Binary Search Recursive Approach in DSA Javascript - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Recursive Binary Search Call

What is the output of the following JavaScript code that uses recursive binary search to find the index of number 7 in the array?

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, left, mid - 1);
  else return binarySearch(arr, target, mid + 1, right);
}

const array = [1, 3, 5, 7, 9, 11];
const result = binarySearch(array, 7, 0, array.length - 1);
console.log(result);

A 3

B 2

C -1

D 4

💡 Hint

Remember that array indexing starts at 0 and the function returns the index where the target is found.

The target 7 is at index 3 in the array [1, 3, 5, 7, 9, 11]. The recursive binary search finds it and returns 3.

❓ Predict Output

intermediate

Result of Searching Missing Element

What will be the output when searching for number 8 in the sorted array using the recursive binary search below?

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, left, mid - 1);
  else return binarySearch(arr, target, mid + 1, right);
}

const array = [2, 4, 6, 10, 12];
const result = binarySearch(array, 8, 0, array.length - 1);
console.log(result);

A 2

B 0

C 3

D -1

💡 Hint

If the target is not found, the function returns -1.

Since 8 is not in the array, the recursive calls eventually cross over and return -1.

🔧 Debug

advanced

Identify the Error in Recursive Binary Search

What error will the following recursive binary search code produce when searching for 7 in the array?

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, left, mid - 1);
  else return binarySearch(arr, target, mid + 1, right);
}

const array = [1, 3, 5, 7];
const result = binarySearch(array, 7, 0, array.length - 1);
console.log(result);

A RangeError: Maximum call stack size exceeded

B -1

C 2

D TypeError: arr is undefined

💡 Hint

Check the base case condition carefully and how it affects the search.

The base case uses 'left >= right' instead of 'left > right'. This causes the function to return -1 even when left equals right, missing the last element. Searching for 7 (at index 3), the recursion reaches left=3, right=3 and returns -1 without checking arr[3].

❓ Predict Output

advanced

Output of Recursive Binary Search with Duplicate Elements

What index will the recursive binary search return when searching for 3 in the array with duplicates?

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, left, mid - 1);
  else return binarySearch(arr, target, mid + 1, right);
}

const array = [1, 2, 3, 3, 3, 4, 5];
const result = binarySearch(array, 3, 0, array.length - 1);
console.log(result);

A 4

B -1

C 3

D 2

💡 Hint

Binary search returns the first found index, not necessarily the first occurrence.

The recursive binary search finds mid = Math.floor((0 + 6) / 2) = 3, arr[3] === 3 and returns 3. It does not guarantee the first or last occurrence in duplicates.

🧠 Conceptual

expert

Why Recursive Binary Search Uses Mid Calculation with Floor?

Why does the recursive binary search calculate the middle index using Math.floor((left + right) / 2) instead of just (left + right) / 2?

A Because array indices must be integers and floor ensures the middle index is an integer

B Because floor rounds up to the nearest integer to avoid missing elements

C Because floor converts the index to a string for comparison

D Because floor is faster than other rounding methods

💡 Hint

Array positions cannot be fractional numbers.

Array indices are whole numbers. Using floor ensures the middle index is an integer so the function accesses a valid position in the array.

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