Find the number with only set bits between Lth and Rth index using C++

In the given problem, we need to find the value of a number that has all the set bits between the given range L, R. For example −

Input: L = 1, R = 5
Output: 62
Explanation: representation of given L and R in binary form is 0..0111110

Input: L = 1, R = 4
Output: 30
Explanation: representation of given L and R in binary form is 0..11110

Methods to find the solution

In the given problem, we will discuss two methods, brute force method and efficient method.

Brute force method

In this method we just iterate over the given range and add up all the powers of 2 in the given range and this will be our answer.

Example

#include<bits/stdc++.h>
using namespace std;
int main() {
   int L = 1, R = 3; // the given range
   int ans = 0; // our answer
   for(int i = L; i <= R; i++) // traversing through the whole range
      ans += pow(2, i); // adding values to the answer.
   cout << ans << "\n";
}

Output

In this approach, we just iterate over the range and simply add the powers of 2 of the numbers in the range. The time complexity of this program is O(N), where N is the size of our range. But we can further improve the time complexity by applying bit knowledge in the given problem.

Efficient Method

In this method we will simply construct a formula to calculate our answer.

Example

#include<bits/stdc++.h>
using namespace std;
int main() {
   int L = 1, R = 3; // the given range
   int ans = 0; // our answer
   for(int i = L; i <= R; i++) // traversing through the whole range
      ans += pow(2, i); // adding values to the answer.
   cout << ans << "\n";
}

Output

In this method, we formulate a formula to calculate the answer.

Explanation of the above code

As you know we need to count the numbers with set bits in the given range, so in this method we find a Bits are set to the number. We then need to subtract a number with all bits set from 1 to (L-1), so we formulate this observation. The overall time complexity of the given code is O(1), which is constant time complexity, which means we can compute any answer in constant time.

Conclusion

This article will write a program for "only numbers with set bits between L-th and R-th indexes". We also learned a C program and complete methods (common and efficient) to solve this problem. We can write the same program in other languages ​​like C, Java, Python and others. Hope you find this article helpful.

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