Python | Count set bits in a range
Last Updated :
15 Jun, 2022
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
We have existing solution for this problem please refer Count set bits in a range link. We can solve this problem quickly in Python. Approach is very simple,
- Convert decimal into binary using bin(num) function.
- Now remove first two characters of output binary string because bin function appends ‘0b’ as prefix in output string by default.
- Slice string starting from index (l-1) to index r and reverse it, then count set bits in between.
Python
def countSetBits(n,l,r):
binary = bin (n)
binary = binary[ 2 :]
binary = binary[ - 1 :: - 1 ]
print ( len ([binary[i] for i in range (l - 1 ,r) if binary[i] = = '1' ]))
if __name__ = = "__main__":
n = 42
l = 2
r = 5
countSetBits(n,l,r)
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Output:
2
Time Complexity : O(log n)
Auxiliary Space : O(log n)
Another Approach:
The set bits in the binary form of a number (obtained using the bin() method) can be obtained using the count() method.
Python3
def countSetBits(n, l, r):
binary = bin (n)[ - 1 : 1 : - 1 ]
print (binary[l - 1 : r].count( "1" ))
n = 42
l = 2
r = 5
countSetBits(n, l, r)
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Time Complexity : O(1)
Auxiliary Space: O(1)
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