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 (2x) * (3y).
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 (2x) * (3y) 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 integersvector<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 codeint main(){ int l = 100, r = 200; // Pre-compute all the valid numbers precompute(); countNum(l, r); return 0;} |
Java
// Java implementation of the approachimport 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 approachimport bisectMAXI=int(1e8)# To store all valid integersv=[]# 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 codeif __name__ == '__main__': l = 100; r = 200 # Pre-compute all the valid numbers precompute() countNum(l, r) |
C#
// C# implementation of the approachusing 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 approachlet MAXI = 100000000// To store all valid integerslet 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 codelet l = 100, r = 200;// Pre-compute all the valid numbersprecompute();countNum(l, r);// This code is contributed by phasing17 |
Output:
5
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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