Binary Search Recursive Approach in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA Javascript

function binarySearch(arr, target, left, right) {
  if (left > right) return -1;
  const mid = Math.floor((left + right) [1] 2);
  if (arr[mid] === target) return mid;
  else if (arr[mid] < target) return binarySearch(arr, target, mid + 1, right);
  else return binarySearch(arr, target, left, mid - 1);
}

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

A+

B-

C*

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using multiplication instead of division.

Forgetting to use Math.floor which can cause decimal indices.

The middle index is found by adding left and right, then dividing by 2 using Math.floor to get an integer.

2fill in blank

medium

Complete the code to check if the target is greater than the middle element.

DSA Javascript

function binarySearch(arr, target, left, right) {
  if (left > right) return -1;
  const mid = Math.floor((left + right) / 2);
  if (arr[mid] === target) return mid;
  else if (arr[mid] [1] target) return binarySearch(arr, target, mid + 1, right);
  else return binarySearch(arr, target, left, mid - 1);
}

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

A>

B<

C===

D!==

Attempts:

3 left

💡 Hint

Common Mistakes

Using equality operator instead of comparison.

Using greater than operator which reverses the logic.

If the middle element is less than the target, search the right half by moving left to mid + 1.

3fill in blank

hard

Fix the error in the base case condition to stop recursion when the search space is invalid.

DSA Javascript

function binarySearch(arr, target, left, right) {
  if (left [1] right) return -1;
  const mid = Math.floor((left + right) / 2);
  if (arr[mid] === target) return mid;
  else if (arr[mid] < target) return binarySearch(arr, target, mid + 1, right);
  else return binarySearch(arr, target, left, mid - 1);
}

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

A>=

B<

C>

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using less than which causes infinite recursion.

Using less than or equal which stops too early.

The recursion should stop when left is greater than right, meaning the target is not found.

4fill in blank

hard

Fill both blanks to complete the recursive calls for searching left and right halves.

DSA Javascript

function binarySearch(arr, target, left, right) {
  if (left > right) return -1;
  const mid = Math.floor((left + right) / 2);
  if (arr[mid] === target) return mid;
  else if (arr[mid] < target) return binarySearch(arr, target, mid [1] 1, right);
  else return binarySearch(arr, target, left, mid [2] 1);
}

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

A+

B-

C*

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using multiplication or division instead of addition or subtraction.

Reversing the signs causing wrong search ranges.

To search the right half, increase mid by 1; to search the left half, decrease mid by 1.

5fill in blank

hard

Fill all three blanks to create a recursive binary search that returns the index or -1 if not found.

DSA Javascript

function binarySearch(arr, target, left, right) {
  if (left [1] right) return -1;
  const mid = Math.floor((left + right) / 2);
  if (arr[mid] === target) return mid;
  else if (arr[mid] [2] target) return binarySearch(arr, target, mid + 1, right);
  else return binarySearch(arr, target, left, mid [3] 1);
}

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

A>

B<

C-

D+

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong comparison operators causing infinite recursion.

Adding instead of subtracting when searching left half.

The base case stops when left is greater than right. If middle element is less than target, search right half. Otherwise, search left half by subtracting 1 from mid.

Practice

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

easy

A. Divide the array into halves and search recursively in the half where the value can be.

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

C. Sort the array first and then search linearly.

D. Use a loop to check elements 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: Recognize recursive approach

   In recursion, the function calls itself on the half where the target value may exist, reducing the search space each time.

3. Final Answer:

   Divide the array into halves and search recursively in the half where the value can be. -> Option A

4. Quick Check:

   Binary search = divide and conquer [OK]

Hint: Remember: binary search splits array, recursion repeats on half [OK]

Common Mistakes:

• Thinking binary search checks every element linearly
• Confusing recursion with iteration
• Not requiring the array to be sorted

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

easy

A. if (array.length == 0) return 0;

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

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

D. if (mid == target) return mid + 1;

Solution

1. Step 1: Identify base case meaning

   The base case stops recursion when the search space is empty, which happens if start index passes end index.

2. Step 2: Match base case condition

   The condition start > end means no elements left to search, so return -1 to indicate not found.

3. Final Answer:

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

4. Quick Check:

   Base case = empty search space [OK]

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

Common Mistakes:

• Using incorrect comparison like start == end as base case
• Returning wrong index or value in base case
• Confusing base case with recursive step

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

function binarySearch(arr, target, start, end) {
  if (start > end) return -1;
  const mid = Math.floor((start + end) / 2);
  if (arr[mid] === target) return mid;
  else if (arr[mid] > target) return binarySearch(arr, target, start, mid - 1);
  else return binarySearch(arr, target, mid + 1, end);
}

const array = [2, 4, 6, 8, 10];
console.log(binarySearch(array, 6, 0, array.length - 1));

medium

A. 2

B. 3

C. -1

D. 1

Solution

1. Step 1: Trace initial call

   Start=0, end=4, mid=2 (element 6). Since arr[mid] == target (6), return mid = 2.

2. Step 2: Confirm output

   The function returns 2 immediately without further recursion.

3. Final Answer:

   2 -> Option A

4. Quick Check:

   Target found at index 2 [OK]

Hint: Check mid index value first to find target quickly [OK]

Common Mistakes:

• Miscomputing mid index
• Confusing index with value
• Forgetting zero-based indexing

4. Identify the error in the following recursive binary search function and choose the correct fix.

function binarySearch(arr, target, start, end) {
  if (start >= end) return -1;
  const mid = Math.floor((start + end) / 2);
  if (arr[mid] === target) return mid;
  else if (arr[mid] > target) return binarySearch(arr, target, start, mid - 1);
  else return binarySearch(arr, target, mid + 1, end);
}

medium

A. Change mid calculation to (start + end + 1) / 2;

B. Remove else if and use only if statements.

C. Return mid + 1 when found instead of mid;

D. Change base case to if (start > end) return -1;

Solution

1. Step 1: Analyze base case condition

   The base case incorrectly uses start >= end, which misses the case when start == end and the element could still be at that index.

2. Step 2: Correct base case

   Changing it to start > end ensures the search includes the last element before stopping.

3. Final Answer:

   Change base case to if (start > end) return -1; -> Option D

4. Quick Check:

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

Hint: Base case should allow checking when start equals end [OK]

Common Mistakes:

• Stopping search too early by wrong base case
• Changing mid calculation unnecessarily
• Returning wrong index on success

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

B. Return mid immediately when arr[mid] == target without further search.

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

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

Solution

1. Step 1: Understand duplicates in binary search

   Standard binary search returns any occurrence, but to find first occurrence, we must check if earlier duplicates exist.

2. Step 2: Modify recursion on match

   When arr[mid] == target, recursively search left half to find if an earlier occurrence exists; if none, mid is first occurrence.

3. Final Answer:

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

4. Quick Check:

   Find first occurrence by searching left on match [OK]

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

Common Mistakes:

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