# 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