Maximum frequency of a remainder modulo 2i

• Last Updated : 30 Apr, 2021

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:
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:

Approach: Find the binary representation of the number by replacing the digit with their 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 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



Output:

2

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