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To write the pseudocode for the brute force method to solve the maximum subarray having run time complexity
0 (n²).

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For each question, an algorithm will be described that operates on N elements, and your answer should include: (a) a big-O expression that describes the total number of operations in the worst case (for ex- ample, O(N³)) (b) a description of how to achieve the same effect as the algorithm described, but achieved with a better big-O time bound (for example, "use mergesort instead of insertion sort") (c) the big-O time bound for your improved approach. Your improved algorithm does not need to be provably the best possible, but it should have a different and better big-O bound. (It may not be as simple as substituting one named algorithm for another; consider what is redundant about the work done by the existing algorithm.) You don't need to use pseudocode to describe your algorithms - the style used in the problem descriptions is also sufficient for your solutions. You can use pseudocode if you like. Do not write real code. If you wish to use an algorithm described in class, you can name…
I keep getting different answers which is the true answer
No plagarism please! Correct and detailed answer will be Upvoted else downvoted. Thank you!
Get the time complexity function from the pseudocode for the addition of the 2 matrices below, and prove whether the big-oh is O(n^2) so that it satisfies the rule f(n) <= c g(n) for(i = 0 ; i < rows; i++){for(j = 0; j < columns; j++)matrix2[i][j] = matrix1[i][j] + matrix2[i][j];} for(i = 0 ; i < rows; i++){for(j = 0; j < columns; j++){printf("%d ", matrix2[i][j]);}printf("\n");}
The pseudo-code to create the matrix:Procedure AdjacencyMatrix(N): //N represents the number of nodesMatrix[N][N]for i from 1 to N for j from 1 to N Take input -> Matrix[i][j] endforendforWe can also populate the Matrix using this common way: Procedure AdjacencyMatrix(N, E): // N -> number of nodesMatrix[N][E] // E -> number of edgesfor i from 1 to E input -> n1, n2, cost Matrix[n1][n2] = cost Matrix[n2][n1] = costendfor make a Java code implemented using above pseudo-code?
python code only, thanks
READ AND SOLVE VERY CAREFULLY WITH PYTHONthanks in advance Given a jungle matrix N*M: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: (Use Python)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 4*4 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 move in two directions: forward and down.- You can only pass thru spaces on the jungle matrix marked ‘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 matrixElse Mark current cell in the solution matrix Move forward horizontally and recursively check if this leads to…
What is the complexity of the following recurrence: T'п) — 4T (п/2) +п, T(1) —D1 %3D
A robber is planning to rob the houses along a street. The constraint is, that he cannot rob adjacent houses. If done, it will alert the police. For the below-given algorithm and input, your task is to draw a DP matrix AND return the maximum earnings in a single night: nums = [4,3,6,2,7,10,5] Algorithm: def rob(self, nums: List[int]) -> int: dp = [0]*len(nums) // Initialize dp array n = len(dp) if len(nums) == 1: return nums[0] elif len(nums) == 2: return max(nums[0],nums[1]) dp[0]=nums[0] dp[1]=max(nums[1],nums[0]) for i in range(2,len(nums)): dp[i] = max(nums[i]+dp[i-2],dp[i-1]) return dp[n-1]