Why Sorting Matters and How It Unlocks Other Algorithms in DSA Java - Test Your Knowledge

Practice - 5 Tasks

1fill in blank

easy

Complete the code to sort the array using Arrays.sort.

DSA Java

int[] numbers = {5, 3, 8, 1};
java.util.Arrays.[1](numbers);

Drag options to blanks, or click blank then click option'

AcopyOf

Breverse

Cshuffle

Dsort

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'reverse' which does not exist in Arrays class.

Using 'shuffle' which is for lists, not arrays.

The method Arrays.sort sorts the array in ascending order.

2fill in blank

medium

Complete the code to find the index of 7 in the sorted array using binary search.

DSA Java

int[] sorted = {1, 3, 5, 7, 9};
int index = java.util.Arrays.binarySearch(sorted, [1]);

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A5

B7

C3

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Passing an index instead of the value to search.

Passing a value not in the array.

Binary search looks for the exact value. Here, we want to find 7.

3fill in blank

hard

Fix the error in the code to correctly sort the list of strings.

DSA Java

java.util.List<String> names = java.util.Arrays.asList("Bob", "Alice", "Eve");
java.util.Collections.[1](names);

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Asort

Bsorted

Corder

Darrange

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'sorted' which is not a method in Collections.

Using 'order' or 'arrange' which do not exist.

The correct method to sort a list is Collections.sort.

4fill in blank

hard

Fill both blanks to create a map of word lengths for words longer than 3 characters.

DSA Java

java.util.List<String> words = java.util.Arrays.asList("cat", "house", "tree", "a");
java.util.Map<String, Integer> lengths = words.stream()
    .filter(word -> word.[1]() > 3)
    .collect(java.util.stream.Collectors.toMap(word -> word, word -> word.[2]()));

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Alength

Bsize

Clength()

Dsize()

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'length' without parentheses which is invalid for String.

Using 'size()' which is for collections, not strings.

Strings use length() to get their length.

5fill in blank

hard

Fill all three blanks to create a sorted map of uppercase words longer than 2 characters.

DSA Java

java.util.List<String> words = java.util.Arrays.asList("dog", "cat", "elephant", "ox");
java.util.Map<String, String> result = words.stream()
    .filter(w -> w.[1]() > 2)
    .sorted()
    .collect(java.util.stream.Collectors.toMap(
        w -> w.[2](),
        w -> w.[3]()
    ));

Drag options to blanks, or click blank then click option'

Alength()

BtoUpperCase

CtoLowerCase

Dsize()

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'size()' which is not valid for strings.

Using 'toLowerCase' instead of 'toUpperCase'.

Use length() to check word length, and toUpperCase() to convert words to uppercase.

Practice

1. Why is sorting an important step before applying many algorithms in data processing?

easy

A. Because sorting organizes data to make searching and comparisons faster

B. Because sorting removes duplicate data automatically

C. Because sorting encrypts the data for security

D. Because sorting compresses the data to save space

Solution

1. Step 1: Understand sorting's role in data organization

   Sorting arranges data in a specific order, such as ascending or descending, which helps algorithms work efficiently.

2. Step 2: Connect sorting to faster searching and comparisons

   When data is sorted, algorithms like binary search can quickly find elements, and comparisons become simpler.

3. Final Answer:

   Because sorting organizes data to make searching and comparisons faster -> Option A

4. Quick Check:

   Sorting helps speed up search and comparison = D [OK]

Hint: Sorting arranges data to speed up search and comparison [OK]

Common Mistakes:

• Thinking sorting removes duplicates automatically
• Confusing sorting with encryption or compression
• Assuming sorting changes data content

2. Which of the following is the correct Java syntax to sort an integer array named arr?

easy

A. arr.sort();

B. sort(arr);

C. Arrays.sort(arr);

D. sortArray(arr);

Solution

1. Step 1: Recall Java's built-in sorting method

   Java provides the Arrays.sort() method to sort arrays efficiently.

2. Step 2: Match syntax with the correct method call

   The correct syntax is Arrays.sort(arr); where arr is the array to sort.

3. Final Answer:

   Arrays.sort(arr); -> Option C

4. Quick Check:

   Java array sorting uses Arrays.sort() = A [OK]

Hint: Use Arrays.sort(array) to sort arrays in Java [OK]

Common Mistakes:

• Using sort(arr) without specifying the class
• Calling sort as a method on the array object
• Using non-existent methods like sortArray

3. What will be the output of the following Java code?

int[] arr = {5, 3, 8, 1};
Arrays.sort(arr);
System.out.print(Arrays.toString(arr));

medium

A. [5, 3, 8, 1]

B. Compilation error

C. [8, 5, 3, 1]

D. [1, 3, 5, 8]

Solution

1. Step 1: Understand the initial array and sorting

   The array starts as {5, 3, 8, 1}. After calling Arrays.sort(arr);, it will be sorted in ascending order.

2. Step 2: Determine the sorted array and output format

   The sorted array is {1, 3, 5, 8}. Using Arrays.toString(arr) prints it as "[1, 3, 5, 8]".

3. Final Answer:

   [1, 3, 5, 8] -> Option D

4. Quick Check:

   Sorted array output = [1, 3, 5, 8] [OK]

Hint: Arrays.sort sorts ascending; Arrays.toString prints array [OK]

Common Mistakes:

• Assuming original order remains after sort
• Confusing output format with array literal
• Expecting descending order without specifying

4. Identify the error in the following Java code snippet that tries to sort a list of strings:

List<String> list = Arrays.asList("banana", "apple", "cherry");
list.sort();

medium

A. The list cannot be sorted because it is immutable

B. The sort method requires a Comparator argument

C. The sort method is not available for List

D. The Arrays.asList method returns an array, not a list

Solution

1. Step 1: Check the list creation and mutability

   Arrays.asList returns a fixed-size list, but it supports sorting since it allows element replacement.

2. Step 2: Analyze the sort method usage

   The list.sort() method requires a Comparator argument. There is no parameterless version, so list.sort(); causes a compilation error.

3. Final Answer:

   The sort method requires a Comparator argument -> Option B

4. Quick Check:

   List.sort() requires Comparator argument = A [OK]

Hint: Use list.sort(Comparator.naturalOrder()) to sort lists [OK]

Common Mistakes:

• Calling list.sort() without providing a Comparator
• Thinking Arrays.asList returns an array
• Assuming list is immutable and cannot be sorted

5. You have two sorted arrays of integers, a = {1, 4, 7} and b = {2, 5, 6}. Which approach best uses sorting to efficiently merge them into a single sorted array?

hard

A. Use two pointers to merge both arrays in one pass without extra sorting

B. Sort each array individually again before merging

C. Concatenate both arrays and then sort the combined array

D. Reverse both arrays and then merge

Solution

1. Step 1: Understand the input arrays are already sorted

   Both arrays a and b are sorted, so we can merge them efficiently without extra sorting.

2. Step 2: Use two-pointer technique to merge

   By using two pointers, one for each array, we compare elements and build a merged sorted array in one pass, which is faster than concatenating and sorting.

3. Final Answer:

   Use two pointers to merge both arrays in one pass without extra sorting -> Option A

4. Quick Check:

   Two-pointer merge uses sorting advantage = C [OK]

Hint: Merge sorted arrays with two pointers for efficiency [OK]

Common Mistakes:

• Concatenating then sorting wastes time
• Sorting arrays again unnecessarily
• Reversing arrays breaks sorted order