Given a non-negative number n and two values l and r. The problem is to check whether all the bits are unset 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 = 17, l = 2, r = 4 Output : Yes (17)10 = (10001)2 The bits in the range 2 to 4 are all unset. Input : n = 36, l = 3, r = 5 Output : No (36)10 = (100100)2 The bits in the range 3 to 5 are all not unset.
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 new_num == 0, return “Yes” (all bits are unset in the given range).
- Else return “No” (all bits are not unset in the given range).
C++
// C++ implementation to check whether all // the bits are unset in the given range // or not#include <bits/stdc++.h>using namespace std;// function to check whether all the bits// are unset in the given range or notstring 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 new num is 0, then all bits // are unset in the given range if (new_num == 0) return "Yes"; // else all bits are not unset return "No"; }// Driver program to test aboveint main(){ unsigned int n = 17; unsigned int l = 2, r = 4; cout << allBitsSetInTheGivenRange(n, l, r); return 0;} |
Java
// Java implementation to // check whether all the// bits are unset in the // given range or notimport java.io.*;class GFG { // function to check whether // all the bits are unset in// the given range or notstatic 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 new num is 0, then // all bits are unset in // the given range if(new_num == 0) return "Yes"; // else all bits // are not unset return "No"; }// Driver Codepublic static void main (String[] args){ int n = 17; int l = 2; int r = 4; System.out.println( allBitsSetInTheGivenRange(n, l, r));}}// This code is contributed by akt_mit |
Python 3
# Python 3 implementation to check whether # all the bits are unset in the given range # or not# function to check whether all the bits# are unset in the given range or notdef 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 new num is 0, then all bits # are unset in the given range if (new_num == 0): return "Yes" # else all bits are not unset return "No"# Driver Codeif __name__ == "__main__": n = 17 l = 2 r = 4 print(allBitsSetInTheGivenRange(n, l, r))# This code is contributed by ita_c |
C#
// C# implementation to // check whether all the // bits are unset in the // given range or not using System;public class GFG{ // function to check whether // all the bits are unset 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 new num is 0, then // all bits are unset in // the given range if(new_num == 0) return "Yes"; // else all bits // are not unset return "No"; } // Driver Code static public void Main (){ int n = 17; int l = 2; int r = 4; Console.WriteLine( allBitsSetInTheGivenRange(n, l, r)); } } // This code is contributed by k_mit |
PHP
<?php// PHP implementation to check// whether all the bits are // unset in the given range // or not// function to check whether // all the bits are unset in// the given range or notfunction 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 & $num; // if new num is 0, then // all bits are unset in // the given range if ($new_num == 0) return "Yes"; // else all bits // are not unset return "No"; }// Driver Code$n = 17;$l = 2; $r = 4;echo allBitsSetInTheGivenRange($n, $l, $r);// This code is contributed// by ajit?> |
Javascript
<script>// Javascript implementation to check whether all // the bits are unset in the given range // or not// function to check whether all the bits// are unset in the given range or notfunction 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 new num is 0, then all bits // are unset in the given range if (new_num == 0) return "Yes"; // else all bits are not unset return "No"; }// Driver program to test abovevar n = 17;var l = 2, r = 4;document.write( allBitsSetInTheGivenRange(n, l, r));</script> |
Output:
Yes
Time Complexity : O(1)
Auxiliary Space: O(1)
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