Trace this code show the work Public static int funcQ1(int m, int n) { if (m...
Trace this code
show the work
Public static int funcQ1(int m, int n) {
if (m < n)
return 0;
else return 1 + funQ1(m-n, n); }
• What is the value of funcQ1(6, 3), based on the code above?
• What is the value of funcQ1(9, 7), based on the code above?
• What is the value of funcQ1(-5, 1), based on the code above?
Homework Answers
1) funcQ1(6, 3) = 1 + funcQ1(3, 3) = 1 + 1 + funcQ1(0, 3) = 1 + 1 + 0 = 2 Answer: 2 2) funcQ1(9, 7) = 1 + funcQ1(2, 7) = 1 + 0 = 1 Answer: 1 1) funcQ1(-5, 1) = 0 (because -5 < 1) Answer: 0
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Trace this code
show the work
Public static int funcQ1(int m, int n) {
if (m...
b) Consider the following code. public static int f(int n) if (n == 1) return 0;...
b) Consider the following code. public static int f(int n) if (n == 1) return 0; else if (n % 2 == 0). return g(n/2); else return g(n+1); public static int g(int n) int r = n % 3; if (r == 0) return f(n/3); else if (r == 1) return f(n+2); else return f(2 * n); // (HERE) public static void main(String[] args) { int x = 3; System.out.println(f(x)); (1) (5 points) Draw the call stack as it would...
Use one of these methods for your trace. public static int binarySearchIterative(int[] numbers, int target) {...
Use one of these methods for your trace. public static int binarySearchIterative(int[] numbers, int target) { boolean found = false; int first = 0; int last = numbers.length - 1; while (first <= last && !found) { int mid = (first + last) / 2; if (numbers[mid] == target) { targetLocation = mid; found = true; } else if (numbers[mid] < target) { first = mid + 1; } else { // numbers[mid] > target last = mid - 1;...
(b) Trace the following code. // Precondition: n > 0 // <method description here> public static...
(b) Trace the following code. // Precondition: n > 0 // <method description here> public static int dothing (int n) { int ans = ; for (int x 1; x <= n; x++) ( for (int y = x; y <= n; y++) ( ans+ } return ans; } (ii) If you were asked to write a one-line description above this method, what would it be? (iii) Could you rewrite the method so that it runs more efficiently?
Consider the following method. public static ArrayList<Integer mystery(int n) ArrayList<Integer seg - new ArrayList<IntegerO; for (int...
Consider the following method. public static ArrayList<Integer mystery(int n) ArrayList<Integer seg - new ArrayList<IntegerO; for (int k = n; k > 0; k--) seq.add(new Integer(k+3)); return seq What is the final output of the following Java statement? System.out.println(mystery(3)); a. [1,2,4,5) b. [2,3,4,5) c. [6,5,4,3] d. [7, 7, 8, 8] e. [7,8,9, 8) O Consider the following method: public static int mystery(int[] arr, int k) ifk-0) return 0; }else{ return arr[k - 1] + mystery(arr, k-1):) The code segment below is...
Consider the following method: Linel: public static int mystery(int n) { Line2: if (n < 10)...
Consider the following method: Linel: public static int mystery(int n) { Line2: if (n < 10) { ine3: return n; Line4: } else { Line5: int a = n/10; Line 6: int b = n % 10; Line 7: return mystery(a + b); Line 8: } Line 9: } What is the result of the following call? System.out.println(mystery(648)); 18 8 12
Java, how would i do this public static void main(String[] args) { int n = 3;...
Java, how would i do this
public static void main(String[] args) { int n = 3; int result; result = factorial(n); + public static int factorial(int n) public static int factorial(int n) n = 3 { returns 3* 2 * 1 { if (n == 1) return n; else return (n * factorial(n-1)); if (n == 1) return n; else return (3 * factorial(3-1)); ܢܟ } public static int factorial(int n) n = 2 public static int factorial(int n) returns...
What is printed by running the following code? public static void main(String[] args) { int[] nums...
