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Count number of set bits in a range using bitset
• Last Updated : 05 Jul, 2019

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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