import numpy as np import random n = 10 X = np.random.randint(1, 20, n*n).reshape(n,n) Y = np.random.randint(1, 20, n*n).reshape(n,n) result = np.zeros(X. shape) for i in range (len(X)): for j in range (len(Y[0])): for k in range (len (Y)): result[i][j] += X[i][k] * Y[k][j] result= result.reshape(-1,1) a = 0 for r in result: a = a + r sum_value = a[0] print("Total sum of all elements is {}".format(sum_value))     Complete analysis of this code and Big O

in java ecplise . Implement a program that randomly generates 10 integers from -100 to 100,stores them in a 1D array, and finds their maximum value. Calculate theexecution time of finding a maximum. Repeat the test for 10,000 and10,000,000 numbers. Provide your results in the form of a table below andprovide a small summary. Implement a Java program that will populate a 5x5 matrix with randomlygenerated integers from 0 to 100.(1) print your matrix in a table form.(2) modify your code to multiply all even numbers by 10 and print the matrix.Your output of questions (1) and (2) should match the format of the followingsample outputs:

import numpy as npfrom scipy.optimize import minimizeimport matplotlib.pyplot as pltimport pickle #Implement the function regressionObjVal def regressionObjVal(w, X, y):​    # compute squared error (scalar) with respect    # to w (vector) for the given data X and y          #    # Inputs:    # w = d x 1    # X = N x d    # y = N x 1    # Output:    # error = scalar value (which is the error calcualted using objective function)​    # IMPLEMENT THIS METHOD - REMOVE THE NEXT LINE    error = 0​    return error #Implement the function regressionGradient def regressionGradient(w, X, y):​    # compute gradient of squared error (scalar) with respect    # to w (vector) for the given data X and y           # Inputs:    # w = d x 1    # X = N x d    # y = N x 1    # Output:    # gradient = d length vector (not a d x 1 matrix)​    if len(w.shape) == 1:        w = w[:,np.newaxis]    # IMPLEMENT THIS METHOD - REMOVE THE NEXT LINE     error_grad = np.zeros((X.shape[1],))        return error_grad…

for given A and B s : import randomA = random.sample(range(0, 101), random.randint(1,100))A.sortprint(A)B = random.sample(range(0, 101), random.randint(1,100))B.sort(reverse=True)print(B) Conduct doubling experiments to compare the run time of selection vs. insertion sort (for both A and B). Plot the results (both case of array) for array sizes n = 2, 4, ... 1024. insertion and selection sort are provided below import timeimport numpy as np ### Selection Sort ###def selection_sort(A):    for i in range(len(A)):        min_idx = i        for j in range(i + 1, len(A)):            if A[min_idx] > A[j]:                min_idx = j        A[i], A[min_idx] = A[min_idx], A[i]    return A ### Insertion Sort ###def insertion_sort(A):    n = len(A)    count = 0    for i in range(1,n):        k = i        while k > 0 and A[k] < A[k-1]:            A[k], A[k-1] = A[k-1], A[k]        k = k - 1        count += 1    return count, A   provide code and result