If nums[mid] < nums[mid + 1], where should the search continue?
ARight half
BLeft half
CAt mid
DSearch ends
✗ Incorrect
The peak lies in the right half because the slope is rising.
What is the minimum number of elements needed to have a peak element?
A3
B1
C2
D0
✗ Incorrect
A single element is trivially a peak.
What is the worst-case time complexity of the binary search approach to find a peak?
AO(n)
BO(n log n)
CO(log n)
DO(1)
✗ Incorrect
Binary search reduces the search space by half each time.
Does the binary search for peak element require the array to be sorted?
AOnly if array length > 10
BYes
COnly partially sorted
DNo
✗ Incorrect
The array does not need to be sorted to find a peak.
Explain how binary search can be used to find a peak element in an unsorted array.
Think about the slope and which side might contain a peak.
You got /5 concepts.
Describe the steps and conditions to move left or right in the binary search for a peak element.
Focus on the relationship between mid and mid+1 elements.
You got /4 concepts.
Practice
(1/5)
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
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).
Step 2: Relate to binary search goal
Binary search finds such an element by comparing middle elements with neighbors to decide search direction.
Final Answer:
An element that is greater than its neighbors -> Option B
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
Step 1: Identify binary search loop condition
Binary search typically uses while left < right: to narrow down the search space until left meets right.
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.
Final Answer:
while left < right: -> Option D
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
Step 1: Identify peak candidates
Check elements: 20 at index 2 is greater than neighbors 3 and 4, so index 2 is a peak.
Step 2: Binary search logic
Binary search will find index 2 by comparing mid elements and moving left or right accordingly.
Final Answer:
2 -> Option A
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
Step 1: Check index usage in condition
The code accesses nums[mid + 1] without checking if mid + 1 is within array bounds.
Step 2: Identify potential error
If mid is at the last index, mid + 1 causes an out of range error.
Final Answer:
Index mid + 1 may cause out of range error -> Option A
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
Step 1: Identify all peak elements
Peaks are at index 1 (20 > 10 and 15), index 5 (90 > 23 and 67).
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.
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.
Final Answer:
5 -> Option C
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
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