Unit 10 Quizziz Review

Consider the following recursive method, which is intended to display the binary equivalent of a decimal number. For example, toBinary(100) should display 1100100.

Which of the following can replace /* missing code */ so that toBinary works as intended?

• System.out.print(num % 2);

  toBinary(num / 2);

• System.out.print(num / 2);

  toBinary(num % 2);

• toBinary(num % 2);

  System.out.print(num / 2);

• toBinary(num / 2);

  System.out.print(num % 2);

• toBinary(num / 2);

  System.out.print(num / 2);

Answer explanation

This will recursively call toBinary on the value of num / 2 until the value of the parameter is less than 2. Then either 0 or 1 is displayed depending on whether the parameter is even or odd. This represents the leftmost digit of the binary equivalent. After the base case is reached, either 0 or 1 will continue to be displayed for each of the previous recursive calls to toBinary.

Consider the following recursive method, which is intended to return a String with any consecutive duplicate characters removed. For example, removeDupChars("aabcccd") returns "abcd".

Which of the following can replace /* missing code */ so that removeDupChars works as intended?

• return removeDupChars(str.substring(2));

• return removeDupChars(str.substring(1)) + str.substring(0, 1);

• return removeDupChars(str.substring(2)) + str.substring(1, 2);

• return str.substring(0, 1) + removeDupChars(str.substring(1));

• return str.substring(1, 2) + removeDupChars(str.substring(2));

Answer explanation

This is executed when two consecutive characters are not the same. The first character in str is concatenated to the string returned from the recursive call to removeDupChars. Each time removeDupChars is recursively called, it uses the string that begins at the second character of str, thus working down to the base case. When the base case is reached, the string that is returned from each recursive call to removeDupChars represents an accumulation of the previously returned strings and is concatenated to the single character that was extracted before the recursive call.

Consider the following method, which is intended to return the sum of all the even digits in its parameter num. For example, sumEvens(15555234) should return 6, the sum of 2 and 4.

Which of the following can be used as a replacement for /* missing statement */ so that the sumEvens method works as intended?

• return sumEvens(num % 10);

• return sumEvens(num / 10);

• return num % 10 + sumEvens(num % 10);

• return num % 10 + sumEvens(num / 10);

• return num / 10 + sumEvens(num % 10);

Answer explanation

This statement is executed in the case where num has more than one digit and the rightmost digit is even. In this case, the method should return the rightmost digit (num % 10) plus the result of calling sumEvens recursively on all digits of num but the rightmost digit (sumEvens(num / 10)).

Consider the following method.

Which of the following best describes the output produced by the method call mystery(val) ?

• All integers from 1 to val, separated by spaces

• All integers from val to 1, separated by spaces

• The digits of val in their original order, separated by spaces

• The digits of val in reverse order, separated by spaces

• The last digit of val, then a space, then the first digit of val

Answer explanation

Each method call prints the current value of n and calls mystery recursively with the parameter n - 1. The base case is when the parameter n has the value 1; in this case, 1 is printed and no further recursive calls are executed.

Consider the following recursive method.

Which of the following best describes the result of the call doSomething(myString) ?

• The method call returns a String containing the contents of myString unchanged.

• The method call returns a String containing the contents of myString with the order of the characters reversed from their order in myString.

• The method call returns a String containing all but the first character of myString.

• The method call returns a String containing only the first and second characters of myString.

• The method call returns a String containing only the first and last characters of myString.

Answer explanation

When there is at least one character in str, the method returns the first character of str concatenated with the value returned by a recursive call. The parameter of the recursive call is the substring of str starting at the second character. As a result, the original string is reconstructed recursively.

Consider the following recursive method.

Which of the following best describes the value returned by the method call calc(num) ?

• The int value 9

• The leftmost digit of num

• The rightmost digit of num

• The number of digits in num

• The result of the integer division of num by 10

Answer explanation

In each recursive call, the rightmost digit of the parameter n is truncated. In the final call to calc, n contains only a single digit. The value of n, or the leftmost digit of the parameter num in the original call to calc, is returned.

Consider the following code segment, which appears in a method in the same class as mergeSortHelper and merge.

int[] arr1 = {9, 1, 3, 5, 4};

int[] temp = new int[arr1.length];

mergeSortHelper(arr1, 0, arr1.length - 1, temp);

Which of the following represents the arrays merged the first time the merge method is executed as a result of the code segment above?

• {9} and {1} are merged to form {1, 9}.

• {1, 9} and {3} are merged to form {1, 3, 9}.

• {1, 9} and {5, 4} are merged to form {1, 4, 5, 9}.

• {1, 3, 9} and {5} are merged to form {1, 3, 5, 9}.

• {1, 3, 9} and {4, 5} are merged to form {1, 3, 4, 5, 9}.

Answer explanation

The merge method appears after both recursive calls in the mergeSortHelper method, so it is not called until the condition from < to evaluates to false and there are no more recursive calls to execute. This occurs when the two parts of the array to merge each contain one element. The original array is split into {9, 1, 3} and {5, 4}. The array segment {9, 1, 3} is split into {9, 1} and {3}. The array segment {9, 1} is split into {9} and {1} and these two array segments are merged to form {1, 9}.

The following code segment appears in a method in the same class as binarySearch.

int[] arr = {2, 3, 12, 34, 54};

int result = binarySearch(arr, 0, arr.length - 1, 5);

If the first call to binarySearch is the call in the code segment above, with low = 0 and high = 4, which, if any, of the following shows the values of low and high when binarySearch is called for the third time?

• low = 0, high = 1

• low = 0, high = 2

• low = 1, high = 1

• low = 2, high = 1

• The method returns to the calling code segment before the third call to binarySearch.

Answer explanation

In the first call, middle is assigned the value 2 and target is compared to arr[2]. Since target is less than arr[2], the second call to binarySearch is executed with low = 0 and high = 1. The variable middle is assigned the value 0 and target is compared to arr[0]. Since target is greater than arr[0], the third call to binarySearch is executed with low = 1 and high = 1.

The following code segment appears in a method in the same class as bSearch.

int target = 10;

int[] arrWithDups = {2, 3, 7, 8, 10, 10, 10, 20};

int arrIndex = bSearch(arrWithDups, 0, arrWithDups.length - 1, target);

What is the value of arrIndex after the code segment has been executed?

• 4

• 5

• 6

• 7

• 10

Answer explanation

The first time bSearch is executed, left has the value 0, right has the value 7, and the value of arr[mid], or arr[3], is less than 10. The second time bSearch is executed, left has the value 4, right has the value 7, and the value 5 is returned since arr[mid], or arr[5], is equal to 10.

The following code segment appears in a method in the same class as binarySearch.

ArrayList<Integer> theList = new ArrayList<Integer>();

for (int k = 10; k < 65; k = k + 5)

{

theList.add(k);

}

int result = binarySearch(theList, 0, theList.size() - 1, 45);

Including the call to binarySearch in the last statement of the given code segment, how many times will binarySearch be called before a value is returned?

• 1

• 2

• 3

• 4

• 8

Answer explanation

The ArrayList contains eleven elements, whose values are 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, and 60. The first time binarySearch is executed, low has the value 0, high has the value 10, and middle has the value 5. The value at index middle, which is 35, is less than target. The second time binarySearch is executed, low has the value 6, high has the value 10, and middle has the value 8. The value at index middle, which is 50, is greater than target. The third time binarySearch is executed, low has the value 6, high has the value 7, and middle has the value 6. The value at index 6, which is 40, is less than target. The fourth and final time binarySearch is executed, low, high, and middle all have the value 7. The value at index 7 is equal to target and the value 7 is returned.