PLEASE HELP: Looking to make this code more efficient without losing functionality, just seeing if its possible, must utilize only recursion and one loop import java.util.Scanner; public class Assignment2 { public static void main (String[] args) { System.out.println("Enter the number of items to read"); Scanner s = new Scanner (System.in); int n = s. nextInt(); int[] a = new int[n]; System.out.println("Enter the numbers:"); for(int i = 0;i = 1 && n2>= 1) { if(h1 [n1 - 1] = =h2[n2-1]) { f2 = f 2+1; return findExact(h1, h2, n1 - 1, h2. length,f2); } else {return find Exact (h1, h2, n1, n2 - 1, f2);}} return f2; }} private static int find Mirror(int[] h1, int[] h2, int n1, int n 2, int f1) { if(f1 = = h1. length) { return f1; } else {if(n 1 > = 1 && n2>= 1) { if(h1[n1 - 1] = = h2[n2-n1]) { f 1 = f1 + 1; return find Mirror(h1, h2, n1-1, n2, f1); } else { return f1;}} return f1;}}}
PLEASE HELP: Looking to make this code more efficient without losing functionality, just seeing if its possible, must utilize only recursion and one loop import java.util.Scanner; public class Assignment2 { public static void main (String[] args) { System.out.println("Enter the number of items to read"); Scanner s = new Scanner (System.in); int n = s. nextInt(); int[] a = new int[n]; System.out.println("Enter the numbers:"); for(int i = 0;i = 1 && n2>= 1) { if(h1 [n1 - 1] = =h2[n2-1]) { f2 = f 2+1; return findExact(h1, h2, n1 - 1, h2. length,f2); } else {return find Exact (h1, h2, n1, n2 - 1, f2);}} return f2; }} private static int find Mirror(int[] h1, int[] h2, int n1, int n 2, int f1) { if(f1 = = h1. length) { return f1; } else {if(n 1 > = 1 && n2>= 1) { if(h1[n1 - 1] = = h2[n2-n1]) { f 1 = f1 + 1; return find Mirror(h1, h2, n1-1, n2, f1); } else { return f1;}} return f1;}}}
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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I need answer typing clear urjent no chatgpt used i will give 5 upvotes
![PLEASE HELP: Looking to make this code more efficient
without losing functionality, just seeing if its possible,
must utilize only recursion and one loop import
java.util.Scanner; public class Assignment2 { public
static void main(String[] args) { System.out.println("Enter
the number of items to read"); Scanner s = new Scanner
(System.in); int n = s. nextInt(); int[] a = new int[n];
System.out.println("Enter the numbers:"); for(int i = 0; i
= 1 && n2 > = 1 ) { if(h1[n1 - 1] = = h2[n2 - 1]) { f2 = f
2 + 1; return find Exact(h1, h2, n1 -1, h2. length,f2); } else
{return findExact(h1, h2, n1, n2 - 1, f2); } } return f2;} }
private static int findMirror(int[] h1, int[] h2, int n1, int n
2, int f1) {if(f1 = = h1.length) { return f1; } else { if(n
1 > = 1 && n2> = 1) { if(h1[n1 - 1] = =h2[n2-n1]) { f
1 = f1 + 1; return find Mirror(h1, h2, n1 - 1, n2, f1); } else {
return f1; } } return f1; }}}](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3dc03ec9-3be6-4418-b604-aed31dc0d8ce%2F811e6575-45f5-465f-9099-27e85df0d9b3%2Fg574fbm_processed.jpeg&w=3840&q=75)
Transcribed Image Text:PLEASE HELP: Looking to make this code more efficient
without losing functionality, just seeing if its possible,
must utilize only recursion and one loop import
java.util.Scanner; public class Assignment2 { public
static void main(String[] args) { System.out.println("Enter
the number of items to read"); Scanner s = new Scanner
(System.in); int n = s. nextInt(); int[] a = new int[n];
System.out.println("Enter the numbers:"); for(int i = 0; i
= 1 && n2 > = 1 ) { if(h1[n1 - 1] = = h2[n2 - 1]) { f2 = f
2 + 1; return find Exact(h1, h2, n1 -1, h2. length,f2); } else
{return findExact(h1, h2, n1, n2 - 1, f2); } } return f2;} }
private static int findMirror(int[] h1, int[] h2, int n1, int n
2, int f1) {if(f1 = = h1.length) { return f1; } else { if(n
1 > = 1 && n2> = 1) { if(h1[n1 - 1] = =h2[n2-n1]) { f
1 = f1 + 1; return find Mirror(h1, h2, n1 - 1, n2, f1); } else {
return f1; } } return f1; }}}
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