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.