Find the Big-Oh of the following. Explain your answer. 1- 1 for (int a : arra) ( 2 print (a); 5 for (int b i arre) { 6 print (b); 2- 1 for (int a ! arra) ( for (int bi arrs) print(a ,"+ b); 4 } 5} 2 3- 1 boolean isPrine(int n) { for (int x - 2; x*x - n; x++) { 3 2 if (n X X -- e) { return false; 7 return true;
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- Write a map-reduce algorithm to come up with a reliable machine learning model that can predict the likelihood of a bank credit card user “churning” – that is, no longer being a paying customer – from a bank. Show steps of your algorithm on sample input. Write a map-reduce program in java to implement the above algorithm. Show the screenshots of the steps of compilation and output.Implement two methods (using iterative and recursive approaches) to compute the sum of the reciprocals of the first n natural numbers. Mathematically, it can be denoted as follows: Sum(n) = 1 + 1/2 + 1/3+ ... +1/n Example 1: Input: [1] Output: [1.0] Explanation: n == 1, so sum(1)=1.0 Example 2: Input: [2] Output: [1.5] Explanation: n == 2, so sum(2)=1+1/2=1.5 Iterative Approach: public static float sum_iterative (int n) { } Recursive Approach: Hint for recursive implementation: The prototype for the recursive approach should be as follows: Sum(n) ={1.0 ifn== 1 Sum(n−1) + 1/n if n >1 public static float sum_recursive (int n) { }Correct answer will be upvoted else downvoted. Computer science. stage is a succession of n integers from 1 to n, in which every one of the numbers happen precisely once. For instance, [1], [3,5,2,1,4], [1,3,2] are stages, and [2,3,2], [4,3,1], [0] are not. Polycarp was given four integers n, l, r (1≤l≤r≤n) and s (1≤s≤n(n+1)2) and requested to find a stage p of numbers from 1 to n that fulfills the accompanying condition: s=pl+pl+1+… +pr. For instance, for n=5, l=3, r=5, and s=8, the accompanying stages are reasonable (not all choices are recorded): p=[3,4,5,2,1]; p=[5,2,4,3,1]; p=[5,2,1,3,4]. However, for instance, there is no change reasonable for the condition above for n=4, l=1, r=1, and s=5. Help Polycarp, for the given n, l, r, and s, find a stage of numbers from 1 to n that fits the condition above. In case there are a few appropriate changes, print any of them. Input The primary line contains a solitary integer t (1≤t≤500). Then, at that point,…
- JAVA Write a static recursive method evenFactors that takes as input two positive integers and prints the even factors of the first integer that are greater than or equal to the second integer. For example, evenFactors(18,1) prints 2, 6, and 18, since the even factors of 18 that are greater than or equal to 1 are 2,6, and 18.Show the derivation of the asymptotic run time, O(n), for each of the following methods. You may use the text highlighting technique shown in class, or list the individual Java statements along with the number of times each is executed as a function of the input size, n. public static void checkerboard(int n){ for (int row = 0; row < n; row++) { for (int col = 0; col < n; col++){ if ((row+col)%2 == 0) { // Print a single hashtag System.out.print("#"); } else { // Print a single space System.out.print(" "); } } // Complete the line System.out.println(); } }Rewrite these Jave methods to recursion methods (no for looops) public static void rotateL(int[] arr, int length, int num) {for (int i = 0; i < num; i++) {int j, first;first = arr[0];for (j = 0; j < arr.length - 1; j++) {arr[j] = arr[j + 1];}arr[j] = first;}} static boolean checkAscSort(int a[], boolean flag, int i, int n) {for (i = 0; i < n - 1; i++) {if (a[i] > a[i + 1]) {flag = false;break;}}return flag;} static boolean checkDesSort(int a[], boolean flag, int i, int n) {for (i = 0; i < n - 1; i++) {if (a[i] < a[i + 1]) {flag = false;break;}}return flag;} static boolean isEqual(int a[], boolean flag, int i) {if (i == 0) {for (i = 0; i < a.length - 1; i++) {if (a[i] != a[i + 1]) {flag = false;}}return flag;} else {for (int k : a) {if (k % 2 != 0) {return false;}}return true;}
- It is possible to represent human concepts (such as mathematics) in a programming language, so that we can get machines to obey specific rules. Because of this, a parser must convert OUR text and symbols into tokens. How come the machine doesn't simply do what we tell it to?The correct statements are: Group of answer choices In terms of computability, DFSM = NDFSM < DPDA < NDPDA < DTM = NDTM. In terms of computability, DFSM = NDFSM < DPDA = NDPDA < DTM = NDTM. In terms of computability, DFSM < NDFSM < DPDA < NDPDA < DTM < NDTM.Write a program that will contain the following methods: printSub1, printSub2, printSub3, printSub4 – print all substrings of S using recursion. For example, if S = "abcd": printSub1 will print "a", "ab", "abc", "abcd", "b", "bc", "bcd", "c", "cd", "d" printSub2 will print "d", "cd", "bcd", "abcd", "c", "bc", "abc", "b", "ab", "a" printSub3 will print "a", "b", "c", "d", "ab", "bc", "cd", "abc", "bcd", "abcd" printSub4 will print "abcd", "bcd", "abc", "cd", "bc", "ab", "d", "c", "b", "a" Note that the actual output will not have quotation marks around the substrings. The substrings must print in the order shown for each method. You may only use these String methods in your work: length(), substring() Your main method in this program should provide an appropriate menu which allows the user the enter the string of their choice and pick any of the print substring options they want. write the methods recursively and in java
- Write a program that will contain the following methods: printSub1, printSub2, printSub3, printSub4 – print all substrings of S using recursion. For example, if S = "abcd": printSub1 will print "a", "ab", "abc", "abcd", "b", "bc", "bcd", "c", "cd", "d" printSub2 will print "d", "cd", "bcd", "abcd", "c", "bc", "abc", "b", "ab", "a" printSub3 will print "a", "b", "c", "d", "ab", "bc", "cd", "abc", "bcd", "abcd" printSub4 will print "abcd", "bcd", "abc", "cd", "bc", "ab", "d", "c", "b", "a" Note that the actual output will not have quotation marks around the substrings. The substrings must print in the order shown for each method. You may only use these String methods in your work: length(), substring() Your main method in this program should provide an appropriate menu which allows the user the enter the string of their choice and pick any of the print substring options they want. IN JAVAGiven is Parser.java. Methods must accept tokens appropriately and generates OperationNode or returns partial result. I need the methods to be created for the following: -Post Increment/Decrement -Exponents -Factor -Term -Expression -Concatenation -Boolean Compare -Match -Array Membership -And -Or -Ternary -Assignment Parser.java import java.text.ParseException; import java.util.LinkedList; import java.util.List; import java.util.Optional; import javax.swing.ActionMap; public class Parser { private TokenHandler tokenHandler; private LinkedList<Token> tokens; public Parser(LinkedList<Token> tokens) { this.tokenHandler = new TokenHandler(tokens); this.tokens = tokens; } public boolean AcceptSeparators() { boolean foundSeparator = false; while (tokenHandler.MoreTokens()) { Optional<Token> currentToken = tokenHandler.getCurrentToken(); if (currentToken.isPresent() &&…case #5 int i , p =1; for(i=1;i<=n;i++) p = p*i case #6 int rec(int r) { if(r==0) return 1; else return r*(rec(r-1); } Compare the two methods (#5 and #6 above) and answer the following: What is the “stopping” case for each (what causes the methods to “end”)? How do you guarantee that the methods will “stop” (infinite loop, infinite recursion)? Which method one is “better”? Why?