Senior Analyst

Understand the problem:
Find two distinct indices i and j such that nums[i] + nums[j] = target.

Naive approach:
Check every pair (i, j) to see if they add up to target. This takes O(n²) time.

Optimized approach using a hash map:

Initialize an empty hash map (dictionary) to store numbers and their indices.

Iterate over the array:

For the current number num, calculate complement = target - num.

Check if complement exists in the hash map.

If yes, return indices [hash_map[complement], current_index].

If no, store the current number and index in the hash map.

Why this works:
Hash map lookup is O(1), so total time complexity is O(n).

Understand the problem:
You need to reverse the direction of the next pointers in the linked list so the list is reversed.

Approach:
Use three pointers:

prev initially None

current starting at the head

next_node to store the next node temporarily

Algorithm:

Iterate through the list while current is not None:

Store current.next in next_node.

Point current.next to prev (reverse the link).

Move prev to current.

Move current to next_node.

When done, prev will be the new head.

Time Complexity: O(n), where n is the number of nodes.

Tip 1: Try to initiate the GD and also give others a chance to speak.
Tip 2: Address others by their name and try to pick the conversation.

Step 1: Understand the Requirement
We can’t use extra space (e.g., hash sets).

We must detect if a cycle exists.

If there’s no cycle, traversal will end at NULL.

✅ Step 2: Initialize Two Pointers
Create two pointers:

slow pointer — starts at head

fast pointer — starts at head

✅ Step 3: Traverse the List
Move slow one step at a time (slow = slow.next)

Move fast two steps at a time (fast = fast.next.next)

✅ Step 4: Detect the Cycle
If at any point slow == fast, a cycle exists.

If fast or fast.next becomes NULL, there is no cycle.

✅ Step 5: Return the Result
If a cycle is found: return True

If traversal ends: return False