Maximum frequency of a remainder modulo 2i

Given an octal number N, the task is to convert the number to decimal and then find the modulo with every power of 2 i.e. 2ⁱ such that i > 0 and 2ⁱ < N, and print the maximum frequency of the modulo.
Examples:

Input: N = 13 
Output: 2 
Octal(13) = decimal(11) 
11 % 2 = 1 
11 % 4 = 3 
11 % 8 = 3 
3 occurs the most i.e. 2 times.

Input: N = 21 
Output: 4 

Approach: Find the binary representation of the number by replacing the digit with its binary representation. Now it is known that every digit in the binary representation represents a power of 2 in increasing order. So the modulo of the number with a power of 2 is the number formed by the binary representation of its preceding bits. For Example:

Octal(13) = decimal(11) = binary(1011) 
So, 
11(1011) % 2 (10) = 1 (1) 
11(1011) % 4 (100) = 3 (11) 
11(1011) % 8 (1000) = 3 (011) 

Here, it can be observed that when there is a zero in the binary representation of the modulo, the number remains the same. So the maximum frequency of the modulo will be 1 + the number of consecutive 0’s in the binary representation (not the leading zeroes) of the number. As the modulo starts from 2, remove the LSB of the number.

Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Binary representation of the digits
const string bin[] = { "000", "001", "010", "011",
                       "100", "101", "110", "111" };

// Function to return the maximum frequency
// of s modulo with a power of 2
int maxFreq(string s)
{

    // Store the binary representation
    string binary = "";

    // Convert the octal to binary
    for (int i = 0; i < s.length(); i++) {
        binary += bin[s[i] - '0'];
    }

    // Remove the LSB
    binary = binary.substr(0, binary.length() - 1);

    int count = 1, prev = -1, i, j = 0;

    for (i = binary.length() - 1; i >= 0; i--, j++)

        // If there is 1 in the binary representation
        if (binary[i] == '1') {

            // Find the number of zeroes in between
            // two 1's in the binary representation
            count = max(count, j - prev);
            prev = j;
        }

    return count;
}

// Driver code
int main()
{
    string octal = "13";

    cout << maxFreq(octal);

    return 0;
}

Java

// Java implementation of the approach
class GFG
{

// Binary representation of the digits
static String bin[] = { "000", "001", "010", "011",
                        "100", "101", "110", "111" };

// Function to return the maximum frequency
// of s modulo with a power of 2
static int maxFreq(String s)
{

    // Store the binary representation
    String binary = "";

    // Convert the octal to binary
    for (int i = 0; i < s.length(); i++)
    {
        binary += bin[s.charAt(i) - '0'];
    }

    // Remove the LSB
    binary = binary.substring(0,
             binary.length() - 1);

    int count = 1, prev = -1, i, j = 0;

    for (i = binary.length() - 1;
         i >= 0; i--, j++)

        // If there is 1 in the binary representation
        if (binary.charAt(i) == '1')
        {

            // Find the number of zeroes in between
            // two 1's in the binary representation
            count = Math.max(count, j - prev);
            prev = j;
        }
    return count;
}

// Driver code
public static void main(String []args)
{
    String octal = "13";

    System.out.println(maxFreq(octal));
}
}

// This code is contributed by 29AjayKumar

Python3

# Python3 implementation of the approach

# Binary representation of the digits
bin = [ "000", "001", "010", "011",
        "100", "101", "110", "111" ];

# Function to return the maximum frequency
# of s modulo with a power of 2
def maxFreq(s) :

    # Store the binary representation
    binary = "";

    # Convert the octal to binary
    for i in range(len(s)) :
        binary += bin[ord(s[i]) - ord('0')];

    # Remove the LSB
    binary = binary[0 : len(binary) - 1];

    count = 1; prev = -1;j = 0;

    for i in range(len(binary) - 1, -1, -1) :

        # If there is 1 in the binary representation
        if (binary[i] == '1') :

            # Find the number of zeroes in between
            # two 1's in the binary representation
            count = max(count, j - prev);
            prev = j;

        j += 1;

    return count;

# Driver code
if __name__ == "__main__" :

    octal = "13";

    print(maxFreq(octal));

# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;

class GFG
{

// Binary representation of the digits
static String []bin = { "000", "001", "010", "011",
                        "100", "101", "110", "111" };

// Function to return the maximum frequency
// of s modulo with a power of 2
static int maxFreq(String s)
{

    // Store the binary representation
    String binary = "";

    // Convert the octal to binary
    for (int K = 0; K < s.Length; K++)
    {
        binary += bin[s[K] - '0'];
    }

    // Remove the LSB
    binary = binary.Substring(0,
             binary.Length - 1);

    int count = 1, prev = -1, i, j = 0;

    for (i = binary.Length - 1;
         i >= 0; i--, j++)

        // If there is 1 in the binary representation
        if (binary[i] == '1')
        {

            // Find the number of zeroes in between
            // two 1's in the binary representation
            count = Math.Max(count, j - prev);
            prev = j;
        }
    return count;
}

// Driver code
public static void Main(String []args)
{
    String octal = "13";

    Console.WriteLine(maxFreq(octal));
}
}

// This code is contributed by 29AjayKumar

Javascript

<script>

// Javascript implementation of the approach

// Binary representation of the digits
bin = [ "000", "001", "010", "011",
            "100", "101", "110", "111" ];

// Function to return the maximum frequency
// of s modulo with a power of 2
function maxFreq( s )
{

    // Store the binary representation
    var binary = "";

    // Convert the octal to binary
    for (var i = 0; i < s.length; i++) {
        binary += bin[s.charAt(i) - '0'];
    }

    // Remove the LSB
    binary = binary.substr(0, binary.length - 1);

    var count = 1, prev = -1, i, j = 0;

    for (i = binary.length - 1; i >= 0; i--, j++)

        // If there is 1 in the binary representation
        if (binary.charAt(i) == '1') {

            // Find the number of zeroes in between
            // two 1's in the binary representation
            count = Math.max(count, j - prev);
            prev = j;
        }

    return count;
}

var octal = "13";
document.write( maxFreq(octal));

// This code is contributed by SoumikMondal

</script>

Output

2

Time Complexity: O(N), as we are using a loop to traverse N times. Where N is the length of the string.
Auxiliary Space: O(N), as we are using extra space for the string binary.