find code blocks necessary to insert a value into an ordered array given a position (index) to insert.
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Below you will find code blocks necessary to insert a value into an ordered array given a position (index) to insert. Be careful! Some lines are not needed.
![Move the necessary blocks over into the response area:
return false;
array[i]
array[i-1];
: bool insertElement(char array[], int& numElems, int capacity, int position, char value) {
if (position < 0 || position > numElems || numElems ==
сараcity) {
return true;
}
: сарасity++;
: for (int i = position; i <= numElems; i++) {
for (int i = numElems-1; i >= position; i--) {
: bool insertElement(char array[], int numElems, int &capacity, int position, char value) {
if (position > 0 || position <= numElems & & numElems
== capacity) {
array[i+l]
array[i];
%D
}
numElems++;
array[position]
= value;](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F31ac961c-fe39-44eb-a9cc-55b50615d34f%2Fdba22548-7437-4693-98a2-8a54ad6d4007%2F36z92k8_processed.png&w=3840&q=75)
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- // function to generate all subsequences of given arrayfunction generate_subseq(arr): // length of arr n = len(arr) // total subsequences for array of length n m = 2**n // to store all subsequences seqs = [] // running a loop for m times for i in range(1, m): // creating an array of zeros of length n a = [0]*n num = i // to use as an index for 'a' j = n-1 // run this loop till num > 0 while num > 0: if num is odd: a[j] = 1 // divide num by 2 num = num/2 // subtract 1 from 'j' j -= 1 // to store current subsequence seq = [] // iterating for n times for i in range(n): // add ith index value to 'seq' if a[i] == 1: seq.append(arr[i]) // add 'seq' to 'seqs' seqs.append(seq) // return seqs return seqs // given listsS1 = ['B','C','D','A','A','C','D']S2 = ['A','C','D','B','A','C']…Suppose an array sorted in ascending order is rotated at some pivot unknownto you beforehand. (i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).Find the minimum element. The complexity must be O(logN)You may assume no duplicate exists in the array."""def find_min_rotate(array): """ Finds the minimum element in a sorted array that has been rotated. """ low = 0 high = len(array) - 1 while low < high: mid = (low + high) // 2 if array[mid] > array[high]: low = mid + 1 else: high = mid return array[low] def find_min_rotate_recur(array, low, high): """ Finds the minimum element in a sorted array that has been rotated. """ mid = (low + high) // 2 if mid == low:.ALGO1(A)// A is an integer array, an index that starts at 11): for i=1 to n-1 do2): minIndex = findSmallest(A,i)3): exchange A[i] with A[minIndex] The sub-routine find the smallest(A, i) in Line 2, and returns the index of the smallest element in thesub-array A[i:n]Suppose the array below is provided as input to ALGO1 2 5 6 7 3 8 1 4 Fill in the Blanks At the end of the first iteration of the for loop (i.e. with i=1)1) a) the element at index 1 is : b) the element at index 4 is : c) the element at index 7 is :
- 9.). integer array nums and an integer k, return the length of the shortest non-empty subarray of nums with a sum of at least k. If there is no such subarray, return -1. A subarray is a contiguous part of an array. Example 1: Input: nums = [1], k = 1 Output: 1 Example 2: Input: nums = [1,2], k = 4 Output: -1 Example 3: Input: nums = [2,-1,2], k = 3 Output: 3.Write 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:(in java) 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) {array unordered Arr has unsorted integers. SortedArr is an integer array. SortedArr performs which task better than unsortedArr? Use the fastest algorithms. Inserting a new element II Searching for a given element III Calculating the mean of the elements (A) I alone (B) II alone (C) III alone (D) I and II alone (E) I, II, and III
- Write 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) {import java.util.Scanner;import java.util.ArrayList; public class UserIDSorting {// TODO: Write the partitioning algorithm - pick the middle element as the // pivot, compare the values using two index variables l and h (low and high), // initialized to the left and right sides of the current elements being sorted,// and determine if a swap is necessarypublic static int partition(ArrayList<String> userIDs, int i, int k) {} // TODO: Write the quicksort algorithm that recursively sorts the low and // high partitionspublic static void quicksort(ArrayList<String> userIDs, int i, int k) {} public static void main(String[] args) {Scanner scnr = new Scanner(System.in); ArrayList<String> userIDList = new ArrayList<String>(); String userID; userID = scnr.next();while (!userID.equals("-1")) {userIDList.add(userID);userID = scnr.next();}// Initial call to quicksort quicksort(userIDList, 0, userIDList.size() - 1); for (int i = 0; i < userIDList.size(); ++i)…Given an array, find the next greater element for each element in the array, ifavailable. If not available, print the element itself. The next greater element y for anelement x in the array is the first element that is greater than x and occurs on its rightside. The next greater element of the right most element in an array is the elementitself.Example: Given A = [ 6 8 4 3 9] the next greater element listB = [8 9 9 9 9]
- What are the advantages and disadvantages of using an unordered array as opposed to an ordered one?Using JAVA, write a method that swaps two rows of a two-dimensional array. The row indices to be swapped are passed as parameters to the method (Assume row indices are valid indices). Write a test program to demonstrate correct operation of the method.In a non-empty array of integers, the priority of a number is determined by thefrequency of its occurrence. Elements are added as they come but the most frequentelement is deleted first. If two elements have the same priority, then the elementwhich came first will be deleted first. For example: nums = [1,2,3,1,3,1,3,2,3]. Afterthe first deletion now the array will be num={1,2,1,1,2}. Implement a priority queuefor the aforementioned scenario.
