answer:California's parole board has once again denied release to Leslie Van Houten, a former member of Charles Manson's cult, who was involved in the 1969 murders of Leno and Rosemary LaBianca. This marks the 19th time Van Houten, now 60, has been denied parole. The board concluded that she still poses a risk to society, a decision praised by Los Angeles Deputy District Attorney Patrick Sequeira, who emphasized the brutal nature of the crimes and Van Houten's lack of full responsibility acceptance. Van Houten, known as Lulu in the Manson family, was 19 when she helped hold down Rosemary LaBianca during the murders. She has been incarcerated at the California Institution for Women since 1978, following her final conviction on first-degree murder charges. Despite being described as a model inmate and mentor, the parole board found her continued imprisonment necessary for public safety. Van Houten's lawyer, Brandie Devall, argued for her release, citing recent California Supreme Court rulings that emphasize the importance of rehabilitation and the need for the parole board to prove current danger. Van Houten has acknowledged her role in the murders and expressed remorse,

answer:```python def find_second_highest(arr): if len(arr) < 2: return None max_num = second_max = float('-inf') for num in arr: if num > max_num: second_max = max_num max_num = num elif num > second_max and num != max_num: second_max = num if second_max == float('-inf'): return None return second_max # Test find_second_highest def test_find_second_highest(): test_cases = [ {input: [1, 2, 3, 4, 5], expected: 4}, {input: [2, 2, 2, 2, 2], expected: None}, {input: [5, 5, 5, 5, 5], expected: None}, {input: [10, 20, 20, 20, 40, 40], expected: 20}, {input: [], expected: None}, {input: [1], expected: None}, {input: [1, 1], expected: None}, {input: [1, 1, 2], expected: 1}, ] for test_case in test_cases: result = find_second_highest(test_case[input]) if result != test_case[expected]: print(fTest case failed: {test_case['input']}. Expected {test_case['expected']}, got {result}.) test_find_second_highest() ``` In this implementation, we initialize `max_num` and `second_max` to negative infinity. We then iterate through the array. If a number is greater than `max_num`, we update `second_max` to the old `max_num` and update `max_num` to the current number. If a number is greater than `second_max` but not equal to `max_num`, we update `second_max`. If `second_max` remains negative infinity after the iteration, we return None, indicating that there is no second-highest number. Otherwise, we return `second_max`.

answer:```python def flatten_list(list_of_lists): return [item for sublist in list_of_lists for item in sublist] # Example usage: input_list = [[1, 2, 3], [4, 5, 6], [7, 8, 9]] print(flatten_list(input_list)) # Output: [1, 2, 3, 4, 5, 6, 7, 8, 9] ``` This function uses nested list comprehension to iterate over each sublist in the input list and then over each item in the sublist, effectively flattening the list of lists into a single list. Alternatively, you can use the `itertools.chain` function to achieve the same result: ```python import itertools def flatten_list(list_of_lists): return list(itertools.chain(*list_of_lists)) ``` This function uses the `*` operator to unpack the sublists into separate arguments to the `chain` function, which then concatenates them into a single iterable that is converted to a list.

answer:```python def find_first_duplicate(arr): seen = set() for num in arr: if num in seen: return num seen.add(num) return None # Example usage: input_list1 = [1, 2, 3, 4, 5, 6, 7, 8, 9, 1] print(find_first_duplicate(input_list1)) # Output: 1 input_list2 = [2, 3, 4, 5, 6, 7, 8, 9] print(find_first_duplicate(input_list2)) # Output: None ``` This function uses a set to keep track of the elements it has seen so far. It iterates over the input array, and for each element, it checks if it is already in the set. If it is, the function returns the element as the first duplicate. If it is not, the element is added to the set. If the function finishes iterating over the array without finding a duplicate, it returns None. This solution has a time complexity of O(n), where n is the length of the input array, because it performs a constant-time operation for each element in the array. The space complexity is also O(n), as in the worst case, the set will contain all elements from the input array.