Open In App

Check whether all the bits are set in the given range

Improve
Improve
Like Article
Like
Save
Share
Report

Given a non-negative number n and two values l and r. The problem is to check whether all the bits are set or not in the range l to r in the binary representation of n
Constraint: 1 <= l <= r <= number of bits in the binary representation of n.
Examples: 
 

Input : n = 22, l = 2, r = 3
Output : Yes
(22)10 = (10110)2
The bits in the range 2 to 3 are all set.

Input : n = 47, l = 2, r = 5 
Output : No
(47)10 = (101111)2
The bits in the range 2 to 5 are all not set.

 

Approach: Following are the steps:
 

  1. Calculate num = ((1 << r) – 1) ^ ((1 << (l-1)) – 1). This will produce a number num having r number of bits and bits in the range l to r are the only set bits.
  2. Calculate new_num = n & num.
  3. If num == new_num, return “Yes” (all bits are set in the given range).
  4. Else return “No” (all bits are not set in the given range).

 

C++




// C++ implementation to check whether all the bits
// are set in the given range or not
#include <bits/stdc++.h>
 
using namespace std;
 
// function to check whether all the bits
// are set in the given range or not
string allBitsSetInTheGivenRange(unsigned int n,
                                 unsigned int l, unsigned int r)
{
    // calculating a number 'num' having 'r'
    // number of bits and bits in the range l
    // to r are the only set bits
    int num = ((1 << r) - 1) ^ ((1 << (l - 1)) - 1);
     
    // new number which will only have one or more
    // set bits in the range l to r and nowhere else
    int new_num = n & num;
     
    // if both are equal, then all bits are set
    // in the given range
    if (num == new_num)
        return "Yes";
         
    // else all bits are not set   
    return "No";   
}
 
// Driver program to test above
int main()
{
    unsigned int n = 22;
    unsigned int l = 2, r = 3;
    cout << allBitsSetInTheGivenRange(n, l, r);
    return 0;
}


Java




// Java implementation to check whether all
// the bits are set in the given range or not
class GFG {
         
    // function to check whether all the bits
    // are set in the given range or not
    static String allBitsSetInTheGivenRange(int n,
                                    int l,int r)
    {
         
        // calculating a number 'num' having 'r'
        // number of bits and bits in the range
        // l to r are the only set bits
        int num = ((1 << r) - 1) ^ ((1 <<
                                  (l - 1)) - 1);
         
        // new number which will only have one
        // or more set bits in the range l to r
        // and nowhere else
        int new_num = n & num;
         
        // if both are equal, then all bits are
        // set in the given range
        if (num == new_num)
            return "Yes";
             
        // else all bits are not set
        return "No";
    }
     
    //Driver code
    public static void main (String[] args)
    {
        int n = 22;
        int l = 2, r = 3;
         
        System.out.print(allBitsSetInTheGivenRange(
                                           n, l, r));
    }
}
 
// This code is contributed by Anant Agarwal.


Python3




# Python3 implementation to check
# whether all the bits are set in
# the given range or not
 
# Function to check whether all the bits
# are set in the given range or not
def allBitsSetInTheGivenRange(n, l, r):
 
    # calculating a number 'num' having 'r'
    # number of bits and bits in the range l
    # to r are the only set bits
    num = ((1 << r) - 1) ^ ((1 << (l - 1)) - 1)
     
    # new number which will only have
    # one or more set bits in the range
    # l to r and nowhere else
    new_num = n & num
     
    # if both are equal, then all bits
    # are set in the given range
    if (num == new_num):
        return "Yes"
         
    # else all bits are not set
    return "No"
 
# Driver code
n, l, r = 22, 2, 3
print(allBitsSetInTheGivenRange(n, l, r))
 
# This code is contributed by Anant Agarwal.


C#





PHP





Javascript





Output: 
 

Yes

Time Complexity – O(1)

Space Complexity – O(1)

 



Last Updated : 22 Jun, 2022
Like Article
Save Article
Previous
Next
Share your thoughts in the comments
Similar Reads