Check whether all the bits are unset in the given range or not Last Updated : 13 Jun, 2022 Comments Improve Suggest changes Like Article Like Report 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 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 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 above int 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 not import java.io.*; 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 public 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 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 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 if __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 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 & $num; // if new num is 0, then // all bits are unset in // the given range if ($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 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 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 above var n = 17; var l = 2, r = 4; document.write( allBitsSetInTheGivenRange(n, l, r)); </script> Output: Yes Time Complexity : O(1) Auxiliary Space: O(1) Comment More infoAdvertise with us Next Article Analysis of Algorithms A Ayush Improve Article Tags : Bit Magic DSA Practice Tags : Bit Magic Similar Reads Basics & PrerequisitesLogic Building ProblemsLogic building is about creating clear, step-by-step methods to solve problems using simple rules and principles. 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