1. One number for every set of lit LEDs

The watch has 10 LEDs. Give each LED one bit of a 10-bit number, called a mask. Bit $i$ of the mask is 1 when LED $i$ is lit. The code puts the 4 hour LEDs in the top bits, 6 to 9, and the 6 minute LEDs in the low bits, 0 to 5.

A set of 10 items has $2^{10} = 1024$ subsets. Each mask from 0 to $2^{10} - 1 = 1023$ is exactly one of them, because each bit is either in or out. So the loop for mask in range(1 << 10) visits every possible set of lit LEDs once. 1 << 10 is $2^{10}$, the number 1 shifted left by 10 bits.

2. Reading one mask

For one mask, the code needs three numbers:

mask >> 6 shifts the bits right by 6 places, so the 6 minute bits fall off and the hour bits remain. 0b111111 is 63, six 1 bits. mask & 0b111111 keeps only the low 6 bits, which is the minute.

For example, the mask 1011111011 in binary is 763. Its top 4 bits are 1011, which is 11. Its low 6 bits are 111011, which is 59. So it shows 11:59, and it lights $3 + 5 = 8$ LEDs.

3. Why the loop finds every time exactly once

Every time the watch can show has exactly one mask: the hour bits followed by the minute bits. Every mask gives exactly one hour and one minute. So walking all 1024 masks and keeping the ones that pass the checks lists every valid time once, with no repeats.

The count of masks with exactly $t$ lit LEDs is the binomial coefficient $\binom{10}{t}$, the number of ways to choose $t$ bits out of 10. For $t = 1$ that is 10 masks.

4. Masks that do not show a real time

4 bits can hold values up to 15, and 6 bits up to 63. But an hour must be at most 11, and a minute at most 59. So the code keeps a mask only when hour < 12 and minute < 60.

This check also explains Example 2. The hours 0 to 11 have at most 3 set bits (7 is 0111 and 11 is 1011). The minutes 0 to 59 have at most 5 set bits (for example 31 is 011111 and 59 is 111011). So a real time lights at most $3 + 5 = 8$ LEDs. Every mask with 9 or 10 lit LEDs fails the check, and the code returns an empty list.

5. Example 1 by hand

turnedOn = 1. The masks with exactly one 1 bit are the 10 powers of two, from $2^0 = 1$ to $2^9 = 512$. The loop meets them in increasing order:

All 10 pass the check hour < 12 and minute < 60. The format f"{hour}:{minute:02d}" writes the hour with no leading zero and the minute with two digits. The result is the expected output, in the same order.

6. Example 2 by hand

turnedOn = 9. There are $\binom{10}{9} = 10$ masks with 9 lit LEDs. By part 4, none of them shows a real time. For example, the mask with only bit 0 off has hour 1111, which is 15, and fails hour < 12. The code returns [], the expected output.

7. From the math to the code

The loop always visits $2^{10} = 1024$ masks, and counting the bits of each one takes at most 10 steps. This is the $O(B \cdot 2^B)$ time with $B = 10$ from the code's docstring.