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

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++

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// C++ implementation of the approach 
#include <bits/stdc++.h> 
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

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Java

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// 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));
    }
}

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Python3

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# 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] *
    pwrTwo[0] = 1 %
    for i in range(1, n): 
        pwrTwo[i] = pwrTwo[i - 1] * (2 % k) 
        pwrTwo[i] %=
  
    # 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 %=
              
        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

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C#

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// 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

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Output:

13


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