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- in netbeans using java Create a class NestedForPatternB. Use nested for-loops that display the following pattern:Write a for loop to print all elements in courseGrades, following each element with a space (including the last). Print forwards, then backwards. End each loop with a newline. Ex: If courseGrades = {7, 9, 11, 10}, print: 7 9 11 10 10 11 9 7 Hint: Use two for loops. Second loop starts with i = courseGrades.length - 1. import java.util.Scanner; public class CourseGradePrinter {public static void main (String [ ] args) {Scanner scnr = new Scanner(System.in);final int NUM_VALS = 4;int [ ] courseGrades = new int[NUM_VALS];int i; for (i = 0; i < courseGrades.length; ++i) {courseGrades[i] = scnr.nextInt();} /* Your solution goes here */ }}Write a program that asks user to enter number of vertices in an undirected graph and then the adjacency matrix representing the undirected graph. The program, then, must display whether the given graph is connected or not. Sample 1Enter number of vertices: 3Enter adjacency matrix:0 1 11 0 01 0 0The graph is connected.
- Given positive integer numInsects, write a while loop that prints, then doubles, numInsects each iteration. Print values = 1Rewrite as a while loop SUIT = 1 BLACK = 0RED = 1 SUIT_COLOURS = {'S': BLACK,'H': RED,'D': RED,'C': BLACK} def same_colour(cards):for i in range(len(cards)):if i == 0:colour = SUIT_COLOURS[cards[i][SUIT]]elif SUIT_COLOURS[cards[i][SUIT]] != colour:return Falsereturn TrueInstrument FrequencyCounter to use Stopwatch and StdDraw to make a plot where the x-axis is the number of calls on get() or put() and the y-axis is the total running time, with a point plotted of the cumulative time after each call. Run your program for Tale of Two Cities using SequentialSearchST and again using BinarySearchST and discuss the results. Note : Sharp jumps in the curve may be explained by caching, which is beyond the scope of this question.
- Write a for loop to print all elements in courseGrades, following each element with a space (including the last). Print forwards, then backwards. End each loop with a newline. Ex: If courseGrades = {7, 9, 11, 10), print: 7 9 11 10 10 11 9 7 Hint: Use two for loops. Second loop starts with i = NUM_VALS - 1. Note: These activities may test code with different test values. This activity will perform two tests, both with a 4-element array (int courseGrades[4]). Also note: If the submitted code tries to access an invalid array element, such as courseGrades[9] for a 4-element array, the test may generate strange results. Or the test may crash and report "Program end never reached", in which case the system doesn't print the test case that caused the reported message. 1 #include HNm in 10 m DO G 2 3 int main(void) { 4 5 6 const int NUM_VALS = 4; int courseGrades [NUM_VALS]; int i; 7 8 for (i = 0; i < NUM_VALS; ++i) { scanf("%d", &(courseGrades[i])); 9 } 10 11 12 13 14 15 } /*Your solution…CodeWorkout Gym Course Search exercises... Q Search kola shreya@columbusstate.edu X176: count7 Given a non-negative int n, return the count of the occurrences of 7 as a digit, so for example 717 yields 2. (no loops). Note that mod (%) by 10 yields the rightmost digit (126 % 10 is 6), while divide () by 10 removes the rightmost digit (126/10 is 12). Your Answer: 1 public int count7(int n) 4} Check my answer! Reset Next exercise Feedback CodeWorkout © Virginia Tech About License Privacy Contact an 123 45In python. A program that analyzes a set of numbers can be very useful. Create an Analysis application thatprompts the user for numbers in the range 1 through 50, terminated by a sentinel, and then performsthe following analysis on the numbers:• Determine the average number• Determine the maximum number• Determine the range (maximum – minimum)• Determine the median (the number that occurs the most often)• Displays a bar graph called a histogram that shows the numbers in each five-unitrange (1–5, 6–10, 11-15, etc.). The histogram may look similar to:
- modify this code to find the required output; graph={ 1: [2,3], 2: [1,3], 3: [1,2,5, 4], 4: [6,11,3,5], 5: [7,4,3], 6: [8,4], 7: [5,9], 8: [6], 9: [7,10], 10: [9,12], 11: [12,4], 12: [10,11] } #perform depth-first search def dfs(graph, start, visited, charge): #current node as visited visited.add(start) #Check the Roomba needs to enter the charging station if charge <= 6: print(f" {start} (Charging, {charge} Wh) -> ") charge =12 else: print(f"{start} (Cleaning, {charge} Wh) ->") charge -= 1 #Take he list of adjacent nodes adj_nodes = graph[start] #Visit the unvisited adjacent nodes for node in adj_nodes: if node not in visited: dfs(graph, node, visited, charge) #user to enter the starting node start_node = int(input ("Enter starting node: ")) # Perform depth-first search starting from the starting node visited = set() dfs(graph, start_node, visited, 12) # Sample outputEnter starting node: 11 (Cleaning, 12 Wh) ->2…Write a for loop to print all elements in courseGrades, following each element with a space (including the last). Print forwards, then backwards. End each loop with a newline. Ex: If courseGrades = {7, 9, 11, 10}, print:7 9 11 10 10 11 9 7 Hint: Use two for loops. Second loop starts with i = NUM_VALS - 1. Note: These activities may test code with different test values. This activity will perform two tests, both with a 4-element array (int courseGrades[4]). Also note: If the submitted code tries to access an invalid array element, such as courseGrades[9] for a 4-element array, the test may generate strange results. Or the test may crash and report "Program end never reached", in which case the system doesn't print the test case that caused the reported message. #include <iostream>using namespace std; int main() { const int NUM_VALS = 4; int courseGrades[NUM_VALS]; int i; for (i = 0; i < NUM_VALS; ++i) { cin >> courseGrades[i]; } /* Your solution goes…Volume of Histogram: Imagine a histogram (bar graph). Design an algorithm to compute the volume of water it could hold if someone poured water across the top. You can assume that each histogram bar has width 1. EXAMPLE Input: {0, 0, 4, 0, 0, 6, 0, 0, 3, 0, 5, 0, 1, 0, 0, 0} (Black bars are the histogram. Gray is water.) Output: 26 0 0 4 0 0 6 0 0 3 0 5 0 1 0 0 0