Check whether all the bits are set in the given range

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++

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// 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;
}

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Java

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// 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.

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Python3

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# 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.

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C#

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// C# implementation to check whether all the bits
// are set in the given range or not
using System;
  
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 ()
    {
        int n = 22;
        int l = 2, r = 3;
        Console.Write(allBitsSetInTheGivenRange(n, l, r));
    }
}
  
// This code is contributed by Anant Agarwal.

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PHP

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<?php
// PHP 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
function 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 = 22;
    $l = 2;
    $r = 3;
    echo allBitsSetInTheGivenRange($n, $l, $r);
      
// This code is contributed by Ajit
?>

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Output:

Yes

This article is contributed by Ayush Jauhari. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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