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excellent

nicely done

When the list of the last node is referencing to first node and rest of cases as well , we can handle like this isLoopDetected =true; if(isLoopDetected) private void removeLoop(Node slow, Node head) { Node curr = head; if (slow == head) { while (slow.next != curr) { slow = slow.next; } } else { while (slow.next != curr.next) { slow = slow.next; curr = curr.next; } } slow.next = null; } If you are using C or C++, use -> instead of dot.

How to solve following case? If we have a list with 5 nodes and loop is from last node to first or head node. Then by detect function slow points to head ,So both arguments passed to remove function becomes head and head->next becomes NULL but we need last node ->next to become NULL.

nicely explained

Sir diagram se achche se smjhme aaya. Good explanation sir.

case :1) The above code is not useful in case of loop is happened at head node. After applying the above concept head node next point to null. case: 2) consider this case, "1->2->3->4->1", if loop is detected at node '4' then in this case we end up in infinite loop if we follow above case to remove the loop. we need to consider above two cases.Below is complete code which covers above cases. void detectAndRemoveLoop(Node* head) { if(!head || !(head->next)) return; auto slow = head, fast = head; while(fast && fast->next){ slow = slow->next; fast = fast->next->next; if(slow == fast){ //case 2: if(slow->next == head){ slow->next = nullptr; return; } // case 1: slow = head; if(slow == fast) { while(fast->next != slow) fast = fast->next; } else { while (slow->next != fast->next) { slow = slow->next; fast = fast->next; } } /* since fast->next is the looping point */ fast->next = NULL; /* remove loop */ } } } Let me know your thoughts.

Wow great topic sir

Wow quite well explanation 🔥

Amazing.. well explained sir.

Great explanation bro. Thanks !!

Grate Explanation sir. I think in one textcase (while passing slow and head it may possible head and slow both pointing the same node)it will give wrong result. Except this case all are perfectly fine.

Good video.You got yourself an extra subscriber

Woww. Nicely explained ..so proud of you 😊

Well explained

can anyone please let me know the time complexity here. the first loop takes O(N) but the what about the other two loops?

Bro in removeLoop() function you are writing head=head->next so without taking value of head in some temp variable and then traversing it you are directly started traversing with head so isn't it change the value of head means our linked list will start from 4 and end on 9🤔

can someone help me understand the case when the loop starts from the starting node itself.. it would really help me to thoroughly understand this problem... :)