You are given an array A[1, ..., n]. Describe an algorithm (proper/efficient pseudocode required) that checks whether the given array is a Heap, and outputs 'yes/no'.

Java: Consider the following algorithm for searching in an unsorted array. If the size of the array is 1, then check if it contains the element to be searched. Otherwise, divide the array into two halves, and  recursively search both halves. Which of (a)–(c) is false? The running time of this algorithm is O(N) The actual running time of this algorithm is likely to be better than sequential search. This is an example of a divide‐and‐conquer algorithm all of the above are true none of the above is true

Let A = [n, n − 1, n − 2, . . . , 3, 2, 1] be an array where the first n positive integers are listed in decreasing order. Determine whether Heapsort or Quicksort sorts this array faster.For this question, assume the Quicksort pivot is always the right-most element.

PYTHON: Please provide a screenshot of your code in Python. Thank you!

15. Consider the problem of finding the first position in which an array b occurs as a subsequence of an array a. Write two nested loops: let result undefined for (let i = 0; i < a.length - b.length; i++) { for (let j = 0; j < b.length; j++) { if (a[i+j] = b[j]) . . . } } Complete with labeled break and continue statements.

Java Quick Sort but make it read the data 10, 7, 8, 9, 1, 5 from a file not an array // Java implementation of QuickSort import java.io.*;    class GFG {        // A utility function to swap two elements     static void swap(int[] arr, int i, int j)     {         int temp = arr[i];         arr[i] = arr[j];         arr[j] = temp;     }        /* This function takes last element as pivot, places        the pivot element at its correct position in sorted        array, and places all smaller (smaller than pivot)        to left of pivot and all greater elements to right        of pivot */     static int partition(int[] arr, int low, int high)     {            // pivot         int pivot = arr[high];            // Index of smaller element and         // indicates the right position         // of pivot found so far         int i = (low - 1);            for (int j = low; j <= high - 1; j++) {                // If current element is smaller             // than the pivot             if…