitRai01

Output:

13

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