AP CSA Algorithms Practise

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• int longest = 0;

• int longest - = words[0].length();

• String longest = " ";

• String longest = words[0];

• String longest = words[1];

Answer explanation

1. 1. Without the initial value being the first value in the array word[0], not all strings length will be considered. words[1] does not work since it will never consider word[0], the first value of the array.

• 2. Since the answer must return a word, A &B are not correct.

• 3. C is incorrect since it does not incorporate any of the values in the array as a starting point to compare it to the next value in the array.

• array values must be sorted in ascending order

• array values must be sorted in descending order

• array values must have only one mode

• array values must not contain values whose sum is not 0

• no precondition is necessary; method always works as intended

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Answer explanation

Explanation of the code:

1. 1. The code defines a function reverse that takes an integer array arr as input.

3. 2. It creates a new integer array newArr with the same size as the original array.

5. 3. It iterates through the elements of the original array using a for loop.

7. 4. The missing statement (/* missing statement */) needs to be placed inside the loop to copy the elements of arr to newArr in the reversed order.

Missing statement:

• newArr[newArr.length - 1 - i] = arr[i];

  • 1. This statement assigns the element at index i of the original array arr to the element at index newArr.length - 1 - i of the new array newArr.

  • 2. By subtracting i from newArr.length - 1, we calculate the corresponding index in the reversed order for the new array.

Example:

Consider the original array arr = {1, 2, 3, 4, 5}.

After filling the newArr using the missing statement:

• newArr[4] = arr[0] (5) (Fills the last element of newArr)

• newArr[3] = arr[1] (2) (Fills the second last element of newArr)

• newArr[2] = arr[2] (3) (Fills the middle element of newArr)

• newArr[1] = arr[3] (4) (Fills the second element of newArr)

• newArr[0] = arr[4] (1) (Fills the first element of newArr)

The resulting newArr will be: newArr = {5, 2, 3, 4, 1} (reversed order).

• for (int k = 0; k < arr.length - 1; k++)

• for (int k = 0; k < arr.length; k++)

• for (int k = 1; k < arr.length; k++)

• for (int k = arr.length - 1; k >= 0; k--)

• for (int k = arr.length - 1; k > 0; k--)

Answer explanation

The loop compares each array element with an array element one position higher. With this header, the loop starts with the element at index 0 and ends with the next-to-last element, ensuring that all consecutive pairs are compared without a bounds error.

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• I & III

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• II & III

• None of the code segments will return an equivalent result

Answer explanation

The given method increases each element of the array numbers by 1. Code segment I does not work as intended. It assigns num a copy of each element of numbers. When num is incremented, it does not change the corresponding value stored in numbers. Code segment II does not compile, as the variable j has not been declared. Code segment III does not work as intended, as it attempts to use num as an index rather than as a copy of a value from the array.

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Answer explanation

The original code segment uses an enhanced for loop to iterate through the elements of array arr, uses variable sum to accumulate the sum of twice the value of the elements in the array, and prints the value of sum. This code segment produces the same output using a regular for loop. As the index k varies from 0 to arr.length - 1, twice the value of arr[k] accumulates in the variable sum.

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• III

• I & II

• I & III

• ALL

Answer explanation

The given code segment prints the array elements in order from left to right using an enhanced for loop. Code segment I uses elements as indices. The first element of arr is 1, and arr[1] is 2; the second element is 2, and arr[2] is 4; the third element is 4, and arr[4] is 3; etc. Code segment II prints the indices 0 to 4 in order. Code segment III prints the array elements in order from left to right using a for loop.

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• values[0]

• values[1]

• Integer.MIN_VALUE

• Integer.MAX_VALUE

Answer explanation

0 does not work as it will not represent a value in the array. Integer.MIN_VALUE does not work if the largest integer in the array is located at index 0. Likewise the max version does not work as it will always print out Integer.MAX_ VALUE. The only one which gives a starting point and allows for full array traverse is values[0]