Search in Rotated Sorted Array in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to return -1 if the array is empty.

DSA Go

if len(nums) == 0 {
    return [1]
}

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

A1

B-1

C0

Dlen(nums)

Attempts:

3 left

💡 Hint

Common Mistakes

Returning 0 which could be confused with a valid index.

Returning length of the array which is out of index range.

Returning -1 indicates the target is not found when the array is empty.

2fill in blank

medium

Complete the code to calculate the middle index correctly.

DSA Go

mid := [1] + (high - low) / 2

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

Ahigh

Blen(nums)

Clow

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using high directly which can cause index out of range.

Using 0 which ignores the current search range.

The middle index is calculated by adding low to half the distance between high and low.

3fill in blank

hard

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

DSA Go

if nums[[1]] <= nums[mid] {

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

Ahigh

B0

Cmid

Dlow

Attempts:

3 left

💡 Hint

Common Mistakes

Using high or mid index incorrectly in the condition.

Using 0 which may not be the current low index.

We compare nums[low] with nums[mid] to check if the left half is sorted.

4fill in blank

hard

Fill both blanks to check if the target is in the left sorted half.

DSA Go

if target >= nums[[1]] && target < nums[[2]] {
    high = mid - 1
} else {
    low = mid + 1
}

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

Alow

Bmid

Chigh

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using high instead of mid in the second comparison.

Using 0 which may not be the current low index.

We check if target is between nums[low] and nums[mid] to decide the search range.

5fill in blank

hard

Fill all three blanks to check if the target is in the right sorted half.

DSA Go

if target > nums[[1]] && target <= nums[[2]] {
    [3] = mid + 1
} else {
    high = mid - 1
}

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

Alow

Bmid

Chigh

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using low instead of mid or high in comparisons.

Updating wrong variable in the assignment.

We check if target is between nums[mid+1] and nums[high], then update low to mid+1.