Arithmetic Coding Bits Needed

Problem

After arithmetic coding, the final interval has some width. Return the number of bits needed to distinguish a point in that interval, which is ceil(-log2(width)). Assume 0 < width <= 1, so an interval of width 1.0 needs 0 bits.

Hint

Halving the interval width adds exactly one bit to the description.

Test cases

case 1

in: [0.125]

expected: 3

case 2

in: [0.3]

expected: 2

case 3

in: [1]

expected: 0

case 4

in: [0.01]

expected: 7

case 5

in: [0.5]

expected: 1

Starter codePython

import math
def arithmetic_bits_needed(width):
    # Your code here
    pass

5 test cases · run your code against them in the editor