Write a loop that sets each array element to the sum of itself and the next element, except for the last element which stays the same. Be careful not to index beyond the last element. Ex: Initial scores: 10, 20, 30, 40 Scores after the loop: 30, 50, 70, 40 The first element is 30 or 10 + 20, the second element is 50 or 20 + 30, and the third element is 70 or 30 + 40. The last element remains the same. 324758.2040686.qx3zqy7 1 #include 2 3 int main(void) { const int SCORES_SIZE = 4; int bonusScores[SCORES_SIZE]; int i; 4 6. 7 for (i = 0; i < SCORES_SIZE; ++i) { scanf("%d", &(bonusScores[i])); } 8 10 11 12 /* Your solution goes here */ 13 for (i = 0; i < SCORES_SIZE; ++i) { printf("%d ", bonusScores[i]); } printf("\n"); 14 15 16 17 18
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- given an array of integer values , return true if 6 appears as either the first or last element in the array. firstLast6([1,2,36]) true firstLast([6,1,2,3]) true firstLast([13,6,1,2,3]) falseWrite method findZero, as started below. findZero should return the index of the first element of array A that contains the value zero, starting from position 0 up through position pos. If no element of A from position 0 to position pos contains the value zero, then findZero should return –1. For example: Array (A) Position (pos) Value returned by findZero(A, pos) 1 0 2 5 6 0 -1 1 0 2 5 6 1 1 1 0 2 5 6 2 1 1 0 2 0 6 2 1 1 0 2 0 6 4 1 1 2 3 4 5 4 -1 Complete method findZero below. Assume that it is called only with values that satisfy its assumptions as stated below. // Description: Returns smallest index k // such that (0 <= k <= pos) and (A[k] == 0) // Assumptions: 0 <= pos < A.length public int findZero(int[] A, int pos) {In java, please read the text file as array. Then, replace the null with 0 and calculate the sum of the new students in ABC University.Example of text file: (university,year,total_students,new_students,new_staff,ABC, 2000, 3457, , ,ABC, 2001, 3675,218, ,EFG, 2000, 2389,602, ,EFG, 2001, 2785,396, 65,)
- We can create a numpy array from separate lists by specifying a format for each field: roster = np.zeros(4, dtype={'names': ('name', 'age', 'major', 'gpa'), 'formats': ('U50', 'i4', 'U4', 'f8')}) roster['name'] = ['Alice', 'Bob', 'Carol', 'Dennis'] roster['age'] = [21, 25, 18, 29] roster['major'] = ['CS', 'Math', 'Chem', 'Phys'] roster['gpa'] = [3.8, 3.2, 4.0, 3.5] where U50 is a string of max length 50, i4 is a 4-byte integer, U4 is a string of max length 4, and f8 is an 8-byte floating-point number. Then roster has value array([('Alice', 21, 'CS', 3.8), ('Bob', 25, 'Math', 3.2), ('Carol', 18, 'Chem', 4. ), ('Dennis', 29, 'Phys', 3.5)], dtype=[('name', '<U50'), ('age', '<i4'), ('major', '<U4'), ('gpa', '<f8')]) Then, for example, roster['name'] will give an array of just the student names. Write a program that reads the numpy array roster we just created that has this format: name,age,major,gpa Convert this to a numpy array as shown above. Then use that…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 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 */ P.S. The answer came out wrong the first try 7 9 11 10 10 11 9 7EnterIf an array is given, identify the next bigger element for each element in the array, if one exists. If the element is not accessible, print the element itself. The next bigger element y in the array for an element x is the first element that is greater than x and appears on its right side. The element itself is the next bigger member of the array's rightmost element.For instance, if A = [6 8 4 3 9], the next bigger element list B = [8 9 9 9 9].
- For this exercise, you need to shift all the elements in the array one space to the right (increase the index of the element by 1). The last element will wrap around and become the new first element. For example, given: moveUp({"first", "second", "third", "fourth"}); You would return: {"fourth", "first", "second", "third"} in javaCreate a Numpy List named as marks with six float numbers, showing the final marks of a student in six subjects. Round the marks from marks array from Q-3 a) up to two decimal numbers What if you need to create the same list as Q-3 a), but for five students. Can you still use 1-D array? How many dimensions should be there? Find the highest marks a student has obtained using marks array from Q-3 a) Use seaborn to show marks array from Q-3 a)You are given an int variable k, an int array zipcodeList that has been declared and initialized, an int variable nZips that contains the number of elements in zipcodeList, and a bool variable duplicates. Write some code that assigns true to duplicates if there are two adjacent elements in the array that have the same value, and that assigns false to duplicates otherwise. Use only k, zipcodeList, nZips, and duplicates. duplicates = false; for (k=0; !duplicates && k<nZips-1; k++)if (zipcodeList[k] == zipcodeList[k+1]) duplicates = true;
- Create a Numpy List named as marks with six float numbers, showing the final marks of a student in six subjects. Round the marks from marks array from Q-3 a) upto two decimal numbers What if you need to create the same list as Q-3 a), but for five students. Can you still use 1-D array? How many dimensions should be there? Find the highest marks a student has obtained using marks array from Q-3 a) Use seaborn to show marks array from Q-3 a)# the method take another circular array and returns a linear array containing the common elements between the two circular arrays. def intersection(self, c2): # To Do print("==========Test 10==========") lin_arr9 = [10, 20, 30, 40, 50, None, None, None] c10 = CircularArray(lin_arr9, 5, 5) c10.printFullLinear() # This should print: 40, 50, None, None, None, 10, 20, 30 lin_arr10 = [5, 40, 15, 25, 10, 20, 5, None, None, None, None, None] c11 = CircularArray(lin_arr10, 8, 7) c11.printFullLinear() # This should print: 10, 20, 5, None, None, None, None, None, 5, 40, 15, 25 output = c10.intersection(c11) print(output) # This should print: [10, 20, 40]Write a program that deletes the first element in an array which matches a search key. Advance thanks <3