Find Peak Element Using Binary Search in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to start the binary search with the correct left boundary.

DSA Go

left := [1]

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

A-1

B0

C1

Dlen(nums)

Attempts:

3 left

💡 Hint

Common Mistakes

Starting left at 1 instead of 0

Using negative indices

Setting left to the length of the array

The binary search starts from the first index, which is 0.

2fill in blank

medium

Complete the code to calculate the middle index correctly.

DSA Go

mid := left + (right - left) [1] 2

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

A-

B*

C+

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using multiplication instead of division

Adding instead of dividing

Subtracting instead of dividing

We divide the difference by 2 to find the middle index between left and right.

3fill in blank

hard

Fix the error in the condition to check if mid is a peak element.

DSA Go

if nums[mid] > nums[mid+[1]] {

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

A0

B2

C1

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing with nums[mid] itself

Using mid+0 which is the same element

Using mid-1 which is the previous element

We compare nums[mid] with the next element nums[mid+1] to decide the search direction.

4fill in blank

hard

Fill both blanks to update the search boundaries correctly.

DSA Go

if nums[mid] > nums[mid+[1]] {
    right = [2]
} else {
    left = mid + 1
}

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

A1

Bmid

C+1

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Setting right = mid - 1 instead of mid

Using mid + 0 instead of mid + 1

Not updating boundaries correctly causing infinite loop

If nums[mid] is greater than nums[mid+1], the peak must be at mid or to the left, so we move the right boundary to mid.

5fill in blank

hard

Fill all three blanks to return the peak element index after the loop.

DSA Go

return [1] // The peak element index is at [2] or [3]

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

Aleft

Bright

Cmid

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Returning mid which may not be updated after loop

Returning 0 which is not necessarily the peak

Returning only right without considering left

After the loop, left and right converge to the peak element index. Returning left is correct.