answer:Here's a Python solution that prints the spiral path and the elements in the spiral order. ```python def spiral_traversal(matrix): rows, cols = len(matrix), len(matrix[0]) visited = [[False] * cols for _ in range(rows)] directions = [(0, 1), (1, 0), (0, -1), (-1, 0)] # right, down, left, up dir_idx = 0 row, col = 0, 0 spiral_path = [] spiral_elements = [] for _ in range(rows * cols): visited[row][col] = True spiral_elements.append(matrix[row][col]) spiral_path.append((row, col)) next_row, next_col = row + directions[dir_idx][0], col + directions[dir_idx][1] if 0 <= next_row < rows and 0 <= next_col < cols and not visited[next_row][next_col]: row, col = next_row, next_col else: dir_idx = (dir_idx + 1) % 4 row, col = row + directions[dir_idx][0], col + directions[dir_idx][1] return spiral_elements, spiral_path def print_spiral_path(matrix, spiral_path): rows, cols = len(matrix), len(matrix[0]) for row in range(rows): for col in range(cols): if (row, col) in spiral_path: print(f{matrix[row][col]:2d}*, end=' ') else: print(f{matrix[row][col]:2d} , end=' ') print() matrix = [ [1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16] ] spiral_elements, spiral_path = spiral_traversal(matrix) print(Spiral Elements:, spiral_elements) print(Spiral Path:) print_spiral_path(matrix, spiral_path) ``` This code works by maintaining a current direction and a current position. It moves in the current direction until it hits a visited cell or goes out of bounds, then it turns right. The visited cells are tracked using a boolean 2D list. The elements in the spiral order are appended to the `spiral_elements` list, and their positions are stored in the `spiral_path` list. The `print_spiral_path` function is used to print the matrix with the spiral path marked. It prints the elements in the matrix, appending an asterisk (*) to the elements that are in the spiral path. This solution has a time complexity of O(n*m), where n and m are the number of rows and columns in the matrix, respectively, since it visits each cell once. The space complexity is also O(n*m) due to the visited matrix and the spiral path list.

answer:Here is a Python solution that uses a dictionary to count the frequency of each element, then sorts the elements based on their frequency and value. ```python def sort_by_frequency(nums): freq = {} for num in nums: if num in freq: freq[num] += 1 else: freq[num] = 1 return sorted(nums, key=lambda x: (-freq[x], x)) # Example usage: nums = [1, 1, 2, 2, 3, 3, 3, 4, 4, 4, 4] print(sort_by_frequency(nums)) # Output: [4, 4, 4, 4, 3, 3, 3, 2, 2, 1, 1] # Explanation of the lambda function: # The sorted function takes a key function as an argument, which it uses to determine the sort order. # The lambda function here returns a tuple for each element in the list. # The first element of the tuple is the negative frequency of the element (-freq[x]), which is used as the primary sort key. # The second element of the tuple is the element itself (x), which is used as the secondary sort key. # By using the negative frequency, we sort in descending order of frequency (i.e., the most frequent elements come first). # If two elements have the same frequency, they are sorted in ascending order of their values. ``` Alternatively, you can use the `Counter` class from the `collections` module to count the frequency of each element, which is a bit more concise. ```python from collections import Counter def sort_by_frequency(nums): freq = Counter(nums) return sorted(nums, key=lambda x: (-freq[x], x)) ``` Both of these solutions have a time complexity of O(n log n) due to the sorting, where n is the length of the input list. The space complexity is O(n) due to the dictionary used to count the frequencies.

answer:Here is a Python solution that uses the built-in `max` function with a custom key to find the row with the maximum number of 1's. If there are multiple rows with the same maximum number of 1's, it returns the row index with the smallest value. ```python def max_ones_row(matrix): return max(range(len(matrix)), key=lambda i: (sum(matrix[i]), -i)) matrix = [ [0, 1, 1, 1], [0, 0, 1, 0], [0, 0, 1, 1], [0, 0, 0, 0], [1, 1, 1, 1] ] print(max_ones_row(matrix)) # Output: 4 ``` Explanation of the code: * `range(len(matrix))` generates a range of row indices. * `max` function finds the maximum row index based on the custom key. * `lambda i: (sum(matrix[i]), -i)` is the custom key function. * `sum(matrix[i])` calculates the number of 1's in the row (since `sum` treats `True` as 1 and `False` as 0, and 1 is treated as `True` in a boolean context). * `-i` is used to break ties in favor of the row index with the smallest value (since Python's `max` is stable, it will prefer the smallest index when the primary key is equal). Alternatively, you can use a list comprehension to calculate the sum of each row and then find the index of the maximum sum. ```python def max_ones_row(matrix): row_sums = [sum(row) for row in matrix] max_sum = max(row_sums) return row_sums.index(max_sum) matrix = [ [0, 1, 1, 1], [0, 0, 1, 0], [0, 0, 1, 1], [0, 0, 0, 0], [1, 1, 1, 1] ] print(max_ones_row(matrix)) # Output: 4 ``` However, this solution has a problem when there are multiple rows with the same maximum number of 1's. It returns the index of the first occurrence, not the smallest index. To fix this, you can use the `enumerate` function to get both the index and the value of each row sum, and then find the index with the smallest value. ```python def max_ones_row(matrix): row_sums = [sum(row) for row in matrix] max_sum = max(row_sums) return min(i for i, x in enumerate(row_sums) if x == max_sum) matrix = [ [0, 1, 1, 1], [0, 0, 1, 0], [0, 0, 1, 1], [0, 0, 0, 0], [1, 1, 1, 1] ] print(max_ones_row(matrix)) # Output: 4 ``` This solution has a time complexity of O(n*m) where n is the number of rows and m is the number of columns in the matrix, because it calculates the sum of each row. The space complexity is O(n) for storing the row sums.

answer:Since y=f(x) is an even function and y=f(x+1) is an odd function, we have f(-x)=f(x) and f(-x+1)=-f(x+1). Thus, f(x-1)=-f(x+1) and f(x-1)=f(x+3). From this, we can infer that f(x) is a periodic function with a period of 4. Therefore, f(10)=f(2)=-f(1) < 0, f(100)=f(0) > 0, and f(1) > 0. Given that 0 leqslant x leqslant 1, we have f(x) > 0 and f'(x) < 0, which means that the function f(x) is monotonically decreasing on [0,1]. Thus, f(0) > f(1) > 0 > -f(1), which implies that c > a > b. Hence, the answer is: boxed{D}. The fact that y=f(x) is an even function and y=f(x+1) is an odd function implies that f(x) is a periodic function with a period of 4. Also, y=f(x) is an even function, and for any 0 leqslant x leqslant 1, we have f'(x) < 0, which means that y=f(x) is a decreasing function on (0,1). By utilizing these properties, we can simplify the three numbers into function values with their variables in 0 leqslant x leqslant 1, and then compare their sizes using monotonicity. This problem tests the application of function's parity. It requires a comprehensive understanding of function properties derived from parity and the use of a function's monotonicity to compare sizes. In comparing the sizes of the three numbers in this problem, a technique of converting them into a monotonic interval is employed. Pay attention to the application of this technique when using monotonicity to compare sizes.