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

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the search boundaries for binary search.

DSA Python

left, right = 0, [1]

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

Alen(nums) - 1

Blen(nums) + 1

Clen(nums)

Dlen(nums) // 2

Attempts:

3 left

💡 Hint

Common Mistakes

Setting right to len(nums) causes index out of range errors.

Using len(nums) // 2 as right limits the search incorrectly.

We set right to len(nums) - 1 because list indices go from 0 to length minus one.

2fill in blank

medium

Complete the code to find the middle index in binary search.

DSA Python

mid = left [1] (right - left) // 2

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 causes wrong mid calculation.

Using multiplication or floor division here is incorrect.

We add half the distance between left and right to left to find the middle index.

3fill in blank

hard

Fix the error in the condition to decide which half to search next.

DSA Python

if nums[mid] < nums[[1]]:

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

Amid - 1

Bmid

Cmid + 1

Dleft

Attempts:

3 left

💡 Hint

Common Mistakes

Using mid - 1 causes index errors or wrong comparisons.

Using mid or left does not compare the correct neighbor.

We compare with the element right after mid, so use mid + 1.

4fill in blank

hard

Fill both blanks to update the search boundaries correctly.

DSA Python

if nums[mid] < nums[mid + 1]:
    [1] = mid + 1
else:
    [2] = mid

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

Aleft

Bright

Cmid

Dmid + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping left and right updates causes infinite loops.

Assigning mid to left or mid + 1 to right is incorrect.

If the middle element is less than its right neighbor, move left to mid + 1 to search right half; else move right to mid to search left half.

5fill in blank

hard

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

DSA Python

def findPeakElement(nums):
    left, right = 0, len(nums) - 1
    while left < right:
        mid = left + (right - left) // 2
        if nums[mid] < nums[mid + 1]:
            left = mid + 1
        else:
            right = mid
    return [1]  # Peak index

peak_index = findPeakElement([2])
print(peak_index, nums[[3]])

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

Aleft

Bnums

Cpeak_index

Dright

Attempts:

3 left

💡 Hint

Common Mistakes

Returning right instead of left is acceptable but less clear here.

Printing nums[left] directly without storing index causes confusion.

Using wrong variable names causes errors.

The peak index is where left and right meet, so return left. We call the function with nums and print the value at peak_index.

Practice

1. What is a peak element in an array when using binary search to find it?

easy

A. The first element of the array

B. An element that is greater than its neighbors

C. The smallest element in the array

D. An element equal to its neighbors

Solution

1. Step 1: Understand the definition of a peak element

   A peak element is one that is strictly greater than its neighbors on both sides (or just one side if at the boundary).

2. Step 2: Relate to binary search goal

   Binary search finds such an element by comparing middle elements with neighbors to decide search direction.

3. Final Answer:

   An element that is greater than its neighbors -> Option B

4. Quick Check:

   Peak element = greater than neighbors [OK]

Hint: Peak element is bigger than neighbors, not smallest [OK]

Common Mistakes:

• Thinking peak is smallest element
• Confusing peak with first or last element always
• Assuming peak equals neighbors

2. Which of the following is the correct Python syntax to start a binary search loop for finding a peak element in array nums?

easy

A. while left <= right:

B. for i in nums:

C. if left < right:

D. while left < right:

Solution

1. Step 1: Identify binary search loop condition

   Binary search typically uses while left < right: to narrow down the search space until left meets right.

2. Step 2: Compare options

   while left <= right: is common in classic binary search but for peak finding, while left < right: is preferred to avoid infinite loops.

3. Final Answer:

   while left < right: -> Option D

4. Quick Check:

   Binary search loop uses < not <= [OK]

Hint: Use while left < right for peak element binary search [OK]

Common Mistakes:

• Using for loop instead of while
• Using <= which may cause infinite loop
• Using if instead of loop

3. What will be the output index of the peak element for the array nums = [1, 3, 20, 4, 1, 0] using binary search peak finding?

medium

A. 2

B. 1

C. 3

D. 0

Solution

1. Step 1: Identify peak candidates

   Check elements: 20 at index 2 is greater than neighbors 3 and 4, so index 2 is a peak.

2. Step 2: Binary search logic

   Binary search will find index 2 by comparing mid elements and moving left or right accordingly.

3. Final Answer:

   2 -> Option A

4. Quick Check:

   Peak at index 2 with value 20 [OK]

Hint: Peak is local max, here 20 at index 2 [OK]

Common Mistakes:

• Choosing first or last element incorrectly
• Confusing index with value
• Not checking both neighbors

4. Given this code snippet to find a peak element, what is the error?

def find_peak(nums):
    left, right = 0, len(nums) - 1
    while left <= right:
        mid = (left + right) // 2
        if nums[mid] < nums[mid + 1]:
            left = mid + 1
        else:
            right = mid - 1
    return left

medium

A. Index mid + 1 may cause out of range error

B. Return should be right instead of left

C. Loop condition should be while left < right

D. Mid calculation is incorrect

Solution

1. Step 1: Check index usage in condition

   The code accesses nums[mid + 1] without checking if mid + 1 is within array bounds.

2. Step 2: Identify potential error

   If mid is at the last index, mid + 1 causes an out of range error.

3. Final Answer:

   Index mid + 1 may cause out of range error -> Option A

4. Quick Check:

   Accessing nums[mid + 1] risks out of range [OK]

Hint: Check array bounds before accessing mid+1 [OK]

Common Mistakes:

• Ignoring array boundary checks
• Using <= in loop causing infinite loop
• Returning wrong index

5. You have an array nums = [10, 20, 15, 2, 23, 90, 67]. Using binary search, which index will be returned as a peak element? Consider multiple peaks exist.

hard

A. 1

B. 4

C. 5

D. 6

Solution

1. Step 1: Identify all peak elements

   Peaks are at index 1 (20 > 10 and 15), index 5 (90 > 23 and 67).

2. Step 2: Understand binary search behavior

   Binary search may return any peak, but typically it finds the one in the middle or right side depending on comparisons.

3. Step 3: Trace binary search on given array

   Mid starts near center; since nums[mid] < nums[mid+1] at some point, search moves right, finally returning index 5.

4. Final Answer:

   5 -> Option C

5. Quick Check:

   Binary search finds peak at index 5 with value 90 [OK]

Hint: Binary search may return any peak; here index 5 is found [OK]

Common Mistakes:

• Assuming only first peak is returned
• Ignoring binary search direction logic
• Confusing index with value