Count of integers of the form (2^x * 3^y) in the range [L, R]

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Given a range [L, R] where 0 ? L ? R ? 10⁸. The task is to find the count of integers from the given range that can be represented as (2^x) * (3^y).
Examples:

Input: L = 1, R = 10
Output: 7
The numbers are 1, 2, 3, 4, 6, 8 and 9
Input: L = 100, R = 200
Output: 5
The numbers are 108, 128, 144, 162 and 192

Approach: Since the numbers, which are powers of two and three, quickly grow, you can use the following algorithm. For all the numbers of the form (2^x) * (3^y) in the range [1, 10⁸] store them in a vector. Later sort the vector. Then the required answer can be calculated using an upper bound. Pre-calculating these integers will be helpful when there are a number of queries of the form [L, R].
Below is the implementation of the above approach:

CPP

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
#define MAXI (int)(1e8)
 
// To store all valid integers
vector<int> v;
 
// Function to find all the integers of the
// form (2^x * 3^y) in the range [0, 1000000]
void precompute()
{
 
    // To store powers of 2 and 3
    // initialized to 2^0 and 3^0
    int x = 1, y = 1;
 
    // While current power of 2
    // is within the range
    while (x <= MAXI) {
 
        // While number is within the range
        while (x * y <= MAXI) {
 
            // Add the number to the vector
            v.push_back(x * y);
 
            // Next power of 3
            y *= 3;
        }
 
        // Next power of 2
        x *= 2;
 
        // Reset to 3^0
        y = 1;
    }
 
    // Sort the vector
    sort(v.begin(), v.end());
}
 
// Function to find the count of numbers
// in the range [l, r] which are
// of the form (2^x * 3^y)
void countNum(int l, int r)
{
    cout << upper_bound(v.begin(), v.end(), r)
                - upper_bound(v.begin(), v.end(), l - 1);
}
 
// Driver code
int main()
{
    int l = 100, r = 200;
 
    // Pre-compute all the valid numbers
    precompute();
 
    countNum(l, r);
 
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
 
class GFG {
  static int MAXI = 100000000;
 
  // To store all valid integers
  static ArrayList<Integer> v = new ArrayList<Integer>();
 
  static int upper_bound(ArrayList<Integer> arr, int elem)
  {
    for (var i = 0; i < arr.size(); i++)
      if (elem < arr.get(i))
        return i;
    return arr.size();
  }
  // Function to find all the integers of the
  // form (2^x * 3^y) in the range [0, 1000000]
  static void precompute()
  {
 
    // To store powers of 2 and 3
    // initialized to 2^0 and 3^0
    int x = 1, y = 1;
 
    // While current power of 2
    // is within the range
    while (x <= MAXI) {
 
      // While number is within the range
      while (x * y <= MAXI) {
 
        // Add the number to the vector
        v.add(x * y);
 
        // Next power of 3
        y *= 3;
      }
 
      // Next power of 2
      x *= 2;
 
      // Reset to 3^0
      y = 1;
    }
 
    // Sort the vector
    Collections.sort(v);
  }
 
  // Function to find the count of numbers
  // in the range [l, r] which are
  // of the form (2^x * 3^y)
  static void countNum(int l, int r)
  {
    System.out.println(upper_bound(v, r)
                       - upper_bound(v, l - 1));
  }
 
  // Driver code
  public static void main(String[] args)
  {
    int l = 100, r = 200;
 
    // Pre-compute all the valid numbers
    precompute();
 
    countNum(l, r);
  }
}
 
// This code is contributed by phasing17

Python3

# Python3 implementation of the approach
import bisect
MAXI=int(1e8)
 
# To store all valid integers
v=[]
 
# Function to find all the integers of the
# form (2^x * 3^y) in the range [0, 1000000]
def precompute():
 
    # To store powers of 2 and 3
    # initialized to 2^0 and 3^0
    x = 1; y = 1
 
    # While current power of 2
    # is within the range
    while (x <= MAXI) :
 
        # While number is within the range
        while (x * y <= MAXI) :
 
            # Add the number to the vector
            v.append(x * y)
 
            # Next power of 3
            y *= 3
         
 
        # Next power of 2
        x *= 2
 
        # Reset to 3^0
        y = 1
     
 
    # Sort the vector
    v.sort()
 
 
# Function to find the count of numbers
# in the range [l, r] which are
# of the form (2^x * 3^y)
def countNum(l, r):
    print(bisect.bisect_right(v, r) - bisect.bisect_right(v,l - 1))
 
 
# Driver code
if __name__ == '__main__':
    l = 100; r = 200
 
    # Pre-compute all the valid numbers
    precompute()
 
    countNum(l, r)

C#

// C# implementation of the approach
using System;
using System.Collections.Generic;
 
class GFG
{
    static int MAXI = 100000000;
     
    // To store all valid integers
    static List<int> v = new List<int>();
     
    static int upper_bound(List<int> arr, int elem)
    {
        for (var i = 0; i < arr.Count; i++)
            if (elem < arr[i])
                return i;
        return arr.Count;
    }
    // Function to find all the integers of the
    // form (2^x * 3^y) in the range [0, 1000000]
    static void precompute()
    {
     
        // To store powers of 2 and 3
        // initialized to 2^0 and 3^0
        int x = 1, y = 1;
     
        // While current power of 2
        // is within the range
        while (x <= MAXI) {
     
            // While number is within the range
            while (x * y <= MAXI) {
     
                // Add the number to the vector
                v.Add(x * y);
     
                // Next power of 3
                y *= 3;
            }
     
            // Next power of 2
            x *= 2;
     
            // Reset to 3^0
            y = 1;
        }
     
        // Sort the vector
        v.Sort();
    }
     
    // Function to find the count of numbers
    // in the range [l, r] which are
    // of the form (2^x * 3^y)
    static void countNum(int l, int r)
    {
        Console.WriteLine(upper_bound(v, r)
                    - upper_bound(v, l - 1));
    }
     
    // Driver code
    public static void Main(string[] args)
    {
        int l = 100, r = 200;
     
        // Pre-compute all the valid numbers
        precompute();
     
        countNum(l, r);
     
    }
}
 
// This code is contributed by phasing17

Javascript

// JavaScript implementation of the approach
let MAXI = 100000000
 
// To store all valid integers
let v = [];
 
function upper_bound(arr, elem)
{
    for (var i = 0; i < arr.length; i++)
        if (elem < arr[i])
            return i
    return arr.length;
}
 
// Function to find all the integers of the
// form (2^x * 3^y) in the range [0, 1000000]
function precompute()
{
 
    // To store powers of 2 and 3
    // initialized to 2^0 and 3^0
    let x = 1, y = 1;
 
    // While current power of 2
    // is within the range
    while (x <= MAXI) {
 
        // While number is within the range
        while (x * y <= MAXI) {
 
            // Add the number to the vector
            v.push(parseInt(x * y));
 
            // Next power of 3
            y *= 3;
        }
 
        // Next power of 2
        x *= 2;
 
        // Reset to 3^0
        y = 1;
    }
 
    // Sort the vector
    v.sort(function(a, b) {
  return a - b;
});
}
 
// Function to find the count of numbers
// in the range [l, r] which are
// of the form (2^x * 3^y)
function countNum(l, r)
{
    console.log(upper_bound(v, r) - upper_bound(v, l - 1));
}
 
// Driver code
let l = 100, r = 200;
 
// Pre-compute all the valid numbers
precompute();
 
countNum(l, r);
 
// This code is contributed by phasing17

Output:

Time Complexity: O(N * log(N)), where N = logX + logY
Auxiliary Space: O(N), as N extra space has been taken.

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