Find Peak Element
A peak element is an element that is strictly greater than its neighbors. Find a peak element and return its index. Use a modified Binary Search.
Why Interviewers Ask This
Salesforce asks this to evaluate a candidate's ability to optimize time complexity beyond linear scanning. They specifically test if you can apply binary search logic to non-sorted data by recognizing that local gradients guarantee a peak exists, demonstrating strong algorithmic intuition and problem-solving under constraints.
How to Answer This Question
1. Clarify Requirements: Confirm edge cases like single elements or peaks at boundaries where only one neighbor exists. 2. Analyze Constraints: Note the O(log n) requirement implies binary search is mandatory, not linear iteration. 3. Define Logic: Explain the core insight that if nums[mid] < nums[mid+1], a peak must exist in the right half; otherwise, it is in the left half including mid. 4. Draft Code: Implement the modified binary search with low, high, and mid pointers, ensuring strict inequality checks. 5. Trace Example: Walk through an array like [1, 2, 3, 1] to show how the search space halves each step until convergence. 6. Validate Complexity: Explicitly state why this achieves O(log n) time and O(1) space, aligning with Salesforce's focus on scalable solutions.
Key Points to Cover
- Recognizing that global sorting is unnecessary due to local gradient properties
- Explicitly stating the O(log n) time complexity constraint early in the discussion
- Correctly handling boundary conditions without separate if-statements for every edge case
- Demonstrating the logic of eliminating half the array based on a single comparison
- Providing a clear trace of the algorithm with a concrete numerical example
Sample Answer
To solve the Peak Element problem efficiently, I first clarify that we need any valid peak index, not necessarily all of them, and handle boundary conditions where the array has one element. Since the interviewer requested a modified binary search, I know we cannot iterate through the entire array. The key insight here is that even though the array isn't sorted globally, local trends allow us to eliminate half the search space. If the middle element is smaller than its right neighbor, we know a peak definitely exists to the right because the values are rising towards the right edge or will eventually fall. Conversely, if the middle is greater than or equal to the right neighbor, a peak must exist on the left side or be the middle element itself. I implement this by initializing low at zero and high at the last index. In the loop, I calculate mid and compare it with mid plus one. If nums[mid] is less than nums[mid+1], I move low to mid plus one. Otherwise, I set high to mid. This process continues until low equals high, returning that index. For example, with input [1, 2, 1, 3, 5, 6, 4], starting at index 3 (value 3), since 3 is less than 5, we move right. Eventually, we converge on index 5 (value 6). This approach guarantees O(log n) time complexity, which is critical for large datasets typical in enterprise environments like Salesforce.
Common Mistakes to Avoid
- Attempting a linear scan which results in O(n) time complexity and fails the optimization requirement
- Forgetting to check strictly greater neighbors, leading to incorrect identification of plateaus as peaks
- Failing to handle the case where the peak is at the very beginning or end of the array
- Confusing the condition for moving left versus right, causing the binary search to miss the peak
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