Ys bro its absolutely correct because I m also facing same issue but now I got it 👍

@@anubhabray5254 class Solution { public: //Function to remove a loop in the linked list. bool detectCycle(Node* head) { Node* fast = head; Node* slow = head; while(fast != NULL && fast -> next != NULL) { slow = slow -> next; fast = fast -> next -> next; if(slow == fast) { return true; } } return false; } void removeLoop(Node* head) { // code here // just remove the loop without losing any nodes if(detectCycle(head) == false) { return; } Node* fast = head; Node* slow = head; do { slow = slow -> next; fast = fast -> next -> next; }while(fast != slow); if(fast == head) { while(slow -> next != head) { slow = slow -> next; } slow -> next = NULL; return; } fast = head; while(slow -> next != fast -> next) { slow = slow -> next; fast = fast -> next; } slow -> next = NULL; } };

@@anubhabray5254 yeah bcz if you dry run it , it will go inside an infinite loop if suppose you have 5 node from 1 to 5 and your cycle is also from head means(1) to the last node (5) then there cycle detection point comes out to be 2 and after that for removing cycle you are rearranging the fast pointer to the head and slow at same point means at 2 only so if both pointers are traversing by only one node then their next will not meet anywhere so it runs to an infinite loop they did not cover the case if cycle detected at head node to last node

@@Rohan-ov3fr can pls upload the code here

@@anubhabray5254 probably you don't have all pointer checks in your conditional statement

In which year u r ?

I am in first year second sem

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yeah you are right the code is not working for this corner case and when try to submit it on geekforgeeks it gives error at this case

it works dear.

@@jayprakashsaini611 no the remove cycle doesnt work, tey urself u will lose all nodes except head after removing the cycle

@@jayprakashsaini611 lit bit code u can add if(slow==fast)//when my circle is from first to the last node for that aka circular linked list. { while(fast->next!=slow) { fast=fast->next; } fast->next=NULL; return ; }

@@anuragpandey3341 hey, this code worked ,share your code i will tell you why u got error void removeLoop(Node* head) { Node *fast=head,*slow=head,*sp; while(fast!=NULL && fast->next!=NULL){ sp = slow; slow = slow->next; fast = fast->next->next; if(fast==slow) goto l1; } return; l1: if(slow==head){ sp->next = NULL; return; } fast = head; while(fast->next!=slow->next){ fast = fast->next; slow = slow->next; } slow->next=NULL; }

@@kushagra4401 My code was: void removeLoop(Node* head) { // code here // just remove the loop without losing any nodes Node* fast = head; Node* slow = head; while(fast->next != NULL && fast != NULL) { fast = fast->next->next; slow = slow->next; if(fast == slow) { break; } } if(fast->next == NULL || fast == NULL) return; fast = head; if(fast == slow) { while(slow->next != fast) { slow = slow->next; } Node* todelete = slow->next; slow->next=NULL; delete todelete; return; } else { while(fast->next != slow->next) { fast = fast->next; slow = slow->next; } Node* todelete = slow->next; slow->next = NULL; delete todelete; return; } }

yes

you should check for that case . keep a prev pointer to slow and when slow == fast at head prev->next ==NULL;

@@pendyalaabhishek8866 It seems correct to me and working fine in IDE but idk why GFG isn't accepting this solution either it's throwing a nullpointer exception.

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yeah!

Exactly

ya man

wahi bhai test case pass nhi ho rha isse

you are right @adarsh mishra ,I also tried

I also stuck in the same situation...help me please

@@008abhishekvishwakarma9 bro you can add an extra condition regarding this situation or you can solve this problem using map to..

This is not to find middle element.its just to detect if a cycle is present

There is one more thing that while allocating memory to our nodes dynamically, we are not deallocating memory of nodes after use so it will cause to memory leakage at every exution of same code

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