Delete nodes which have a greater value on right side | LinkedList | GFG | Love Babbar DSA Sheet🔥🔥

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Hey, Guys in this video I have explained with code how we can solve the problem 'Delete nodes which have a greater value on right side '.
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Опубликовано:

17 апр 2021

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Комментарии : 37

@nandiniagarwal9116 Год назад

Amazing bhaiya ....... Aise hi aap questions practice karate rahiye bhot helpful hai

@sahasransupanda8169 3 года назад

What will happen if we use three pointers and start traversing from left without reversing the linkedlist ...what will be the time complexity??

@kushkumar9936 3 года назад

your videos is very helpful

@_hulk748 Год назад

Thankyou sir❤🙇‍♂🙏

@ashvinimeshram5242 3 года назад

Nice explaination

@sumedhdvs8739 3 года назад

good explanation bro

@sanathbhardwaj919 3 года назад

Best explanation

@adityapandey8245 3 года назад

Thanks vro

@1piecegaming622 3 года назад

next question clone a linked list with random and next pointer please

@vashishthsharma4411 Год назад

thank you bhai

@bihariboyzyt826 3 года назад

@MohammadImran-bb7sj Год назад

I think, we can do this problem without reversing the linked list. Right?

@pinkkitty6553 11 месяцев назад

Thabks

@jayeshnayee6735 2 года назад

How to solve Using recursion???

@anjandey6089 3 года назад

BST ka upr video bana do vai

@govinds8745 Год назад

Bhai, node delete karne ka invitation du kya. Worst case ke hisab se memory leak ki problem hogi

@rishavraj9284 5 месяцев назад

Ye chutiye log code ko rat lete hai isiliye delete nhi kiya😅 memory leak hone se iska code baccha lega

@shubhenducomic352 2 года назад

code ka link bhejdo plz

@rohitchilhorkar8114 2 года назад

Can you tell me why this code is giving TLE. Node *compute(Node *head) { // your code goes here while(head->data < head->next->data && head->next!=NULL) head = head->next; struct Node* prev = head; if(head->next == NULL) return head;//as in 10 20 30.. struct Node* mid = head->next; struct Node* last = head->next->next; while(last!=NULL) { if(mid->data < last->data && last->next == NULL) prev->next = last; else if(mid->data < last->data) { mid = mid->next; last = last->next; } if(mid->data >= last->data) { prev->next = mid; prev = mid; mid = mid->next; last = last->next; } } prev->next = NULL; return head; }

@SurajChoubeyoppa 2 года назад

Short and fast solution: class Solution { public: Node *compute(Node *head) { if(head == NULL || head -> next == NULL) return head; head -> next = compute(head->next); if(head -> next -> data > head -> data) return head -> next; return head; } };

@debasishphukon5345 2 года назад

Nice solution re :D

@rahulsaini2084 2 года назад

Ya bro no doubt short solution but it's depend on conditions as you use recursion so tha space complexity will be o(n) .

@priyanshusingh2454 2 года назад

Node* fun(Node* head, int &mx) { if(head->next == NULL) { mx = head->data; return head; } head->next = fun(head->next, mx); if(head->data >= mx) { mx = head->data; return head; } return head->next; } Node *compute(Node *head) { int mx = 0; return fun(head, mx); }

@akshaysuryavanshi9420 8 месяцев назад

if(second!=NULL){ second= second->next; }. second is the pointer for head's next element. Put this condition in ur reverse function.

@prasannapm3220 2 года назад

good

@anubhabray5254 3 года назад

I have done it using brute force approach but got TLE for the same: Node* deleteNode(Node **head,int x) { if(x==1){ Node* temp = *head; if(temp->next==NULL){ free(*head); return NULL; } *head=temp->next; free(temp); return *head; } int i=1; Node* temp = *head; while(inext; i++; } Node* nextnode=temp->next; temp->next=nextnode->next; free(nextnode); return *head; } Node *compute(Node *head) { Node* temp=head; vector v; int pos=0; while(temp->next!=NULL){ pos++; Node* check=temp->next; while(check!=NULL){ if(check->data>temp->data){ v.push_back(pos); break; } check=check->next; } temp=temp->next; } while(v.size()>0){ int x = v.back(); v.pop_back(); head = deleteNode(&head,x); } return head; }

@Man_of_Culture. 3 года назад

bro I have solved this problem without reversing the list Node *compute(Node *head) { Node *c=head,*n=NULL,*p=NULL; c=head; int i,flag=0; while(c) { i=0; if(c->next) { n=c->next; if(c->datadata) { if(c==head) { head=n; free(c); c=n; i=1; } else { p->next=n; free(c); c=n; i=1; } flag=1; } } if(i==0) { p=c; c=c->next; } if(!c and flag==1) { c=head; flag=0; } else if(!c and flag==0) break; } return(head); }