Given an unsorted array of integers, write a function in Python to find the length of the longest increasing subsequence (LIS) in the array. For example, given the array [10, 9, 2, 5, 3, 7, 101, 18], the LIS is [2, 3, 7, 101], which has a length of 4. Your solution should have a time complexity of O(n log n), where n is the length of the input array. Here's some code to get you started: def longest increasing_subsequence(arr): # TODO: implement function pass # example usage arr [10, 9, 2, 5, 3, 7, 101, 18] print(longest_increasing_subsequence(arr)) # should print 4

Call a sequence X[1 · · n] of numbers oscillating if X[i] < X[i + 1] for all even i, and X[i] > X[i + 1] for all odd i. Describe an efficient algorithm to compute the length of the longest oscillating subsequence of an arbitrary array A of integers.

Using DASK in Python to solve this problem. Given a dask array, compute the difference along axis 0 using mean squared error This can be done without loops -- think outside of the box! This also needs to work for 1D, 2D, 3D, 4D, etc. def sequential_MSE_axis0 (anArray): # this does use the earlier MSE implementation n_elements_axis0 = anArray. shape [0] - 1 anArray.shape[0] results = da.zeros(n_elements_axis0) for i in range(n_elements_axis0): results[i] = MSE (anArray[i], anArray[i+1]) return results Complete the function below. Don't call . compute() Test Result oneDda.arange(4); print (all (isclose(a, b) for a, b in zip (MSE_axis0 (oneD), [1., 1., 1.]))) twoD = True da.arange(4).reshape((2,2)); print (all(isclose(a, b) for a, True b in zip (MSE_axis 0 (twoD), [4.]))) = threeD da.arange(8).reshape((2, 2,2)); print (all(isclose(a, b). for a, b in zip (MSE_axis0 (threeD).compute(), [16.]))) True def MSE_axis0 (anArray): ## WRITE YOUR CODE HERE

In Python: Code two versions of Quicksort algorithms, one using the Lomuto method and one using the Hoare method. Clearly document your code so that it is clear how and why your algorithms are correct. For each of the two algorithms the output must be the following three items: the total numberof comparisons, the total number of swaps, and the total running time. Demonstrate your two Quicksort algorithms on the input array A= [2, 8, 7, 1, 3, 5, 6, 4].

You are given a 0-indexed array nums of size n consisting of non-negative integers. You need to apply n - 1 operations to this array where, in the ith operation (0-indexed), you will apply the following on the ith element of nums: • If nums[i] == nums [1 + 1], then multiply nums [1] by 2 and set nums [i+1] to 0. Otherwise, you skip this operation. After performing all the operations, shift all the 0's to the end of the array. • For example, the array [1,0,2,0,0,1] after shifting all its 0's to the end, is [1,2,1,0,0,0]. Return the resulting array. Note that the operations are applied sequentially, not all at once. Example 1: Input: nums = [1,2,2,1,1,0] Output: [1,4,2,0,0,0]

Can someone help me turn this in python code, please?

Given two sorted arrays nums1 and nums2 of size m and n respectively, return the median of the two sorted arrays. The overall run time complexity should be o(log (m+n)). Example 1: Input: nums1 [1,3], nums2 = [2] Output: 2.00000 Explanation: merged array = [1,2,3] and median is 2. Example 2: Input: nums1 [1,2], nums2 = [3,4] Output: 2.50000 Explanation: merged array = [1,2,3,4] and median is (2 + 3) / 2 = 2.5. Constraints: nums1.length == m • nums2.length == n 0 <= m <= 1000 0 <= n <= 1000 1 <= m + n <= 2000 -10° <= nums1[i], nums2[i] <= 10°

Help only for b and c, not question a

it's not a programming question.

implement QuickSort of ints that sorts the numbers in the non-decreasing order. Implement the rearrange function using QuickSort ( such that the pivot is set on the extreme left and the rearrangement is carried on on two pointers) using the O(n) time algorithmThe function gets as input an array, and index of the pivot.The function rearranges the array, and returns the index of the pivot after the rearrangement. int rearrange(int* A, int n, int pivot_index); Implement the QuickSort algorithm. - For n<=2 the algorithm just sorts the (small) array (smaller number first). - For n>=3 the algorithm uses the rearrange function with the pivot chosen to be the median of A[0], A[n/2], A[n-1]. void quick_sort(int* A, int n);

Given two sorted arrays nums1 and nums2 of size m and n respectively, return the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)). Example 1: Input: nums1 = [1,3], nums2 = [2] Output: 2.00000 Explanation: merged array = [1,2,3] and median is 2. Example 2: Input: nums1 = [1,2], nums2 = [3,4] Output: 2.50000 Explanation: merged array = [1,2,3,4] and median is (2 + 3) / 2 = 2.5. Constraints: nums1.length == m nums2.length == n 0 <= m <= 1000 0 <= n <= 1000 1 <= m + n <= 2000 -106 <= nums1[i], nums2[i] <= 106 Write the whole code in python language Attach the code outputs also and explain the implementation.

Implement a function in Java to find the largest and smallest elements in an array of integers, and return them in a pair. The function should have a time complexity of O(n) and a space complexity of O(1).

how to solve this problem in python