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Find Peak Element

Certification: Intermediate Level Accuracy: 0% Submissions: 0 Points: 10

Write a JavaScript program to find a peak element in an array. A peak element is an element that is strictly greater than its neighbors. Given a 0-indexed integer array nums, find a peak element and return its index. If the array contains multiple peaks, return the index to any one of the peaks. You may imagine that nums[-1] = nums[n] = -∞. The algorithm should run in O(log n) time complexity.

Example 1
    
      
  • Input: nums = [1,2,3,1]
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  • Output: 2
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  • Explanation: 
    • Check element at index 2, which has value 3. 
    • Compare with left neighbor: 3 > 2 (at index 1). 
    • Compare with right neighbor: 3 > 1 (at index 3). 
    • Since 3 is greater than both neighbors, index 2 is a peak.
    •
Example 2
    
      
  • Input: nums = [1,2,1,3,5,6,4]
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  • Output: 5
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  • Explanation: 
    • The array has multiple peaks at indices 1 and 5. 
    • Element at index 1 has value 2: 2 > 1 (left) and 2 > 1 (right). 
    • Element at index 5 has value 6: 6 > 5 (left) and 6 > 4 (right). 
    • Either index 1 or 5 can be returned as both are valid peaks.
    •
Constraints
    
      
  • 1 ≤ nums.length ≤ 1000
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  • -2^31 ≤ nums[i] ≤ 2^31 - 1
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  • nums[i] != nums[i + 1] for all valid i
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  • Time Complexity: O(log n)
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  • Space Complexity: O(1)
    •
ArraysKPMGD. E. Shaw
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Solution

Solution Hints

    
      
  • Use binary search to achieve O(log n) time complexity
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  • Compare the middle element with its right neighbor to determine search direction
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  • If nums[mid] < nums[mid + 1], the peak must be on the right side
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  • If nums[mid] > nums[mid + 1], the peak could be at mid or on the left side
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  • Continue binary search until you find a peak element
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  • Handle edge cases for single element array and boundary elements
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Steps to solve by this approach:

 Step 1: Initialize left and right pointers to define the search range (0 to nums.length - 1).

 Step 2: Calculate the middle index using the formula mid = (left + right) / 2.

 Step 3: Compare the middle element with its right neighbor to determine the search direction.

 Step 4: If nums[mid] < nums[mid + 1], move the left pointer to mid + 1 (search right half).

 Step 5: If nums[mid] > nums[mid + 1], move the right pointer to mid (search left half including mid).

 Step 6: Continue the binary search until left equals right, which gives the peak element index.

 Step 7: Return the final index as it guaranteed to be a peak element.

Submitted Code :