Question 2Choose the correct answer after considering the following declarations:map m;unordered_map um;O None of these answers.O m.find("X") runs in the same Big-Oh time as find(m.begin(),m.end(), "X")O m.find("X") runs in O(1)O um.find("X") runs in O(n log n)O find(m.begin(),m.end(), "X") runs in O(n log n)

Explanation: C++ Source Code: map: #include <iostream> #include <map> using namespace…

Choose the correct answer after considering the following declarations: map m; unordered_map um; None of these answers. O m.find("X") runs in the same Big-Oh time as find(m.begin().m.end(), "X") O m.find("X") runs in O(1) um.find("X") runs in O(n log n) find(m.begin(),m.end(), "X") runs in O(n log n)

Choose the correct answer after considering the following declarations: map m; unordered_map um; None of these answers. O m.find("X") runs in the same Big-Oh time as find(m.begin().m.end(), "X") O m.find("X") runs in O(1) um.find("X") runs in O(n log n) find(m.begin(),m.end(), "X") runs in O(n log n)

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

See similar textbooks

Similar questions

In this problem you will implement a function called triangle_countwhich will take as input a graph object G, representing an undirectedgraph G, and will return the number of triangles in G. Do not use anyimports for this problem. To simplify the problem you may assume thateach edge is stored twice in G. That is if an edge goes from u to v then vwill be in u’s collection and u will be in v’s collection. If you would prefer,you may assume that an edge is only stored once. In either case G willneed to be regarded as undirected.
import java.util.HashMap; import java.util.Map; public class LinearSearchMap { // Define a method that takes in a map and a target value as parameters public static boolean linearSearch(Map<String, Integer> map, int target) { // Iterate through each entry in the map for () { // Check if the value of the current entry is equal to the target if () { // If the value is equal to the target, return true } } // If no entry with the target value is found, return false } public static void main(String[] args) { // Create a HashMap of strings and integers Map<String, Integer> numbers = new HashMap<>(); // Populate the HashMap with key-value pairs numbers.put("One", 1); numbers.put("Two", 2); numbers.put("Three", 3); numbers.put("Four", 4); numbers.put("Five", 5); // Set the target value to search for…
Implementation of largest-cover difference #23@title Implementation of largest-cover difference def rectangle_difference(r, t): "Computes the rectangle difference rt, and outputs the result as a list of rectangles." assert len(r) == len(t), "Rectangles have different dimensions" ### YOUR SOLUTION HERE In the above code, we notice how it paid off to encapsulate intervals in their own abstraction. Now that we need to reason about rectangles, we do not need to get bogged down into complicated considerations of how to perform difference; the comparisons between endpoints are all done in the context of intervals, where it is easier to reason about them. It is often the case that problems become easy, once you think at them in the appropriate context. It is much easier to develop the code for the difference of intervals, as we have done, and then move on to difference of rectangles, rather than trying to write code for rectangle difference directly. The result is also far more elegant. # Let…
Given the following sets: U= {1,2,3,4,5,6,7,8,}, A={1,4,5,7}, B= {2,5,6,7,}, and C= {3,4,6,7} Complete the following set operations: a. A U(BUC) b. (A N (B N C))' . c. (A N B) U ( A N C) d. (A N B')U (A N C')
4. Show the output of the following code: public class Test { public static void main(String[] args) { Map map = new LinkedHashMap(); map.put("123", "John Smith"); map.put ("111", "George Smith"); map.put("123", "Steve Yao"); map.put("222", "Steve Yao"); System.out:println("(1) System.out.println(" (2) } uutr wtni " + map); + new TreeMap (map)) ; }
Let A={1,2,3). B={3,4,5,6}, and C={6,7,8}.Find (AUB)ᑎC.Write your answer using set notation, i.e. list all elements separated by comma inside braces{} or use the symbol of empty set, if needed Ø. Do not insert any spaces
Write the function calcWordFrequencies() in script.js that uses the JavaScript prompt() function to read a list of words (separated by spaces). The function should output those words and their frequencies to the console. Ex: If the prompt input is: hey hi Mark hi mark the console output is: hey 1 hi 2 Mark 1 mark 1 Hint: Place unique words in a map, where the key is the word, and the associated value is the word's frequency. To test the JavaScript in your web browser, call the calcWordFrequencies() function from the JavaScript console. ------ <!DOCTYPE html><html> <script src="script.js"></script> <body> <p>To test your function, call calcWordFrequencies() from the JavaScript console in the browser.</p> </body></html> -------------------------------- function calcWordFrequencies() { let wordList.split(' ')=[ ]; wordList.push(prompt("words?")) result = { }; for(let i = 0; i <…
In AList class, we have the get function to return an item at a particular position in list indexed by i. For example, let L=[1,2,3]. A call to L.get (1) will return 2. Write in the answer an enhanced get function to accept negative indices. Specifically, the last item has index -1, the second last item has index -2, and the i-th last item has index -i. For example, let L=[1,2,31. A call to L.get (-1) will return 3. A call to L.get (-2) will return 2. A call to L.get (-3) will return 1. Below is a skeleton of the AList class: template class AList { private: /** The underlying array. */ ItemType *items; /** Stores the current size of the list. */ int count; /** Max number of items allowed. */ int maxCnt; public: // Other functions omitted... ItemType get (int i) { // Copy this function in the answer and write code below this line. } ;
True or false: a) We can use memory maps with LinkedLists. b) If all of the JUnit tests pass, the program is guaranteed to always work.
Consider the following function:int enigma(NodeInt32* node){if(node == NULL)return 0; elsereturn 1 + enigma(node->next);} Which best describes the enigma function?A. The function performs a binary search of a list. B. The function computes the length of a list. C. The function sorts a list. D. None of the above.
Can someone help me with this code, I not really understanding how to do this? import java.util.ArrayList;import java.util.HashMap;import java.util.Map;/*** @version Spring 2019* @author Kyle*/public class MapProblems {/*** Modify and return the given map as follows: if the key "a" has a value, set the key "b" to* have that value, and set the key "a" to have the value "". Basically "b" is confiscating the* value and replacing it with the empty string.** @param map to be edited* @return map*/public Map<String, String> confiscate(Map<String, String> map) {if (map.containsKey("a") && map.containsValue("a")) {}}/*** Modify and return the given map as follows: if the key "duck" has a value, set the key* "goose" to have that same value. In all cases remove the key "swan", the rest of the map* should not change.** @param map to be edited* @return map*/public Map<String, String> mapBird1(Map<String, String> map) {throw new UnsupportedOperationException("Not…
Description: Given a string, find the first non-repeating character in it and return its index. If it doesn't exist, return -1. Example 1 input: "leetcode" output: 0 Example 2 input: "loveleetcode" output: 2 The solution we are targeting here is a linear time solution since we have to go over the entire string to know which one is unique in it. A straightforward method we can easily think of is having a Dictionary record letter frequency. Subsequently, iterate over the input again and return the index that has a letter frequency of 1.