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:
- 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.
- Calculate new_num = n & num.
- If num == new_num, return “Yes” (all bits are set in the given range).
- 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#
// 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. |
PHP
<?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 ?> |
Javascript
<script> // javascript 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 var 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 var 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 var n = 22; var l = 2, r = 3; document.write(allBitsSetInTheGivenRange( n, l, r)); // This code contributed by Princi Singh </script> |
Output:
Yes
Time Complexity – O(1)
Space Complexity – O(1)
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