added mock interview problem

[krishnadheerajkrovi]

Please review

[super30admin]

The solution correctly identifies the peak of the mountain array using binary search, then performs two separate binary searches - one on the ascending part (with increasing target comparison) and one on the descending part (with decreasing target comparison). This is the standard approach for this problem.

Strengths:

• Clean separation of concerns with helper methods for binary searches
• Correctly handles the two-phase nature of the mountain array
• Uses proper binary search boundaries

Areas for Improvement:

1. Edge Case Handling: The solution doesn't handle edge cases well:

   • If peak == 0 (monotonically decreasing array), the descent search will have peak+1 > length-1, causing issues
   • If peak == length-1 (monotonically increasing array), the descent search will have peak+1 > length-1
   • Empty or single-element mountain arrays aren't handled

2. Binary Search Logic: In binarySearch2, the comparison logic is inverted - when searching descending order, if mid_ele > target, you should move left (r = mid - 1), not right. Currently it does l = mid + 1 when mid_ele > target, which is incorrect for descending search.

3. Peak Finding: The peak finding logic could be more robust. When l == r, the loop ends, but this should be verified to ensure peak is indeed the maximum element.

4. Code Style: Missing docstrings for helper methods, and the code could benefit from more descriptive variable names.

5. API Calls: The solution makes multiple calls to mountain_arr.get() which could be expensive in real scenarios. While not critical for correctness, caching values when possible could be an optimization.