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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

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 <bits/stdc++.h>
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<N> 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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