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# Python | Count unset bits in a range

• Last Updated : 31 Dec, 2017

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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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,

1. Convert decimal into binary using bin(num) function.
2. Now remove first two characters of output binary string because bin function appends ‘0b’ as prefix in output string by default.
3. 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)`

Output:

```2
```

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