Mixed Precision Memory Saving

Problem

Compute the bytes saved by storing parameters in half precision instead of full precision: num_params * (full_bytes - half_bytes).

The signature is mixed_precision_saving(num_params, full_bytes=4, half_bytes=2). Return an integer number of bytes.

Hint

Each parameter drops from 4 bytes to 2 bytes in fp16/bf16.

case 1

in: [1000000]

expected: 2000000

case 2

in: [1]

expected: 2

case 3

in: [7000000000,4,2]

expected: 14000000000

case 4

in: [1000,2,1]

expected: 1000

Starter codePython

def mixed_precision_saving(num_params, full_bytes=4, half_bytes=2):
    # Your code here
    pass

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