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Search inside a Rotated Sorted Array (search_in_sortedunknownsized.java)
Missing Implementation: The student has not provided any Java code solving the rotated sorted array search problem. The file only contains a comment with the filename.
Expected Content: The solution should implement a binary search algorithm with O(log n) time complexity, handling the rotated array by determining which half is sorted at each step.
Recommendation: The student needs to implement the binary search logic that:
Uses two pointers (low and high) to define search space
Calculates mid point and checks if it equals target
Determines which half of the array is sorted
Checks if target lies within the sorted half's range
Adjusts search boundaries accordingly
Returns -1 if target not found
VERDICT: NEEDS_IMPROVEMENT
Search Inside a Sorted Array whose Length is unknown (Rotatearray.java)
**
Read the problem carefully: Make sure you understand what the problem is asking for. This problem specifically mentions the ArrayReader interface and an array of unknown size, which requires an exponential search to find the range first.
Use the provided interface: The problem provides an ArrayReader interface that you must use instead of direct array access.
Two-phase approach: The reference solution shows the correct approach:
Phase 1: Exponentially expand the search range (high *= 2) until you find a value >= target or hit the boundary
Phase 2: Perform standard binary search within that range
Return the result: The problem expects a return value, not a print statement.
Understand the constraints: The problem states you don't know the array size, so you cannot use arr.length.
**
VERDICT: NEEDS_IMPROVEMENT
Search a 2D Matrix (matrix.java)
The student has not submitted any code for this problem. To pass this evaluation, the student needs to:
Implement the binary search solution for the 2D matrix search problem. The key approach is to treat the 2D matrix as a flattened 1D sorted array and apply binary search:
Calculate the total number of elements: m * n
Use binary search with low = 0 and high = m * n - 1
Convert mid index to row and column: row = mid / n, col = mid % n
Compare matrix[row][col] with target and adjust search range accordingly
Follow the reference solution's approach which correctly implements O(log(m*n)) time complexity with O(1) space complexity.
Ensure the code compiles and runs correctly for all test cases including edge cases like single element matrices and targets at boundaries.
VERDICT: NEEDS_IMPROVEMENT
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