Consider the insertion sort algorithm, which sorts an array `A` of length `N` in ascending order. Below is a partial implementation of the insertion sort algorithm in Python: python function insertion_sort(A[O...N-1]) if N > 1: for i in range(1, N): index = i for j in range(i - 1, -1, -1): if A[index] < A[j]: A[index], A[j] = A[j], A[index] index = j else: break Copy code a. Complete the implementation of the 'insertion_sort function in Python, ensuring that it correctly sorts the input array `A` in ascending order. b. Explain the purpose of each loop and conditional statement within the insertion sort algorithm. c. Test your implementation with different input arrays and verify that it produces the expected output. d. Analyze the time complexity of your 'insertion_sort' implementation. Provide a step-by-step explanation of how you determined the time cplexity and discuss its implications.

Implement the algorithm in C++. Algorithm: substring (s, p, len)   Initialize i = p and count =0 Use an array „dest‟ to hold the required substring Repeat steps 4 ,5 and 6 while count <len dest[count] =s[i] count = count+1 i = i +1 Insert string terminator at end of dest Exit

Given an array arr] and an integer K where K is smaller than size of array, the task is to find the Kth smallest element in the given array. It is given that all array elements are distinct. Note :- I and r denotes the starting and ending index of the array. Example 1: Input: N = 6 arr[] = 7 10 4 3 20 15 K = 3 Output: 7

Implement a function that returns an array containing all the consecutive numbers in ascendant order from the given value lowup to the given value high (bounds included).  Example: getSequence (3, 8) → [3, 4, 5, 6, 7, 8] WRITE IN PYTHON PLEASE

For a recursive binary search, put the following code in the correct order by dragging each line to the correct place. 1 2 A 5 60 7 8 ⠀ :::: if (right >= left) { if (array [mid] == item) return mid; int binarySearch(int array [], int left, int right, int item) { return binarySearch(array, left, mid - 1, item); if (array [mid] > item) return binary Search(array, mid+ 1, right, item); int mid=1+ (right - left) / 2; return -1; }

Find the complexity of the function used to find the kth smallest integer in an unordered array of integers: int selectkth(int a[], int k, int n) {int i, j, mini, tmp;   for (i = 0; i < k; i++) {   mini = i;      for (j = i+1; j < n; j++)      if (a[j]<a[mini])         mini = j;   tmp = a[i];   a[i] = a[mini];   a[mini] = tmp;   }return a[k-1];}

java ARRAY[] = [50, 11, 33, 21, 40, 50, 40, 40, 21] ARRAY[] = [11, 21, 33, 40, 50]   ATTN : Further, please be reminded that you cannot use library functions to either sort and or perform the de-duplication operation.   solve the problem in two ways (1) Implement the function in such a way that your solution solves the problem with O(n log2(n)) time complexity overall and O(n) space complexity. Here, n is the length of the list of input integers (array). I believe the sorting routine that can be used here is Merge Sort. Please state as code comment which sorting routine you are using, sort the arrray with that algorithm and solve the de-duplication problem thereafter. De-duplication part of the solution in itself must adhere to O(n) time and O(1) space bounds. we will not use any memory used by recursion.   Important: Take the size of the input array from the User.Prompt user to input the integers in the array.PLEASE MAKE SURE TO EXPLAIN THE SORTING ALGORITHM AND DE-DUPLICATION…

01... ""Implementation of the Misra-Gries algorithm.Given a list of items and a value k, it returns the every item in the listthat appears at least n/k times, where n is the length of the array By default, k is set to 2, solving the majority problem. For the majority problem, this algorithm only guarantees that if there isan element that appears more than n/2 times, it will be outputed. If thereis no such element, any arbitrary element is returned by the algorithm.Therefore, we need to iterate through again at the end. But since we have filtredout the suspects, the memory complexity is significantly lower thanit would be to create counter for every element in the list. For example:Input misras_gries([1,4,4,4,5,4,4])Output {'4':5}Input misras_gries([0,0,0,1,1,1,1])Output {'1':4}Input misras_gries([0,0,0,0,1,1,1,2,2],3)Output {'0':4,'1':3}Input misras_gries([0,0,0,1,1,1]Output None"""..

Select the for-loop which iterates through all even index values of an array.A. for(int idx = 0; idx < length; idx++)B. for(int idx = 0; idx < length; idx%2)C. for(int idx = 0; idx < length; idx+2)D. for(int idx = 0; idx < length; idx=idx+2)

Java Selection Sort but make it read the data 64, 25, 12, 22, 11 from a file not an array   // Java program for implementation of Selection Sort class SelectionSort {     void sort(int arr[])     {         int n = arr.length;           // One by one move boundary of unsorted subarray         for (int i = 0; i < n-1; i++)         {             // Find the minimum element in unsorted array             int min_idx = i;             for (int j = i+1; j < n; j++)                 if (arr[j] < arr[min_idx])                     min_idx = j;               // Swap the found minimum element with the first             // element             int temp = arr[min_idx];             arr[min_idx] = arr[i];             arr[i] = temp;         }     }       // Prints the array     void printArray(int arr[])     {         int n = arr.length;         for (int i=0; i<n; ++i)             System.out.print(arr[i]+" ");         System.out.println();     }       // Driver code to test above     public…