Count set bits in a range Last Updated : 15 Jul, 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 count the number of set bits in the range l to r in the binary representation of n, i.e, to count set bits from the rightmost lth bit to the rightmost rth bit. Constraint: 1 <= l <= r <= number of bits in the binary representation of n.Examples: Input : n = 42, l = 2, r = 5 Output : 2 (42)10 = (101010)2 There are '2' set bits in the range 2 to 5. Input : n = 79, l = 1, r = 4 Output : 4 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.Count number of set bits in the number (n & num). Refer this post. C++ // C++ implementation to count set bits in the // given range #include <bits/stdc++.h> using namespace std; // Function to get no of set bits in the // binary representation of 'n' unsigned int countSetBits(int n) { unsigned int count = 0; while (n) { n &= (n - 1); count++; } return count; } // function to count set bits in the given range unsigned int countSetBitsInGivenRange(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); // returns number of set bits in the range // 'l' to 'r' in 'n' return countSetBits(n & num); } // Driver program to test above int main() { unsigned int n = 42; unsigned int l = 2, r = 5; cout << countSetBitsInGivenRange(n, l, r); return 0; } Java // Java implementation to count set bits in the // given range class GFG { // Function to get no of set bits in the // binary representation of 'n' static int countSetBits(int n) { int count = 0; while (n > 0) { n &= (n - 1); count++; } return count; } // function to count set bits in the given range static int countSetBitsInGivenRange(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); // returns number of set bits in the range // 'l' to 'r' in 'n' return countSetBits(n & num); } // Driver code public static void main(String[] args) { int n = 42; int l = 2, r = 5; System.out.print(countSetBitsInGivenRange(n, l, r)); } } // This code is contributed by Anant Agarwal. Python3 # Python3 implementation to count # set bits in the given range # Function to get no of set bits in the # binary representation of 'n' def countSetBits(n): count = 0 while (n): n &= (n - 1) count = count + 1 return count # function to count set bits in # the given range def countSetBitsInGivenRange(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) # returns number of set bits in the range # 'l' to 'r' in 'n' return countSetBits(n & num) # Driver program to test above n = 42 l = 2 r = 5 ans = countSetBitsInGivenRange(n, l, r) print (ans) # This code is contributed by Saloni Gupta. C# // C# implementation to count set bits in the // given range using System; class GFG { // Function to get no of set bits in the // binary representation of 'n' static int countSetBits(int n) { int count = 0; while (n>0) { n &= (n - 1); count++; } return count; } // function to count set bits in the given range static int countSetBitsInGivenRange(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); // returns number of set bits in the range // 'l' to 'r' in 'n' return countSetBits(n & num); } //Driver code public static void Main() { int n = 42; int l = 2, r = 5; Console.WriteLine(countSetBitsInGivenRange(n, l, r)); } } // This code is contributed by Anant Agarwal. PHP <?php // PHP implementation to count // set bits in the given range // Function to get no of set bits in // the binary representation of 'n' function countSetBits($n) { $count = 0; while ($n) { $n &= ($n - 1); $count++; } return $count; } // function to count set // bits in the given range function countSetBitsInGivenRange($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); // returns number of // set bits in the range // 'l' to 'r' in 'n' return countSetBits($n & $num); } // Driver Code $n = 42; $l = 2; $r = 5; echo countSetBitsInGivenRange($n, $l, $r); // This code is contributed by Ajit. ?> JavaScript <script> // Javascript implementation to // count set bits in the // given range // Function to get no of set bits in the // binary representation of 'n' function countSetBits(n) { let count = 0; while (n > 0) { n &= (n - 1); count++; } return count; } // function to count set // bits in the given range function countSetBitsInGivenRange(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 let num = ((1 << r) - 1) ^ ((1 << (l - 1)) - 1); // returns number of set // bits in the range // 'l' to 'r' in 'n' return countSetBits(n & num); } // driver program let n = 42; let l = 2, r = 5; document.write(countSetBitsInGivenRange(n, l, r)); </script> Output: 2 Time Complexity: O(logn) Auxiliary Space: O(1) Comment More infoAdvertise with us Next Article Count set bits in a range A Ayush Jauhari Improve Article Tags : Bit Magic DSA Practice Tags : Bit Magic Similar Reads Count unset bits in a range Given a non-negative number n and two values l and r. 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