# Python | Count set bits in a range

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,

1. Convert decimal into binary using bin(num) function.
2. Now remove first two characters of output binary string because bin function appends â€˜0bâ€™ as prefix in output string by default.
3. Slice string starting from index (l-1) to index r and reverse it, then count set bits in between.

## Python

 `# Function to count set bits in a range`` ` `def` `countSetBits(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 set bit '1' 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]``=``=``'1'``]))`` ` `# Driver program``if` `__name__ ``=``=` `"__main__":``    ``n``=``42``    ``l``=``2``    ``r``=``5``    ``countSetBits(n,l,r)`

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

 `# Function to count set bits in a range` `def` `countSetBits(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(``"1"``))`  `# Driver Code``n ``=` `42``l ``=` `2``r ``=` `5``countSetBits(n, l, r)`  `#This code is contributed by phasing17`

Output
```2
```

Time Complexity : O(1)

Auxiliary Space: O(1)

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