For this given code, please help me create a recursive case for transpose of matrix using the cache oblivious algorithm. for ( size_t i=0; i
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For this given code, please help me create a recursive case for transpose of matrix using the cache oblivious
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- ## Count the number of unique paths from a[0][0] to a[m-1][n-1]# We are allowed to move either right or down from a cell in the matrix.# Approaches-# (i) Recursion- Recurse starting from a[m-1][n-1], upwards and leftwards,# add the path count of both recursions and return count.# (ii) Dynamic Programming- Start from a[0][0].Store the count in a count# matrix. Return count[m-1][n-1]# T(n)- O(mn), S(n)- O(mn)# def count_paths(m, n): if m < 1 or n < 1: return -1 count = [[None for j in range(n)] for i in range(m)] # Taking care of the edge cases- matrix of size 1xn or mx1 for i in range(n): count[0][i] = 1 for j in range(m): count[j][0] = 1 for i in range(1, m): for j in range(1, n): # Number of ways to reach a[i][j] = number of ways to reach # a[i-1][j] + a[i][j-1] count[i][j] = count[i - 1][j] + count[i][j - 1]…Given an n x n matrix where each of the rows and columns is sorted in ascending order, return the kth smallest element in the matrix. Note that it is the kth smallest element in the sorted order, not the kth distinct element. You must find a solution with a memory complexity better than O(n2). Example 1: Input: matrix = [[1,5,9],[10,11,13],[12,13,15]], k = 8 Output: 13 Explanation: The elements in the matrix are [1,5,9,10,11,12,13,13,15], and the 8th smallest number is 13 Example 2: Input: matrix = [[-5]], k = 1 Output: -5#plea# Count the number of unique paths from a[0][0] to a[m-1][n-1]# We are allowed to move either right or down from a cell in the matrix.# Approaches-# (i) Recursion- Recurse starting from a[m-1][n-1], upwards and leftwards,# add the path count of both recursions and return count.# (ii) Dynamic Programming- Start from a[0][0].Store the count in a count# matrix. Return count[m-1][n-1]# T(n)- O(mn), S(n)- O(mn)# def count_paths(m, n): if m < 1 or n < 1: return -1 count = [[None for j in range(n)] for i in range(m)] # Taking care of the edge cases- matrix of size 1xn or mx1 for i in range(n): count[0][i] = 1 for j in range(m): count[j][0] = 1 for i in range(1, m): for j in range(1, n): # Number of ways to reach a[i][j] = number of ways to reach # a[i-1][j] + a[i][j-1] count[i][j] = count[i - 1][j] + count[i][j - 1]…
- Q3. Compute the cost of r JOINA=B s using the method of index loops with an index of height 2 on B of s. Do the computation for the clustering index case. Assume that r occupies 4,000 blocks with 20 tuples per block, s occupies 10,000 blocks with 5 tuples per block, and the amount of main memory available is 402 blocks. Assume that at most 5 tuples in s match each tuple in r.Question 17. Sorting a data set is an important sub-problem in data science. Given the size n of a data set, which statements are correct? a) Bubble Sort has worst-case run-time complexity O(n).b) Bubble Sort has worst-case run-time complexity O(n log N).c) Bubble Sort has worst-case run-time complexity O(n2).d) Merge Sort has worst-case run-time complexity O(n).e) Merge Sort has worst-case run-time complexity O(n log N).f) Merge Sort has worst-case run-time complexity O(n2).g) Quick Sort has worst-case run-time complexity O(n).h) Quick Sort has worst-case run-time complexity O(n log N).i) Quick Sort has worst-case run-time complexity O(n2).Given a sequence of integers, e.g., \[ 9,-1,45,6,8,21,34,5,55,65,543,18,90,122,132,0,66,100,-12,17 \] design an algorithm/method to re-arrange the order of the data items (i.e., form a new sequence of the integers) so that when the data items are inserted sequentially into an initially empty BST, the newly created BST will be a balanced BST. a) Design the algorithm (in pseudo code) and analyse your algorithm. b) Write a Python function/method to implement your algorithm. c) Write a program that calls the above function, thereby demonstrating your algorithm by (i). Print the re-arranged data items (i.e., a new sequence) generated by the algorithm, and (ii). Build a BST using the newly generated data sequence as input (that is, insert the data items, oneby-one, into an initially empty BST). Print the tree shape of the BST. (Note: The program should use at least two datasets to demonstrate the algorithm, one of which is the dataset given above).
- Prob: Using sequential and dynamic list to implement the Addition, Substraction,or Multiplication of two unary polynomials. For Am(x) and Bn(x). Am(x)=A0+A1x1+A2x2+A3x3+… +Amxm Bn(x)=B0+B1x1+B2x2+B3x3+… +Bnxn Please use program to solve M(x)= Am(x)+Bn(x), M(x)= Am(x)-Bn(x) and M(x)= Am(x)×Bn(x). Requirements： 1)Justify the sparseness of the polynomial. 2)Use array structure and linked list separately. 3)Result polynomial has no Zero coefficients components and same power components； 4)Two different output format: ASC, DESC according to the power. Note: Write the full code in C++. Thanks for your help.#com# Count the number of unique paths from a[0][0] to a[m-1][n-1]# We are allowed to move either right or down from a cell in the matrix.# Approaches-# (i) Recursion- Recurse starting from a[m-1][n-1], upwards and leftwards,# add the path count of both recursions and return count.# (ii) Dynamic Programming- Start from a[0][0].Store the count in a count# matrix. Return count[m-1][n-1]# T(n)- O(mn), S(n)- O(mn)# def count_paths(m, n): if m < 1 or n < 1: return -1 count = [[None for j in range(n)] for i in range(m)] # Taking care of the edge cases- matrix of size 1xn or mx1 for i in range(n): count[0][i] = 1 for j in range(m): count[j][0] = 1 for i in range(1, m): for j in range(1, n): # Number of ways to reach a[i][j] = number of ways to reach # a[i-1][j] + a[i][j-1] count[i][j] = count[i - 1][j] + count[i][j - 1]…Answer the given question with a proper explanation and step-by-step solution. Devise formulas (based on n, p and i) for my_first_i and my_last_i for processor i, in the global sum example (Like above). Assume that the processor numbers and array index start from 0. Remember that each core should be assigned roughly the same number of elements of computations in the loop. Hint: First consider the case when n is evenly divisible by p. n --> total number (1,2,3 .... n)p --> total number of processors We need to calculate sum of numbers.
- Given a jungle matrix NxM: jungle = [ [1, 0, 0, 0], [1, 1, 0, 1], [0, 1, 0, 0], [1, 1, 1, 1,] ] Where 0 means the block is dead end and 1 means the block can be used in the path from source to destination. Task: Starting at position (0, 0), the goal is to reach position (N-1, M-1). Your program needs to build and output the solution matrix – a 4x4 matrix with 1’s in positions used to get from the starting position (0,0) to the ending position (N-1,M-1) with the following constraints: You can only move one space at a time You can only in two directions: forward and down. You can only pass thru spaces on the jungle matrix mark ' 1 ' If you cannot reach the ending position - print a message that you're trapped in the jungle Algorithm: If destination is reached print the solution matrix Else Mark current cell in the solution matrix Move forward horizontally and recursively check if this leads to a solution If there is no solution, move down and recursively check if this leads to…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.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