QWrite the recurrencefindMin (A, left,if (left < right){max1 = findMin (A,left, (right + left) / 2)max2 = findMin (A, (right + left) / 2, right)}elserelation for the following method:right) {return A[left]if (maxl > max2)return max1elsereturn max2

Given the recursive method, findMin(A, left, right) { if(left<right){ max1=findMin(A,…

Answered: QWrite the recurrence relation for the…

Computer Networking: A Top-Down Approach (7th Edition)

7th Edition

ISBN:9780133594140

Author:James Kurose, Keith Ross

Publisher:James Kurose, Keith Ross

Chapter1: Computer Networks And The Internet

Section: Chapter Questions

Problem R1RQ: What is the difference between a host and an end system? List several different types of end...

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The following recursive method called z is created. This method accepts two parameters: A string s, and an integer index The code in the method is: if (index == s.length()) return ""; <------ base case if(index % 2 == 0) return ""+ s.charAt(index) + z(s,index+1); <---- recursive call else return z(s,index+1); <----- recursive call What would be the output with the call: z("javajavajava",0);
Java Programming: Make AST Nodes: IfNode, WhileNode, RepeatNode. All will have BooleanCompare for a condition and a collection of StatementNode. ForNode will have a Node for from and a node for to. This will have to be of type Node because it could be any expression. For example: for a from 1+1 to c-6 Make parsing functions for each. Java won't let you create methods called if(), etc. parseIf() is an example of a way around that; use whatever you like but use good sense in your names. Next let's look at function calls. Each function has a name and a collection of parameters. A parameter can be a VAR variable or something that came from booleanCompare (IntegerNode, VariableReferenceNode, etc). It would be reasonable to make 2 objects - ParameterVariableNode and ParameterExpressionNode. But for this very simple and well-defined case, we can do something simple: ParameterNode has a VariableReferenceNode (for VAR IDENTIFIER) and a Node for the case where the parameter is not a VAR.…
Consider the following function:int enigma(NodeInt32* node){if(node == NULL)return 0; elsereturn 1 + enigma(node->next);}(a)The enigma function is recursive.A. True B. False
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.
First Pass: ( 5 1 4 2 8 ) \to ( 1 5 4 2 8 ), Here, algorithm compares the first two elements, and swaps them. ( 1 5 4 2 8 ) \to ( 1 4 5 2 8 ), Swap since 5 > 4 ( 1 4 5 2 8 ) \to ( 1 4 2 5 8 ), Swap since 5 > 2 ( 1 4 2 5 8 ) \to ( 1 4 2 5 8 ), Now, since these elements are already in order (8 > 5), algorithm does not swap them. Second Pass: ( 1 4 2 5 8 ) \to ( 1 4 2 5 8 ) ( 1 4 2 5 8 ) \to ( 1 2 4 5 8 ), Swap since 4 > 2 ( 1 2 4 5 8 ) \to ( 1 2 4 5 8 ) ( 1 2 4 5 8 ) \to ( 1 2 4 5 8 ) Now, the array is already sorted, but our algorithm does not know if it is completed. The algorithm needs one whole pass without any swap to know it is sorted. Third Pass: ( 1 2 4 5 8 ) \to ( 1 2 4 5 8 ) ( 1 2 4 5 8 ) \to ( 1 2 4 5 8 ) ( 1 2 4 5 8 ) \to ( 1 2 4 5 8 ) ( 1 2 4 5 8 ) \to ( 1 2 4 5 8 ) Finally, the array is sorted, and the algorithm can terminate. Bubble sort is a simple sorting algorithm. It works by repeatedly stepping through the list to be sorted, comparing each pair of…
gcd(5x+7y,2x+3y) = ?
Generate pseudocode for searching for an Employee ID I was given this example but I don't understand what it means an employee ID (123) is hashed into a value of 5. For hDict(5) we store apointer to the first available spot in lValue, which at the time of the insertion was 2. We store a value of123 in lValue(2) and a value of null in lPtr(2) because there are no other elements right now, so weterminate the linked list. Later on another employee ID (135) needs to be inserted. It’s hash value isalso 5. In order to find where to insert it in lValue we need to traverse the linked list to get to the lastelement. The first element we come to is 2. lPtr(2) is null so we know we’re at the end of the linked list.So we get the next empty slot in lValue, which was 3, and assign the value 3 to lPtr(2). We assign thenew Employee ID (135) to next open slot lValue(3) and set lPtr(3) to null because it now becomes theend of the list.

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QWrite the recurrence relation for the following method: findMin (A, left, right) { if (left < right) {max1 = findMin (A, left, (right + left) / 2) max2 = findMin (A, (right + left) / 2, right) } else return A[left] if (max1 > max2)