Search in Rotated Sorted Array in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to start the binary search with correct initial boundaries.

DSA Javascript

function search(nums, target) {
  let left = 0;
  let right = [1];
  while (left <= right) {
    // search logic
  }
}

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

Anums.length - 1

Bnums.length + 1

Cnums.length

Dnums.length / 2

Attempts:

3 left

💡 Hint

Common Mistakes

Setting right to nums.length causes out-of-bound errors.

Using nums.length / 2 limits the search incorrectly.

The right boundary should be the last index, which is length minus one.

2fill in blank

medium

Complete the code to calculate the middle index correctly.

DSA Javascript

while (left <= right) {
  let mid = (left [1] right) / 2;
  // rest of the code
}

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

A*

B+

C-

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using subtraction instead of addition.

Multiplying left and right instead of adding.

To find the middle, add left and right and divide by 2.

3fill in blank

hard

Fix the error in the condition to check if the left half is sorted.

DSA Javascript

if (nums[left] [1] nums[mid]) {
  // left half is sorted
}

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

A>=

B<=

C>

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using greater than instead of less than.

Using less than or equal instead of less than.

If nums[left] is less than nums[mid], the left half is sorted.

4fill in blank

hard

Fill both blanks to update the search boundaries when target is in the left sorted half.

DSA Javascript

if (nums[left] <= target && target [1] nums[mid]) {
  right = [2];
} else {
  left = mid + 1;
}

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

A<

B>

Cmid

Dleft

Attempts:

3 left

💡 Hint

Common Mistakes

Using greater than instead of less than for target comparison.

Setting right to left instead of mid.

Target must be less than mid to be in left half, so right becomes mid.

5fill in blank

hard

Fill all three blanks to update boundaries when the right half is sorted and target is in it.

DSA Javascript

else {
  if (target [1] nums[mid] && target [2] nums[right]) {
    left = [3];
  } else {
    right = mid - 1;
  }
}

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

A>

B>=

C<=

Dmid + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using less than or equal instead of greater than for first comparison.

Setting left to mid instead of mid + 1.

Target must be greater than mid and less or equal to right, so left moves to mid + 1.