?? QUESTION 6Which of the following will correctly declare, construct and initialize a tow dimensional array of integers?O a. int x[]={{};b. int x[][]={1,2};c. int x[][]={{1,2,3},{4,5,6}};O d. int xO[]= new int {123,123};QUESTION 7

Answered: Which of the following will correctly…

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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1) Consider an integer array a of length n with indexing starting at 0, where n is a positive integer.If the elements of array a are to be written out in reverse order, which of the following C++ code fragment does NOT do the job? Question options: a. int i=n-1; while (i>=1){cout << a[i] << endl; i = i-1;} cout << a[i] << endl; b. int i=n-1; while (i>=1){cout << a[i] << endl; i = i-1;} c. int i=n-1; while (i>=0){cout << a[i] << endl; i = i-1;} d. int i=n; while (i>0){cout << a[i-1] << endl; i = i-1;} 2) Assume we use 8-bit cell to store floating point numbers, 1 bit for sign, 3 bits for excessed exponent, and 4 bits for significand. What is the decimal value for a cell with bit pattern 0 111 1101 Question options: a. 125 b. 224…
5. Implement the binary search function below that returns the index of the target value in an integer array. int search(int a[], int t, int l, int r); Given the binary search function in question 5 and the array a below, list all activations when calling search(a, 19, 0 15). -9 -5 -2 0 1 3 7 11 17 19 21 25 27 31 37 41 a index 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
can you solve this please ? Given an array of unsorted integers, you are requested to investigate whether there exist a pair of numbers in this array that has a sum equal to a given key or not. For example, if the arr = [8, 7, 2, 5, 3] and the key = 12, the answer will be "Yes" as the array has a pair (7 and 5) where their sum = 12 = key, otherwise "No" is returned. To solve this problem, the following idea can be used: i. Consider every pair in the given array and check if their sum = key. ii. Sort the array first, keeping two pointers to on the minimum and maximum of the array and then move from both directions toward the center of the array while checking whether the sum = key or not. Answer the following questions: a) Transform each of the above ideas (i) & (ii) into algorithm (pseudocode) and compute their complexities b) Critic these two ideas: [Which is better? Why?] c) Design a better solution with less complexity. [note: your idea should be…
ALGO2(A)// A is an integer array, index starting at 11:for i=1 to n-1 do2: for j=n to i+1 do3: if (A[j] < A[j-1]) then4: 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 constantsOption 2) a*n + b , where a and b are constantsOption 3) a * n * log(n) + b, where a,b are constantsOption 4) a*(n^3) + b*(n^2) + c*n + d, where a,b,c,d are constants
A) Starting with a dynamic array of length = 32 and numElements = 30, the length after we execute 90 insert at end operations is [answer]Group of answer choices 120 122 128 none of the above B) In each of the following cases, we start with a dynamic array of length = 128 and numElements = 64. After we execute 30 delete last operations, the number of elements isand the length isAfter we execute 32 delete last operations, the number of elements isand the length isAfter we execute 60 delete last operations, the number of elements isand the length is no hand written
I have this problem: You are given an array called source, with length n, and a set of m arrays called target, each alsowith length n. The arrays contain only positive integers. The arrays are unsorted. For any numberx, the smallest number in target[x] is guaranteed to be smaller than target[x+1]. All the numbers intarget are unique. The following figure illustrates an example of valid source and target arrays forthis case, where n=5 and m=6.source = [14, 13, 15, 11, 12] target[0] = [4, 2, 1, 3, 5]target[1] = [6, 8, 7, 10, 9]target[2] = [14, 15, 12, 11, 13]target[3] = [17, 20, 16, 18, 19]target[4] = [22, 21, 24, 23, 25]target[5] = [30, 29, 28, 27, 26]The smallest number in each array in target is styled with bold and italic. You can quickly see thatthe smallest number in target[0] is smaller than the smallest number in target[1], the smallestnumber in target[1] is smaller than the smallest number in target[2], and so on. You can also seethat there is no duplicate number within the…
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:.
3. Consider an 1-d array of type int and a pointer pointing variable pointing to the 2nd item of the array. Write down the code for the following: int arr(10) = (5,6,23,4,89,70,12,78,33); int *ptr = arr+1; (a) Print the first item of the array using ptr. (b) Print all prime numbers of the array using value at address. (c) Print the last item of the array using a pointer to an array. You are not allowed to initialize the pointer to the address of the last item. Iterate using a loop to change the value of the pointer. (d) Now print the array in reverse order using a pointer. 101
Suppose we implement the += operator as shown in the following. What goes wrong with b += b ? void bag::operator +=(const bag& addend) // Library facilities used: cassert{ size_type i; // An array index assert(size( ) + addend.size( ) <= CAPACITY); for (i = 0; i < addend.used; ++i) { data[used] = addend.data[i]; ++used; }} Group of answer choices If we activate b += b, then the private member variable size is the same variable as addend.size. The size( ) + addend.size( ) is less than CAPACITY. So the program after the assert statement will be executed. If we activate b += b, then the private member variable size is the same variable as addend.size. The size( ) + addend.size( ) is larger than CAPACITY. So the program after the assert statement will not have chance to be executed. If we activate b += b, then the private member variable used is the same variable as addend.used. Each iteration of the loop adds 1 to used, and hence…
Suppose you have given a circular array of integers where start and size is given. You need to find the second largest even number and second smallest even number of the array. After that, replace the next index value of the second largest even value index with summation of the second largest even value and second smallest odd value, and previous index value of the second smallest even value index with the multiplication of the second largest odd and second smallest even value. Note: You must change the start and size accordingly. Assume that O represents empty spaces in the array. [You are not allowed to use any additional array/list or any built-in functions except lenO, range0].
Implement the function below. void swap(int pos1, int pos2){} Initial code to be completed: class ArrayList : public List { int* array; int index; int capacity; void dyn_all_add(){ int cap = ceil(capacity * 1.5); array = (int*)realloc(array,cap * sizeof(int)); capacity = cap; } void dyn_all_rem(){ int cap = capacity - (capacity/3); array = (int*)realloc(array,cap * sizeof(int)); capacity = cap; } public: // CONSTRUCTOR ArrayList() { capacity = 4; array = (int*)malloc(capacity); index = 0; } int add(int num) { if (index == capacity){ dyn_all_add(); } *(array + index) = num; index++; return index; } int get(int pos){ if (pos-1 < index){ return *(array + pos-1); } return -1; } int size(){ return index; }…
please help me public static void main(String args[]){ int[] x={1,4,88,9,13,77,4,18}; for(int i=0;i<x.length;i++){ for( int j=i+1;j<x.length;j++){ if(x[i]==x[j]){ System.out.println("duplicates exists is : "+x[i]); }}} System.out.println("no duplication"); } The above segment of code is searching the duplicated values inside an array, where the complexity is O(n). Suggest another solution for the above problem with less time complexity.

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Which of the following will correctly declare, construct and initialize a tow dimension O a. int x[]={{}}; b. int x[0]={1,2}; O c. int x[][]={{1,2,3},{4,5,6}}; O d.int x0]= new int {123,123};