Why Binary Search and What Sorted Order Gives You in DSA Java - Test Your Knowledge

Practice - 5 Tasks

1fill in blank

easy

Complete the code to start the binary search with the correct initial middle index.

DSA Java

int mid = (low + [1]) / 2;

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

Ahigh

Blow

Cmid

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using low twice instead of high

Using mid variable before initialization

The middle index is calculated by adding low and high indices and dividing by 2.

2fill in blank

medium

Complete the code to update the low index when the target is greater than the middle element.

DSA Java

if (target > arr[mid]) {
    low = [1];
}

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

Amid

Blow + 1

Cmid + 1

Dhigh

Attempts:

3 left

💡 Hint

Common Mistakes

Setting low to mid (causes infinite loop)

Setting low to high

When target is greater than middle element, search continues in the right half, so low becomes mid + 1.

3fill in blank

hard

Fix the error in the loop condition to correctly continue binary search while low is less than or equal to high.

DSA Java

while (low [1] high) {
    int mid = (low + high) / 2;
    // search logic
}

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

A!=

B<

C>

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using only less than (<) causes missing last element

Using greater than (>) causes no loop

The loop should continue while low is less than or equal to high to include all elements.

4fill in blank

hard

Fill both blanks to complete the binary search update steps for when target is less than or greater than the middle element.

DSA Java

if (target < arr[mid]) {
    high = [1];
} else {
    low = [2];
}

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

Amid - 1

Bmid + 1

Cmid

Dhigh - 1

Attempts:

3 left

💡 Hint

Common Mistakes

Setting high or low to mid causes infinite loop

Using wrong arithmetic on indices

If target is less than middle, high moves to mid - 1; else low moves to mid + 1.

5fill in blank

hard

Fill all three blanks to create a dictionary comprehension that maps words to their lengths only if length is greater than 3.

DSA Java

Map<String, Integer> result = words.stream()
    .filter(word -> word.length() [1] 3)
    .collect(Collectors.toMap(
        word -> word.[2](),
        word -> word.[3]()
    ));

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

A>

BtoUpperCase

Clength

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using less than operator in filter

Using wrong methods for keys or values

Filter words with length greater than 3, map keys to uppercase words, values to their lengths.

Practice

1. Why is binary search only applicable on a sorted array?

easy

A. Because binary search divides the array and compares middle elements, which requires sorted order to decide direction

B. Because binary search works on any array regardless of order

C. Because binary search only works on arrays with unique elements

D. Because binary search requires the array to be reversed

Solution

1. Step 1: Understand binary search mechanism

   Binary search picks the middle element and compares it to the target to decide whether to search left or right.

2. Step 2: Importance of sorted order

   Without sorted order, we cannot decide which half to discard because elements are unordered.

3. Final Answer:

   Because binary search divides the array and compares middle elements, which requires sorted order to decide direction -> Option A

4. Quick Check:

   Sorted array needed for binary search direction [OK]

Hint: Binary search needs sorted data to decide left or right half [OK]

Common Mistakes:

• Thinking binary search works on unsorted arrays
• Believing binary search requires unique elements
• Confusing binary search with linear search

2. Which of the following is the correct syntax to perform binary search on a sorted array arr in Java using Arrays.binarySearch()?

easy

A. int index = Arrays.binarySearch(arr, target);

B. int index = Arrays.searchBinary(arr, target);

C. int index = binarySearch(arr, target);

D. int index = arr.binarySearch(target);

Solution

1. Step 1: Recall Java binary search method

   Java's standard library provides Arrays.binarySearch(array, key) to search sorted arrays.

2. Step 2: Check method call syntax

   The method is static in Arrays class, so call is Arrays.binarySearch(arr, target).

3. Final Answer:

   int index = Arrays.binarySearch(arr, target); -> Option A

4. Quick Check:

   Correct Java binary search syntax = int index = Arrays.binarySearch(arr, target); [OK]

Hint: Use Arrays.binarySearch(arr, target) for Java binary search [OK]

Common Mistakes:

• Using wrong method name like searchBinary
• Calling binarySearch on array instance
• Missing static class Arrays

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

int[] arr = {2, 4, 6, 8, 10};
int index = Arrays.binarySearch(arr, 6);
System.out.println(index);

medium

A. 3

B. -1

C. 0

D. 2

Solution

1. Step 1: Understand array and target

   Array is sorted: {2,4,6,8,10}. Target is 6.

2. Step 2: Find index of 6 in array

   6 is at index 2 (0-based indexing).

3. Final Answer:

   2 -> Option D

4. Quick Check:

   Index of 6 in arr = 2 [OK]

Hint: Index returned is position of target in sorted array [OK]

Common Mistakes:

• Confusing index with value
• Assuming binarySearch returns -1 if found
• Using 1-based indexing

4. Identify the error in the following code snippet that tries to use binary search on an unsorted array:

int[] arr = {5, 2, 9, 1, 7};
int index = Arrays.binarySearch(arr, 7);
System.out.println(index);

medium

A. binarySearch method does not exist in Arrays class

B. The array must be sorted before calling binarySearch

C. The target value 7 is not present in the array

D. The array should be declared as List, not int[]

Solution

1. Step 1: Check array order

   The array {5,2,9,1,7} is unsorted.

2. Step 2: Understand binarySearch requirement

   Binary search requires sorted arrays to work correctly.

3. Final Answer:

   The array must be sorted before calling binarySearch -> Option B

4. Quick Check:

   Binary search needs sorted array [OK]

Hint: Always sort array before binary search [OK]

Common Mistakes:

• Ignoring array sorting requirement
• Assuming binarySearch works on any array
• Thinking target absence causes error

5. Given a sorted array arr = {1, 3, 5, 7, 9, 11}, you want to find the index of the number 8 using binary search. What will be the result and why?

hard

A. -1 because 8 is not in the array

B. 6 because 8 would be inserted at the end

C. -5 because binarySearch returns negative insertion point minus 1

D. 3 because 8 is closest to 7 at index 3

Solution

1. Step 1: Check if 8 is in the array

   Array is {1,3,5,7,9,11}, 8 is not present.

2. Step 2: Understand binarySearch return for missing elements

   Java's binarySearch returns -(insertion_point) - 1 if element not found. 8 would be inserted at index 4 (between 7 and 9), so return is -(4) - 1 = -5.

3. Final Answer:

   -5 because binarySearch returns negative insertion point minus 1 -> Option C

4. Quick Check:

   Missing element returns negative insertion point minus 1 [OK]

Hint: Missing element returns -(insertion_point) - 1 [OK]

Common Mistakes:

• Assuming -1 means not found always
• Returning closest index instead of negative value
• Confusing insertion point calculation