Binary search 2 solution

[shakthinandana]

No description provided.

[super30admin]

Find the First and Last Position of an Element in given Sorted Array (first_last_sorted_array.py)

Your solution is well-structured and meets the O(log n) time complexity requirement. Here are some points for improvement:

1. Variable Naming: Consider using more descriptive variable names such as start and end instead of s and e to enhance readability.

2. Edge Cases: Your code correctly handles the empty array case. However, it's always good to test with other edge cases, such as when the target is the first or last element, or when there are duplicate targets.

3. Code Clarity: The code is clear, but you could add comments explaining the two binary searches: the first one finds the start index by moving left upon finding the target, and the second one finds the end index by moving right.

4. Efficiency: Your approach is efficient. However, note that the reference solution uses two separate helper functions, which might make the code more modular and readable. You could consider breaking the binary searches into helper functions for clarity, but it's not necessary.

5. One Potential Issue: In the first binary search, you update s every time you find the target. This is correct because you are always moving left to find the first occurrence. However, if the array is very large, this might not be a problem. But note: the reference solution checks if the current found index is the first occurrence by looking at the previous element. This allows it to break immediately when the first occurrence is found. Your method continues to search the left half until it is exhausted. While both are O(log n), the reference solution might be slightly faster in some cases because it breaks early. However, your solution is still acceptable.

Overall, good job! Your solution is correct and efficient.

VERDICT: PASS

---

Find the Minimum Element in a Rotated Array(sorted) (find_min_sorted_array.py)

Your solution is on the right track and correctly uses binary search to achieve O(log n) time complexity. However, there are a few issues to address:

• Edge Case Handling: When mid is 0, accessing nums[mid-1] will cause an index error. You should check if mid is 0 before comparing with the left neighbor. Alternatively, you can structure the condition to avoid this, like in the reference solution: (mid == 0 || nums[mid] < nums[mid-1]). In Python, you can write: (mid == 0 or nums[mid] < nums[mid-1]).

• Minimum Element Check: Your condition only checks the left neighbor. However, the minimum element should be smaller than both its neighbors (if they exist). The reference solution checks both left and right neighbors. Although in a rotated sorted array the minimum element is the only one that is smaller than its left (if exists) and right (if exists), your current check might miss some cases? Actually, in a rotated array, the minimum element is the only one that is smaller than its immediate left (if it exists) and also smaller than the last element. But note: when the array is rotated, the minimum element is the start of the original array. It is typically found at the point where the array breaks the ascending order. So checking only the left neighbor might be sufficient because the array is sorted in ascending order before rotation. However, you should also consider the case when the minimum is at the beginning (no left neighbor) or at the end. But your current code does not check the right neighbor. However, the problem states that the array has unique elements and is rotated between 1 and n times, so the minimum element is always the one that is smaller than its left (if exists) and also smaller than the right (if exists). But actually, the condition you have (nums[mid-1] > nums[mid]) is sufficient to identify the minimum because the array was sorted and then rotated. However, you must avoid the index error when mid is 0.

• Alternative Approach: Another common approach is to compare the mid element with the rightmost element in the current segment. If nums[mid] > nums[high], then the minimum must be in the right half. Otherwise, it is in the left half. This approach avoids having to check neighbors directly. For example:
  while low < high:
  mid = (low + high) // 2
  if nums[mid] > nums[high]:
  low = mid + 1
  else:
  high = mid
  return nums[low]
  This method is simpler and handles all edge cases without index errors.

• Code Readability: Your code is clean and easy to follow. Adding comments to explain the conditions would be helpful for others to understand your logic.

To fix your solution, you can modify the condition for checking the minimum element to avoid index errors. For example:
if (mid == 0 or nums[mid] < nums[mid-1]) and (mid == len(nums)-1 or nums[mid] < nums[mid+1]):
return nums[mid]

Alternatively, you can use the alternative approach mentioned above which is more robust and simpler.

VERDICT: NEEDS_IMPROVEMENT

---

Find the Peak Element (find_peak_element.py)

Your solution shows a good understanding of using binary search to achieve logarithmic time complexity. However, there are a few critical issues that need to be addressed:

• Boundary Checks: During binary search, when mid is 0 or the last index, you must avoid comparing with neighbors that don't exist. The reference solution handles this by checking mid == 0 or mid == n-1 in the condition. You should incorporate similar checks in your condition for comparing neighbors. For example, instead of directly comparing nums[mid] with nums[mid-1] and nums[mid+1], you should first check if mid is at the boundaries.

• Initial Checks: Your initial checks for low and high are good for handling arrays of size 1 or when the peak is at the ends. However, these checks are only performed once at the beginning. Since the binary search might set low or high to boundary indices later, you need to handle boundaries within the loop as well.

• Logic Flow: In the while loop, you check if nums[mid] is greater than both neighbors. This is correct, but you need to ensure that when mid is 0, you don't check the left neighbor, and when mid is the last index, you don't check the right neighbor. Also, your subsequent conditions for moving low or high are based only on the right neighbor. The reference solution uses the fact that if the right neighbor is larger, then there must be a peak on the right side (because the array goes to negative infinity at the ends). Similarly, if the left neighbor is larger, there is a peak on the left. Your condition only checks the right neighbor, which might work in some cases but is not complete. For instance, if the left neighbor is larger but the right is not, you should move left, but your code moves right based on the right neighbor.

• Recommendation: Modify the condition in the while loop to include boundary checks. For example:

  if (mid == 0 or nums[mid] > nums[mid-1]) and (mid == len(nums)-1 or nums[mid] > nums[mid+1]):
      return mid

  Then, to decide which way to go, you can compare nums[mid] with nums[mid+1] (if mid < len(nums)-1). If nums[mid+1] > nums[mid], then move right; otherwise, move left. This is similar to the reference solution.

Here is a corrected version of your code:

class Solution(object):
    def findPeakElement(self, nums):
        low = 0
        high = len(nums) - 1
        n = len(nums)
        
        while low <= high:
            mid = low + (high - low) // 2
            # Check if mid is a peak
            left_val = nums[mid-1] if mid > 0 else float('-inf')
            right_val = nums[mid+1] if mid < n-1 else float('-inf')
            if nums[mid] > left_val and nums[mid] > right_val:
                return mid
            # If right neighbor is larger, peak must be on the right
            if mid < n-1 and nums[mid+1] > nums[mid]:
                low = mid + 1
            else:
                high = mid - 1
        return -1

Alternatively, you can use the reference solution's logic without explicitly using float('-inf') by using index checks.

VERDICT: NEEDS_IMPROVEMENT