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

Practice - 5 Tasks

1fill in blank

easy

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

DSA Javascript

let left = [1];

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

A1

B0

C-1

Darray.length

Attempts:

3 left

💡 Hint

Common Mistakes

Starting left at 1 or negative index causes skipping the first element.

The left index in binary search starts at 0, the first position of the sorted array.

2fill in blank

medium

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

DSA Javascript

let mid = Math.floor((left + [1]) / 2);

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

Aright

Barray.length

Cleft

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using array.length instead of right causes wrong mid calculation.

The middle index is the floor of the average of left and right indices.

3fill in blank

hard

Fix the error in updating the left index when the middle element is less than the target.

DSA Javascript

if (array[mid] < target) {
  left = [1];
}

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

Amid

Bright

Cleft + 1

Dmid + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Setting left to mid causes infinite loop if mid doesn't change.

When middle element is less than target, search moves right, so left becomes mid + 1.

4fill in blank

hard

Fill both blanks to update right index and return found index correctly.

DSA Javascript

if (array[mid] === target) {
  return [1];
} else if (array[mid] > target) {
  [2] = mid - 1;
}

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

Amid

Bleft

Cright

Dtarget

Attempts:

3 left

💡 Hint

Common Mistakes

Returning target instead of index; updating wrong boundary variable.

Return mid when target found; update right to mid - 1 when middle is greater than target.

5fill in blank

hard

Fill all three blanks to complete the binary search function.

DSA Javascript

function binarySearch(array, target) {
  let left = 0;
  let right = array.length - 1;
  while (left <= right) {
    let mid = Math.floor((left + right) / 2);
    if (array[mid] === target) {
      return [1];
    } else if (array[mid] < target) {
      left = [2];
    } else {
      right = [3];
    }
  }
  return -1;
}

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

Amid

Bmid + 1

Cmid - 1

Dleft

Attempts:

3 left

💡 Hint

Common Mistakes

Incorrect boundary updates cause infinite loops or missed targets.

Return mid if found; move left to mid + 1 if middle less than target; move right to mid - 1 if middle greater.