What is printed by running the following code? public static void main(String[] args) { int[] nums = {2, 3, 4}; int n = 5; changeMe1(n, nums); System.out.print( n ); System.out.print(nums[0]); changeMe2(n, nums); System.out.print( n ); System.out.print(nums[0]); } public static void changeMe1(int number, int[] list) { number++; list[0]++; } public static void changeMe2(int number, int[] list) { number = 9; list = new int[1]; list[0] = 99; }
please evaluate the following code. this is JAVA a. class Car { public int i =...
please evaluate the following code. this is JAVA a. class Car { public int i = 3; public Car(int i) { this.i = i; } } ... Car x = new Car(7), y = new Car(5); x = y; y.i = 9; System.out.println(x.i); b. class Driver { public static void main(String[] args) { int[] x = {5, 2, 3, 6, 5}; int n = x.length; for (int j = n-2; j > 0; j--) x[j] = x[j-1]; for (int j...
import java.util.Arrays; public class lab { public static void main(String args[]) { int...
import java.util.Arrays; public class lab { public static void main(String args[]) { int arr[] = {10, 7, 8, 9, 1, 5,6,7}; int arr2[] = {9, 8, 7, 6, 5, 4, 3, 2, 1}; int arr3[] = {1, 3, 5, 3, 2, 6, 20}; quicksort(arr,0,arr.length-1); quicksort(arr2,0,arr2.length-1); quicksort(arr3,0,arr3.length-1); System.out.println(Arrays.toString(arr)); System.out.println(Arrays.toString(arr2)); System.out.println(Arrays.toString(arr3)); } private static int partition(int[] items,int low, int high) { int i=0; int j=0;...
3. Consider the mystery method given. public static int mystery ( int n) [ if (n...
3. Consider the mystery method given. public static int mystery ( int n) [ if (n == 0 ) { return 1; How do we get the values for recurse? else if (n%2 == 0 ) { int recurse = mystery ( n - 1); int result = recurse + n; return result; since n =5, we go to the else statement and do int recurse = mystery(5-1) which equals 4? why is 3 written? else { int recurse =...
b) Consider the following code. public static int f(int n) if (n == 1) return 0; else if (n % 2 == 0). return g(n/2); else return g(n+1); public static int g(int n) int r = n % 3; if (r == 0) return f(n/3); else if (r == 1) return f(n+2); else return f(2 * n); // (HERE) public static void main(String[] args) { int x = 3; System.out.println(f(x)); (1) (5 points) Draw the call stack as it would...
(b) Trace the following code. // Precondition: n > 0 // <method description here> public static int dothing (int n) { int ans = ; for (int x 1; x <= n; x++) ( for (int y = x; y <= n; y++) ( ans+ } return ans; } (ii) If you were asked to write a one-line description above this method, what would it be? (iii) Could you rewrite the method so that it runs more efficiently?
Consider the following method. public static ArrayList<Integer mystery(int n) ArrayList<Integer seg - new ArrayList<IntegerO; for (int k = n; k > 0; k--) seq.add(new Integer(k+3)); return seq What is the final output of the following Java statement? System.out.println(mystery(3)); a. [1,2,4,5) b. [2,3,4,5) c. [6,5,4,3] d. [7, 7, 8, 8] e. [7,8,9, 8) O Consider the following method: public static int mystery(int[] arr, int k) ifk-0) return 0; }else{ return arr[k - 1] + mystery(arr, k-1):) The code segment below is...
Consider the following method: Linel: public static int mystery(int n) { Line2: if (n < 10) { ine3: return n; Line4: } else { Line5: int a = n/10; Line 6: int b = n % 10; Line 7: return mystery(a + b); Line 8: } Line 9: } What is the result of the following call? System.out.println(mystery(648)); 18 8 12
Java, how would i do this
public static void main(String[] args) { int n = 3; int result; result = factorial(n); + public static int factorial(int n) public static int factorial(int n) n = 3 { returns 3* 2 * 1 { if (n == 1) return n; else return (n * factorial(n-1)); if (n == 1) return n; else return (3 * factorial(3-1)); ܢܟ } public static int factorial(int n) n = 2 public static int factorial(int n) returns...
3. Consider the mystery method given. public static int mystery ( int n) [ if (n == 0 ) { return 1; How do we get the values for recurse? else if (n%2 == 0 ) { int recurse = mystery ( n - 1); int result = recurse + n; return result; since n =5, we go to the else statement and do int recurse = mystery(5-1) which equals 4? why is 3 written? else { int recurse =...
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