ALGO2(A)// A is an integer array, index starting at 1 1:for i=1 to n-1 do 2: for j=n to i+1 do 3: if (A[j] < A[j-1]) then 4: exchange A[j] with A[j-1] Which function best describes the worst-case running time behaviour of ALGO2 on an array of size n?
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ALGO2(A)// A is an integer array, index starting at 1
1:for i=1 to n-1 do
2: for j=n to i+1 do
3: if (A[j] < A[j-1]) then
4: exchange A[j] with A[j-1]
Which function best describes the worst-case running time behaviour of ALGO2 on an array of size n?
Option 1) a*(n^2) + b*n +c, where a,b,c are constants
Option 2) a*n + b , where a and b are constants
Option 3) a * n * log(n) + b, where a,b are constants
Option 4) a*(n^3) + b*(n^2) + c*n + d, where a,b,c,d are constants
Step by step
Solved in 2 steps
- 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 :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:.(1) The following code transposes the elements of an M × M array, where M is a con- stant defined by # define: void transpose(long A[M][M]) { long i, j; for (i = 0; i < M; i++)for (j = 0; j < i; j++) { long t = A[i][j]; A[i][j] = A[j][i]; A[j][i] = t; } } When compiled, with optimization level −O1, GCC generates the following code for the inner loop of the function. 1 .L1: 234 5 6 7 8 9 movq movq movq movq addq addq cmpq jne .L1 (%rdx), %rcx (%rax), %rsi (%rdx) (%rax) $8, %rdx $120, %rax %rdi, %rax %rsi, %rcx, We can see that GCC has converted the array indexing to pointer code. (a) Which register holds a pointer to array element A[i][j]? (b) Which register holds a pointer to array element A[j][i]? (c) What is the value of M?
- 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 IIIJava Insertion Sort but make it read the data 12, 11, 13, 5, 6 from a file not an array // Java program for implementation of Insertion Sort public class InsertionSort { /*Function to sort array using insertion sort*/ void sort(int arr[]) { int n = arr.length; for (int i = 1; i < n; ++i) { int key = arr[i]; int j = i - 1; /* Move elements of arr[0..i-1], that are greater than key, to one position ahead of their current position */ while (j >= 0 && arr[j] > key) { arr[j + 1] = arr[j]; j = j - 1; } arr[j + 1] = key; } } /* A utility function to print array of size n*/ static void printArray(int arr[]) { int n = arr.length; for (int i = 0; i < n; ++i) System.out.print(arr[i] + " "); System.out.println(); } // Driver method…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.
- 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.Magic Index: A magic index in an array A[ 1 .•. n-1] is defined to be an index such that A[ i] i. Given a sorted array of distinct integers, write a method to find a magic index, if one exists, in array A.FOLLOW UPWhat if the values are not distinct?In a non-empty array of integers, the priority of a number is determined by the frequency of its occurrence. Elements are added as they come but the most frequent element is deleted first. If two elements have the same priority, then the element which came first will be deleted first. For example: nums = [1,2,3,1,3,1,3,2,3]. After the first deletion now the array will be num=[2,3,3,3,2,3]. Implement a priority queue for the aforementioned scenario. give the answer in brief and the correct one.
- Plot the result of (given) selectionsort(A) when A is a perfectly sorted and contains randoms integers ,0 to 100 inclusive, for array sizes n = 2,4,8,16,..1024. please provide code for A and generating the plot for each doubling array, as well as the resulting plot. def selectionsort(A): for i in range(len(A)): min_idx = i for j in range(i + 1, len(A)): if A[min_idx] > A[j]: min_idx = j A[i], A[min_idx] = A[min_idx], A[i] return AFollowing is the function for interpolation search. This searching algorithm estimates the position (index) of a key in array based on the elements in the first position and last position in the array, and the length of array. The array must be sorted in ascending order. Suppose array A contains the following 15 elements: A = [1, 3, 3, 10, 17, 22, 22, 22, 24, 25, 26, 27, 27, 28, 28] At first iteration, at which position (index) the element of 24 is estimated in array A? In which part of array (starting index and ending index) the searching should continue? How many iterations the searching are performed until the element of 24 is found? int InterpolationSearch(int x[], int key, int n) { int mid, min = 0, max = n-1; while(x[min] < key && x[max] > key) { mid = min + ((key-x[min])*(max-min)) / (x[max]-x[min]); if(x[mid] < key) min = mid + 1; else if(x[mid] > key) max = mid - 1; else return mid; } if…// 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']…