# Count number of set bits in a range using bitset

Given a large binary number.The task is to count the number of 1’s in a given range from L to R (1 based indexing).

Examples:

Input : s = “101101011010100000111”, L = 6, R = 15
Output : 5
s [L : R] = “1011010100”
There is only 5 set bits.

Input : s = “10110”, L = 2, R = 5
Output : 2

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

• Convert the string of size len to the bitset of size N.
• There is no need of (N – len) + (L – 1) bits in the left side and (N – R) bits in the right side of the bitset .
• Remove those bits efficiently using left and right shift bitwise operation.
• Now there are all zeroes in the left side of L and right side of R, so just use count() function to get the count of 1’s in the bitset as all positions except [L, R] are ‘0’.

Below is the implementation of above approach:

## C++

 // C++ implementation of above approach #include using namespace std; #define N 32    // C++ function to count  // number of 1's using bitset int GetOne(string s, int L, int R) {        int len = s.length();        // Converting the string into bitset     bitset bit(s);        // Bitwise operations     // Left shift     bit <<= (N - len + L - 1);        // Right shifts     bit >>= (N - len + L - 1);     bit >>= (len - R);        // Now bit has only those bits     // which are in range [L, R]        // return count of one in [L, R]     return bit.count(); }    // Driver code int main() {        string s = "01010001011";        int L = 2, R = 4;        cout << GetOne(s, L, R);        return 0; }

## Python3

 # Python3 implementation of above approach     N = 32     # function for converting binary # string into integer value def binStrToInt(binary_str):     length = len(binary_str)     num = 0     for i in range(length):         num = num + int(binary_str[i])         num = num * 2     return num / 2       # function to count  # number of 1's using bitset  def GetOne(s, L, R) :         length = len(s);         # Converting the string into bitset      bit = s.zfill(32-len(s));            bit = int(binStrToInt(bit))        # Bitwise operations      # Left shift      bit <<= (N - length + L - 1);         # Right shifts      bit >>= (N - length + L - 1);      bit >>= (length - R);         # Now bit has only those bits      # which are in range [L, R]         # return count of one in [L, R]      return bin(bit).count('1');        # Driver code  if __name__ == "__main__" :         s = "01010001011";         L = 2; R = 4;         print(GetOne(s, L, R));         # This code is contributed by AnkitRai01

Output:

2

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Improved By : AnkitRai01