Complete Two-Pointer-2

[dhruvil15]

No description provided.

[super30admin]

Merging of 2 arrays (Problem37.java)

Your solution is excellent! It correctly implements the optimal algorithm with O(m + n) time and O(1) space complexity. The code is clean and efficient.

Strengths:

• Correct handling of the merge process from the end to avoid overwriting.
• Efficient use of pointers to traverse the arrays.
• Properly handles the case when there are remaining elements in nums2.

Areas for improvement:

• Consider adding inline comments to explain the purpose of each pointer and the while loops. This can help others (and yourself) understand the code more quickly.
• For example, you could add a comment explaining why you start from the end (to avoid overwriting elements in nums1 that haven't been processed yet).

Overall, this is a high-quality solution.

VERDICT: PASS

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Search 2D sorted matrix II (Problem38.java)

Your solution is excellent! You have implemented an efficient algorithm that optimally searches the matrix. Here are some strengths and minor suggestions:

Strengths:

• Correctness: Your solution correctly identifies whether the target exists in the matrix.
• Efficiency: The O(m + n) time complexity is optimal for this problem, and O(1) space complexity is the best possible.
• Code Quality: The code is concise, readable, and well-commented. You explained your approach clearly.

Areas for Improvement:

• Although your solution is already optimal, consider adding a check for an empty matrix at the beginning to handle edge cases. However, the constraints state that m and n are at least 1, so it's not strictly necessary.
• You might want to mention why you start from the top-right (or bottom-left) corner in your comments. This helps others understand the intuition behind the approach.

Overall, this is a great solution. Keep up the good work!

VERDICT: PASS

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Edit and Remove Duplicates in an array (Problem39.java)

This is an excellent solution! You have successfully implemented the two-pointer technique with a counter to track duplicates. Your code is efficient and meets all the problem requirements. The comments at the top are also helpful for understanding the approach.

One minor suggestion: While your code is correct, you might consider adding a brief comment inside the while loop to explain the purpose of the count variable and the condition checks. This can help others (and your future self) understand the logic more quickly. For example:

// Track the count of consecutive duplicates
if (fast != 0 && nums[fast] == nums[fast-1]) {
    count++;
} else {
    count = 1; // reset count for new number
}

// If the current element is within the allowed duplicates (at most k), copy it to the slow pointer
if (count <= k) {
    nums[slow] = nums[fast];
    slow++;
}

Overall, great job! Keep up the good work.

VERDICT: PASS