Completed Binary-Search-2

[ashritha0806]

No description provided.

[super30admin]

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

Your solution is well-structured and follows the expected approach. You have correctly implemented two binary search functions to find the first and last occurrences of the target. The code is readable and well-commented.

However, there are a few points to improve:

1. In the lastposition function, you use n which is defined in the outer function. While this works, it makes the function less reusable and relies on the outer scope. Instead, you should use the parameter r (the high index) for the boundary check. Specifically, change the condition to:
   if mid == r or nums[mid+1] != target
   This way, the function does not depend on the outer variable n and is more self-contained.

2. You defined n = len(nums)-1 which is the last index. Then you call firstposition and lastposition with r = n. This is correct. But note that when the array is empty, n = -1. In that case, both functions will be called with l=0 and r=-1, which means the while loop condition l<=r is false, so they return -1. This handles the empty array correctly. So no issue.

3. It is good practice to avoid using outer variables inside inner functions if they can be passed as parameters or derived from parameters. Here, you can avoid using n by using r as described.

4. The code is efficient and meets the time and space complexity requirements.

Overall, the solution is correct and efficient. With a small adjustment to make the lastposition function more robust, it would be perfect.

VERDICT: PASS

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

Strengths:

• The solution is concise and effectively uses binary search to achieve O(log n) time complexity.
• The code is clean and well-commented, explaining the approach clearly.
• The logic is correct: by comparing nums[mid] with nums[r], it correctly narrows down the search to the half that contains the minimum.

Areas for Improvement:

• The code uses integer division for mid, which is standard in Python, but it's important to note that this might cause issues in very large arrays (though not in this problem due to constraints). However, it's acceptable here.
• The variable names l and r are a bit short; using left and right might improve readability without sacrificing much brevity.
• The solution does not handle the case where the array is not rotated (i.e., fully sorted) explicitly, but the algorithm naturally handles it because when nums[mid] <= nums[r], it moves r to mid, which will eventually lead to the minimum element. However, it's worth noting that the code is robust in this regard.
• The solution returns nums[r] at the end, which is correct because the loop ends when l == r, and that index holds the minimum. Alternatively, returning nums[l] would be equivalent.

Overall, the solution is excellent and meets all the problem requirements.

VERDICT: PASS

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Find the Peak Element (Find_peak.py)

Strengths:

• The binary search approach is correctly implemented, with proper calculation of the midpoint and adjustments to the search boundaries.
• The conditions for checking if the midpoint is a peak are correctly implemented, considering edge cases for the first and last elements.
• The time and space complexity are optimal.

Areas for Improvement:

• The comments at the beginning of the code mention using two stacks for storing pushed values and minimum values, which appears to be a copy-paste error from a different problem (like Min Stack). This should be removed or replaced with comments relevant to the peak element problem.

• The condition elif nums[mid + 1] > nums[mid] and mid < n-1 is correct, but the and mid < n-1 is redundant because if mid < n-1 is false, the first part nums[mid+1] would cause an index error. However, in the binary search, when mid is at the end, the initial condition (checking if mid is a peak) would have already handled it. So the condition is safe, but the mid < n-1 is unnecessary because the condition nums[mid + 1] > nums[mid] is only evaluated if mid < n-1 is true? Actually, in Python, if mid is n-1, then mid+1 would be out of bounds. Therefore, the condition mid < n-1 is necessary to avoid an index error. However, note that in the binary search, when mid is n-1, the first condition (mid == n-1 or ...) would have already returned mid if it is a peak. So mid is never n-1 when we reach the elif? Actually, if mid is n-1, the first condition would return mid because it is the last element and greater than its left neighbor (and the right neighbor is considered negative infinity). So in the elif, we are safe because mid is always less than n-1 when we check nums[mid+1]. Therefore, the mid < n-1 in the elif is redundant. You can remove it because the condition mid < n-1 is already implied by the fact that we are in the elif (if mid were n-1, we would have returned in the first condition). Similarly, for the left edge. So the code can be slightly simplified by removing the redundant check.

• The code returns -1 if no peak is found, but according to the problem, there is always at least one peak. So returning -1 is acceptable as a fallback, but it should never occur.

• The variable names l, r, n are concise but could be more descriptive (e.g., low, high, size). However, this is a minor point.

Revised code suggestion:

class Solution(object):
    def findPeakElement(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        low = 0
        high = len(nums) - 1
        
        while low <= high:
            mid = low + (high - low) // 2
            # Check if mid is a peak
            if (mid == 0 or nums[mid] > nums[mid-1]) and (mid == len(nums)-1 or nums[mid] > nums[mid+1]):
                return mid
            # If the right neighbor is greater, search right
            elif nums[mid] < nums[mid+1]:
                low = mid + 1
            # Otherwise, search left
            else:
                high = mid - 1
        return -1

Note: In the elif, we don't need to check mid < len(nums)-1 because if mid were the last index, we would have already returned in the first condition. So it's safe.

VERDICT: PASS