Answer:
B. {1, 2, 2, 3, 3, 4, 5}
Explanation:
Given
The above code segment
Required
Determine which list does not work
The list that didn't work is 
Considering options (A) to (E), we notice that only list B has consecutive duplicate numbers i.e. 2,2 and 3,3
All other list do not have consecutive duplicate numbers
Option B can be represented as:
![nums[0] = 1](https://tex.z-dn.net/?f=nums%5B0%5D%20%3D%201)
![nums[1] = 2](https://tex.z-dn.net/?f=nums%5B1%5D%20%3D%202)
![nums[2] = 2](https://tex.z-dn.net/?f=nums%5B2%5D%20%3D%202)
![nums[3] = 3](https://tex.z-dn.net/?f=nums%5B3%5D%20%3D%203)
![nums[4] = 3](https://tex.z-dn.net/?f=nums%5B4%5D%20%3D%203)
![nums[5] = 4](https://tex.z-dn.net/?f=nums%5B5%5D%20%3D%204)
![nums[6] = 5](https://tex.z-dn.net/?f=nums%5B6%5D%20%3D%205)
if (nums.get(j).equals(nums.get(j + 1)))
The above if condition checks for duplicate numbers.
In (B), when the elements at index 1 and 2 (i.e. 2 and 2) are compared, one of the 2's is removed and the Arraylist becomes:
![nums[0] = 1](https://tex.z-dn.net/?f=nums%5B0%5D%20%3D%201)
![nums[1] = 2](https://tex.z-dn.net/?f=nums%5B1%5D%20%3D%202)
![nums[2] = 3](https://tex.z-dn.net/?f=nums%5B2%5D%20%3D%203)
![nums[3] = 3](https://tex.z-dn.net/?f=nums%5B3%5D%20%3D%203)
![nums[4] = 4](https://tex.z-dn.net/?f=nums%5B4%5D%20%3D%204)
![nums[5] = 5](https://tex.z-dn.net/?f=nums%5B5%5D%20%3D%205)
The next comparison is: index 3 and 4. Meaning that comparison of index 2 and 3 has been skipped.
<em>This is so because of the way the if statement is constructed.</em>
Answer
First part:
The transmitted 8-bit sequence for ASCII character '&' with odd parity will be 00100110. Here leftmost bit is odd parity bit.
Second part:
The invalid bit sequence are option a. 01001000 and d. 11100111
Explanation:
Explanation for first part:
In odd parity, check bit of either 0 or 1 is added to the binary number as leftmost bit for making the number of 1s in binary number odd.
If there are even number of 1s present in the original number then 1 is added as leftmost bit to make total number of 1s odd.
If there are odd number of 1s present in the original number then 0 is added as leftmost bit to keep the total number of 1s odd.
Explanation for second part:
A valid odd parity bit sequence will always have odd number of 1s.
Since in option a and d, total number of 1s are 2 and 6 i.e. even number. Therefore they are invalid odd parity check bit sequences.
And since in option b and c, total number of 1s are 5 and 7 i.e. odd numbers respectively. Therefore they are valid odd parity check bit sequences.