Related Articles

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

• Difficulty Level : Easy
• Last Updated : 09 May, 2019

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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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:

 `// C++ implementation of the approach``#include ``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;``}`
Output:
```5
```

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

My Personal Notes arrow_drop_up