Linked List 1

[YogeshPardeshi]

No description provided.

[super30admin]

Reverse a LinkedList (Problem1.java)

Strengths:

• The solution is clean, concise, and follows the correct iterative approach for reversing a linked list
• Variable naming is clear and descriptive (prev, curr, temp)
• The logic flow is easy to follow
• Handles all edge cases correctly (empty list, single node, multiple nodes)
• In-place reversal with O(1) space complexity

Areas for Improvement:

• The solution is already optimal for this problem. No significant improvements needed.
• Could optionally add comments explaining the reversal logic for better readability, but the code is self-explanatory.

VERDICT: PASS

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Remove nth node (Problem2.java)

Strengths:

• Excellent use of the two-pointer technique, which is the optimal approach for this problem
• Smart use of a dummy node simplifies edge case handling (removing the first node)
• Clean, readable code with meaningful variable names
• Properly cleans up the removed node reference
• Achieves the follow-up requirement of one-pass solution

Areas for Improvement:

• The code is already well-optimized. One minor suggestion: the while loop condition count <= n could be written as count < n + 1 for slightly better clarity, though both are correct.

VERDICT: PASS

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Cycle in linked list (Problem3.java)

Strengths:

• Correct implementation of Floyd's cycle detection algorithm
• Good variable naming conventions
• Clean and readable code structure
• Proper handling of edge cases (null head)
• Achieves O(1) space complexity as required by the follow-up

Areas for Improvement:

• The flag boolean variable is unnecessary. You can determine if a cycle exists by checking if the loop exited via the condition or via the break statement. If slow == fast after the loop, a cycle exists; otherwise, return null.
• Consider adding comments to explain the Floyd's algorithm logic, especially the mathematical relationship that makes finding the cycle start possible.

Minor Optimization:

// Instead of using flag, you can simplify to:
while (fast != null && fast.next != null) {
    slow = slow.next;
    fast = fast.next.next;
    if (slow == fast) break;
}

if (fast == null || fast.next == null) return null; // No cycle

slow = head;
while (slow != fast) {
    slow = slow.next;
    fast = fast.next;
}
return slow;

VERDICT: PASS