Completed Binary Search 1

[MrunaliVaidya98]

No description provided.

[super30admin]

Search in Rotated Sorted Array (search.java)

Strengths:

• Good variable naming (left, right, mid)
• Correct overall binary search structure
• Proper handling of the mid calculation to avoid overflow

Areas for Improvement:

1. Critical Bug Fix: The second if(nums[left] <= nums[mid]) should be else if(nums[left] > nums[mid]) or simply else. This is the main logical error causing incorrect behavior.
2. Code Structure: When using if-else chains for binary search on rotated arrays, ensure proper branching with else clauses to avoid executing multiple branches for the same iteration.
3. Add Braces: Even for single-line blocks, using braces improves readability and prevents bugs during code modification.

Corrected Logic:

if(nums[left] <= nums[mid]) { // left sorted
    if(target >= nums[left] && target < nums[mid]) {
        right = mid - 1;
    } else {
        left = mid + 1;
    }
} else { // right sorted
    if(target > nums[mid] && target <= nums[right]) {
        left = mid + 1;
    } else {
        right = mid - 1;
    }
}

VERDICT: NEEDS_IMPROVEMENT

---

Search in a Sorted Array of Unknown Size (searchMatrix.java)

Your binary search implementation for a 2D matrix is logically sound, but you've solved the wrong problem. The problem requires:

1. Working with an ArrayReader interface (not a 2D array)
2. Handling an array of unknown size by exponentially expanding the search range
3. Returning an index (int) or -1, not a boolean

Key issues to fix:

1. Wrong problem: You implemented "Search a 2D Matrix" instead of "Search in Sorted Array of Unknown Size"
2. Syntax error: matrix[row[col]<target] has a typo - should be matrix[row][col] < target
3. Wrong method signature: Should be int search(ArrayReader reader, int target) not boolean searchMatrix(int[][] matrix, int target)

For the correct problem, you need to:

1. First exponentially expand high until reader.get(high) >= target
2. Then perform standard binary search between low and high
3. Return the index or -1

VERDICT: NEEDS_IMPROVEMENT

---

Search a 2D Matrix (unknown_right _boundary.java)

Looking at this problem, I need to evaluate the student's solution against the reference solution for the "Search a 2D Matrix" problem.

Problem Analysis

The original problem asks to search a 2D matrix where:

• Each row is sorted in non-decreasing order
• The first integer of each row is greater than the last integer of the previous row
• Must be solved in O(log(m * n)) time

Evaluation

1. Correctness:
The student's solution is designed for a completely different problem. The code uses an ArrayReader interface with a .get(index) method, which is from LeetCode problem 702 (Search in a Sorted Array of Unknown Length), not the 2D matrix search problem (LeetCode 74). The student has:

• Used an exponential search to find the right boundary
• Implemented binary search on a 1D array-like structure
• This approach doesn't handle the 2D matrix structure at all

2. Time Complexity:
The exponential search + binary search approach has O(log n) time complexity, which is acceptable for the problem it was designed for, but it's not solving the given problem.

3. Space Complexity:
O(1) space complexity is achieved, which is good.

4. Code Quality:

• The code is well-structured with clear comments
• Variable naming is appropriate
• The binary search implementation is correct for a 1D array

5. Efficiency:
The solution is efficient for its intended problem but completely misses the point of the 2D matrix search problem.

Feedback

Strengths:

• Good implementation of exponential search + binary search pattern
• Clean code structure with helpful comments
• Proper handling of edge cases in binary search
• Correct use of left + (right - left)/2 to prevent overflow

Areas for Improvement:

• The solution addresses a different problem entirely (LeetCode 702 vs LeetCode 74)
• For the 2D matrix problem, you need to either:
  1. Treat the matrix as a flattened 1D array and use binary search (like the reference solution)
  2. First search rows using binary search, then search columns
• The code should be adapted to work with a 2D matrix (matrix[row][col]), not an ArrayReader
• Need to handle the 2D structure by converting 1D index to 2D coordinates: row = mid / n, col = mid % n

Key Takeaway:
The binary search logic is solid, but you need to apply it to the correct problem structure. For a 2D matrix, the key insight is that it can be treated as a single sorted array of length m*n.

---

VERDICT: NEEDS_IMPROVEMENT

The solution is well-written but solves the wrong problem. It needs to be adapted to work with the 2D matrix structure as described in the problem.

VERDICT: NEEDS_IMPROVEMENT