Solving Binary String Modulo Problem

Given a string "s" and an integer "m" your objective is to calculate the remainder "r" when the decimal value of binary string "s" is divided by "m".

Examples:

Input: s = "101", m = 2
Output: 1
Explanation: If we have a string "(101)" its decimal equivalent is "(5)". Therefore if we compute 5 mod 2 the result will be 1.

Input: s = "1000", m = 4
Output: 0
Explanation: If we have a string "(1000)" and m = 4 then r can be calculated as k mod m, which, in this case's 8 mod 4. The final result will be 0.

Approach: To solve the problem

The idea is to calculate the value of each corresponding set bit and use it right away without storing it in an array, as we know the value of every higher set bit is 2x of the previous set bit so we can simply use a variable power and at every bit we multiply power with 2 to get the value of the current i^th bit (value of 2ⁱ).

Steps for Implementing the above Approach:

• Initialize the answer with 0, and power with 1.
• Now run a loop over the binary string from the back side of the string(i.e. from the least significant bit of binary string).
  • Check if the current bit is set or not then add the power(2ⁱ => which is stored in the power variable) into the answer and take its modulo with m.
  • Now multiply power with 2 to get the next set bit value and also take it modulo with m so that it can't overflow the integer limit.
• Return the answer.

Below is the implementation of the above idea:

#include <iostream>
using namespace std;

int modulo(string s, int m)
{
    int ans = 0;
    int power = 1;
    for (int i = s.size() - 1; i >= 0; i--) {
        if (s[i] == '1') {
            ans += power;
            ans %= m;
        }
        power *= 2;
        power %= m;
    }
    return ans;
}

int main()
{
    string s = "101";
    int m = 2;
    int result = modulo(s, m);
    cout << result << endl;
    return 0;
}

public class Main {

    // Function to calculate modulo value
    public static int modulo(String s, int m) {
        int ans = 0;
        int power = 1;

        // Loop through the string in reverse order
        for (int i = s.length() - 1; i >= 0; i--) {
            // Check if the current character is '1'
            if (s.charAt(i) == '1') {
                ans += power;
                ans %= m;
            }
            power *= 2;
            power %= m;
        }
        return ans;
    }

    // Driver code
    public static void main(String[] args) {
        String s = "101";
        int m = 2;
        int result = modulo(s, m);
        System.out.println(result);
    }
}

// This code is contributed by shivamgupta310570

def modulo(s, m):
    ans = 0
    power = 1
    # Loop through the string in reverse order
    for i in range(len(s) - 1, -1, -1):
        # Check if the current digit is '1'
        if s[i] == '1':
            ans += power
            ans %= m
        power *= 2
        power %= m
    return ans
# Driver code
def main():
    s = "101"
    m = 2
    result = modulo(s, m)
    print(result)

 
if __name__ == "__main__":
    main()

using System;

public class GFG
{
    // Function to calculate modulo value
    public static int Modulo(string s, int m)
    {
        int ans = 0;
        int power = 1;

        // Loop through the string in reverse order
        for (int i = s.Length - 1; i >= 0; i--)
        {
            // Check if the current character is '1'
            if (s[i] == '1')
            {
                ans += power;
                ans %= m;
            }
            power *= 2;
            power %= m;
        }
        return ans;
    }

    // Driver code
    public static void Main(string[] args)
    {
        string s = "101";
        int m = 2;
        int result = Modulo(s, m);
        Console.WriteLine(result);
    }
}

// JavaScript Implementation

function modulo(s, m) {
  let ans = 0;
  let power = 1;
  for (let i = s.length - 1; i >= 0; i--) {
    if (s[i] === '1') {
      ans += power;
      ans %= m;
    }
    power *= 2;
    power %= m;
  }
  return ans;
}

let s = "101";
let m = 2;
let result = modulo(s, m);
console.log(result);

// This code is contributed by Sakshi

1

Time Complexity:- O(N), As we are only using a single loop over the size of binary string.
Auxiliary Space:- O(1), As we are not using any extra space.

Anonymous

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• Basic Coding Problems
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• Algorithms-Bit Algorithms
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