Binary Search Recursive Approach in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to calculate the middle index in the recursive binary search.

DSA Java

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

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

Alow

Blength

Cmid

Dhigh

Attempts:

3 left

💡 Hint

Common Mistakes

Using low twice instead of high and low.

Using mid before it is defined.

The middle index is calculated using the high and low indices to avoid overflow.

2fill in blank

medium

Complete the code to check if the middle element is the target in recursive binary search.

DSA Java

if (arr[mid] == [1]) {
    return mid;
}

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

Atarget

Bmid

Clow

Dhigh

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing with mid index instead of target value.

Using low or high instead of target.

We compare the middle element with the target value to find if we found the element.

3fill in blank

hard

Fix the error in the recursive call to search the left half when target is smaller than middle element.

DSA Java

return binarySearch(arr, low, [1], target);

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

Amid - 1

Bmid + 1

Chigh

Dlow

Attempts:

3 left

💡 Hint

Common Mistakes

Using mid + 1 instead of mid - 1.

Using high or low incorrectly.

When target is smaller, search the left half by setting high to mid - 1.

4fill in blank

hard

Fill both blanks to search the right half when target is greater than middle element.

DSA Java

return binarySearch(arr, [1], [2], target);

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

Amid + 1

Blow

Chigh

Dmid - 1

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping low and high values.

Using mid - 1 instead of mid + 1 for low.

To search the right half, set low to mid + 1 and keep high unchanged.

5fill in blank

hard

Fill all three blanks to complete the base case that stops recursion when element is not found.

DSA Java

if ([1] > [2]) {
    return [3];
}

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

Alow

Bhigh

C-1

Dmid

Attempts:

3 left

💡 Hint

Common Mistakes

Using mid instead of low or high in condition.

Returning mid instead of -1.

The recursion stops when low is greater than high, meaning the target is not in the array, so return -1.

Practice

1. What is the main idea behind the binary search recursive approach when searching for a value in a sorted array?

easy

A. Check every element one by one from start to end

B. Sort the array first and then search linearly

C. Divide the array into halves and search recursively in the half where the value may exist

D. Use a stack to store all elements and pop until the value is found

Solution

1. Step 1: Understand binary search logic

   Binary search works by dividing the sorted array into two halves and deciding which half to search next based on comparison.

2. Step 2: Apply recursion in binary search

   The recursive approach calls the same function on the half where the target value could be, reducing the search space each time.

3. Final Answer:

   Divide the array into halves and search recursively in the half where the value may exist -> Option C

4. Quick Check:

   Binary search = divide and conquer [OK]

Hint: Binary search splits array, search half recursively [OK]

Common Mistakes:

• Thinking binary search checks every element
• Not using recursion or iteration to reduce search space
• Assuming array is unsorted

2. Which of the following is the correct base case condition in a recursive binary search function in Java?

easy

A. if (start < end) return -1;

B. if (start == end) return start;

C. if (array[mid] == target) return mid;

D. if (start > end) return -1;

Solution

1. Step 1: Identify base case for recursion end

   The base case ends recursion when the search space is invalid, i.e., start index is greater than end index.

2. Step 2: Confirm correct base case syntax

   Using if (start > end) return -1; correctly signals the target is not found.

3. Final Answer:

   if (start > end) return -1; -> Option D

4. Quick Check:

   Base case = start > end [OK]

Hint: Base case ends when start passes end [OK]

Common Mistakes:

• Using start == end as base case incorrectly
• Returning wrong index when target not found
• Confusing base case with target found condition

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

public class Test {
  public static int binarySearch(int[] arr, int start, int end, int target) {
    if (start > end) return -1;
    int mid = start + (end - start) / 2;
    if (arr[mid] == target) return mid;
    else if (arr[mid] > target) return binarySearch(arr, start, mid - 1, target);
    else return binarySearch(arr, mid + 1, end, target);
  }
  public static void main(String[] args) {
    int[] data = {2, 4, 6, 8, 10};
    System.out.println(binarySearch(data, 0, data.length - 1, 6));
  }
}

medium

A. 2

B. 3

C. -1

D. 1

Solution

1. Step 1: Trace initial call and mid calculation

   Array is [2,4,6,8,10], start=0, end=4, target=6. Mid = 0 + (4-0)/2 = 2.

2. Step 2: Check if arr[mid] equals target

   arr[2] = 6 equals target 6, so return mid = 2.

3. Final Answer:

   2 -> Option A

4. Quick Check:

   Target found at index 2 [OK]

Hint: Calculate mid, compare, return index if match [OK]

Common Mistakes:

• Miscomputing mid index
• Returning wrong index when target found
• Confusing array indices

4. Identify the error in the following recursive binary search code snippet:

public static int binarySearch(int[] arr, int start, int end, int target) {
  if (start >= end) return -1;
  int mid = (start + end) / 2;
  if (arr[mid] == target) return mid;
  else if (arr[mid] > target) return binarySearch(arr, start, mid - 1, target);
  else return binarySearch(arr, mid + 1, end, target);
}

medium

A. Base case condition should be start > end, not start >= end

B. Mid calculation should be start - end instead of start + end

C. Should return mid + 1 when target is greater

D. No error, code is correct

Solution

1. Step 1: Analyze base case condition

   Base case currently is if (start >= end) return -1;. This causes missing the case when start == end and target is at that index.

2. Step 2: Correct base case to allow checking when start == end

   Changing to if (start > end) return -1; ensures the single element at start==end is checked.

3. Final Answer:

   Base case condition should be start > end, not start >= end -> Option A

4. Quick Check:

   Base case must allow start == end check [OK]

Hint: Base case excludes start == end incorrectly [OK]

Common Mistakes:

• Using >= instead of > in base case
• Ignoring single element check
• Incorrect mid calculation

5. Given a sorted array with possible duplicate values, how can you modify the recursive binary search to find the first occurrence of a target value?

hard

A. Return mid immediately when arr[mid] == target

B. When arr[mid] == target, continue searching left half to find earlier occurrence

C. Search only the right half when arr[mid] == target

D. Use linear search after binary search to find first occurrence

Solution

1. Step 1: Understand duplicates and first occurrence

   Duplicates mean target can appear multiple times. To find first occurrence, we must check if earlier indices also have target.

2. Step 2: Modify recursive call when target found

   When arr[mid] == target, recursively search left half to find if target appears earlier. If none found, return current mid.

3. Final Answer:

   When arr[mid] == target, continue searching left half to find earlier occurrence -> Option B

4. Quick Check:

   Search left half on match for first occurrence [OK]

Hint: On match, search left half to find first occurrence [OK]

Common Mistakes:

• Returning first found index without checking left side
• Searching right half on match
• Using linear search unnecessarily