Given a non-negative number n and two values l and r. The problem is to count the number of unset bits in the range l to r in the binary representation of n, i.e, to count unset bits from the rightmost lth bit to the rightmost rth bit. Examples:
Input : n = 42, l = 2, r = 5 Output : 2 (42)10 = (101010)2 There are '2' unset bits in the range 2 to 5. Input : n = 80, l = 1, r = 4 Output : 4
We have existing solution for this problem please refer Count unset 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 unset bits in between.
# Function to count unset bits in a range def unsetBits(n,l,r):
# convert n into it's binary
binary = bin (n)
# remove first two characters
binary = binary[ 2 :]
# reverse string
binary = binary[ - 1 :: - 1 ]
# count all unset bit '0' starting from index l-1
# to r, where r is exclusive
print ( len ([binary[i] for i in range (l - 1 ,r) if binary[i] = = '0' ]))
# Driver program if __name__ = = "__main__":
n = 42
l = 2
r = 5
unsetBits(n,l,r)
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Output:
2
Another Approach:
The set bits in the binary form of a number (obtained using the bin() method) can be obtained using the count() method.
# Function to count set bits in a range def countUnsetBits(n, l, r):
# convert n into its binary form
# using bin()
# and then process it using string
# slice methods
binary = bin (n)[ - 1 : 1 : - 1 ]
# count all set bit '1' starting from index l-1
# to r, where r is exclusive
print (binary[l - 1 : r].count( "0" ))
# Driver Code n = 42
l = 2
r = 5
countUnsetBits(n, l, r) #This code is contributed by phasing17 |
2
Time Complexity: O(1)
Auxiliary Space: O(1)