You are given the head of a linked list.
Remove every node which has a node with a strictly greater value anywhere to the right side of it.
Return the head of the modified linked list.
I have tried the method of recursion using the following code but nevertheless time limit exceeded is what I get
class Solution {
public:
ListNode* solve(ListNode* head)
{
ListNode* temp;
if(head->next=NULL)
return head;
if(head->next->val>head->val)
{
head=head->next;
}
for(temp=head;temp->next!=NULL;temp=temp->next)
{
while(temp->next->next!=NULL)
{
if(temp->next->val<temp->next->next->val)
temp->next=temp->next->next;
}
}
solve(head);
return head;
}
ListNode* removeNodes(ListNode* head) {
return solve(head);
}
};
Related
I am trying to find the time complexity of the recursive function below. I've tried to draw the tree, but it is confusing because in the if condition the function is called once, and otherwise twice.
To give some context, the function is called on nodes of a tree. The task is to calculate the max rating of each node. The rule is that if you add some node to the rating you can't add it's children to the node, but if you don't add it than you can which children to add or don't.
Here is the function:
static int solve(Node node, boolean take) {
int result;
if(take) {
result = node.rating;
for(Node child : node.children) {
result += solve(child, false);
}
return result;
}
result = 0;
for(Node child : node.children) {
result += Math.max(solve(child, true), solve(child, false));
}
return result;
}
Below is my code. I'm trying to return head node back after I insert value to either left or right node. I understood the concept of insertion, but I'm unable to understand how can I return my head node back to that now it is back to original state with addition node added.
Here is exactly I don't understand.
When I insert my node how can I break the loop and return its head node back.
Recursion is stack concept which will output based on LIFO and if it is lifo how can I have head node returned back
Here's my code:
class Node {
int data;
Node left;
Node right;
}
static Node Insert(Node root,int value)
{
return nodeHelper(root,value);
}
static Node nodeHelper(Node root,int value){
Node nodeTracker = root;
Node temp;
if(root!=null){
if(value>root.data){
if(root.right==null){
temp =new Node();
temp.data = value;
root.right = temp;
return nodeTracker;
}
else{
nodeHelper(root.right,value);
}
}
else{
if(root.left==null){
temp=new Node();
temp.data = value;
root.left = temp;
return nodeTracker;
}
else{
nodeHelper(root.left,value);
}
}
}
else{
temp=new Node();
temp.data = value;
return temp;
}
}
}
To return the root of the tree, you need a third parameter that you pass around to keep track of the root. Like this:
Node* nodeHelper(Node* nodeTracker, Node* parent, int value)
Remove the local nodeTracker variable.
Your recursive calls become:
return nodeHelper(nodeTracker, parent.left, value);
(and, of course, same thing for the right branch)
And your initial call in the insert function is:
return nodeHelper(root, root, value);
Here is what I've done so far:
struct rep_list {
struct node *head;
struct node *tail;
}
typedef rep_list *list;
int length(const list lst) {
if (lst->head == NULL) {
return 0;
}
else {
lst->head = lst->head->next;
return 1 + length(lst);
}
}
This works, but the head of the list the function accepts as a parameter gets changed. I don't know how to fix that.
I'm not allowed to change the function definition so it should always accept a list variable.
Any ideas?
EDIT: I tried to do what Tyler S suggested in the comments but I encountered another problem. If I create a node* variable at the beginning, it should point to lst->head. But then every recursive call to the function changes the value back to lst->head and I cannot move forward.
You don't need a local node: just don't change the list head. Instead, pass the next pointer as the recursion head.
int length(const list lst) {
if (lst->head == NULL) {
return 0;
}
else {
return 1 + length(lst->head-next);
}
}
I see. Okay; this gets a bit clunky because of the chosen representation. You need a temporary variable to contain the remaining list. This iscludes changing the head.
int length(const list lst) {
if (lst->head == NULL) {
return 0;
}
else {
new_lst = new(list)
new_lst->head = lst->head->next;
var result = 1 + length(new_lst);
free(new_lst)
return result
}
}
At each recursion step, you create a new list object, point it to the 2nd element of the current list, and continue. Does this do the job for you?
Although this solution is clunky and I hate it, its the only way I can see to accomplish what you're asking without modifying the method signature. We create a temporary node * as member data of the class and modify it when we start.
struct rep_list {
struct node *head;
struct node *tail;
}
node *temp = NULL;
bool didSetHead = false;
typedef rep_list *list;
int length(const list lst) {
if ((didSetHead) && (lst->head != temp)) {
temp = lst->head;
didSetHead = false;
}
if (temp == NULL) {
didSetHead = true;
return 0;
}
else {
temp = temp->next;
return 1 + length(temp);
}
}
Please note, I haven't tested this code and you may have to play with a bit, but the idea will work.
In the following code, I'm trying to get a better hand at understanding how recursion actually work. I've always been a bit confused about it's actual working. I want to know what value does the inorder() function actually return in every step. From where does it get these values of 0,0,11,0,0,11,12,0,0,11 respectively. Could someone tell me the logic? It's a basic inorder tree traversal program.The reason why I'm trying to understand these outputs is because the same logic is somehow used to find the depth of the tree( I think) where with every recursion the value of depth increases without initialization.
#include<stdio.h>
#include<stdlib.h>
struct node
{
int data;
struct node *left;
struct node *right;
};
struct node* newNode(int data)
{
struct node* node=(struct node*)malloc(sizeof(struct node));
node->data=data;
node->left=NULL;
node->right=NULL;
return node;
}
int inorder(struct node *temp) {
if (temp != NULL) {
printf("\nleft %d\n",inorder(temp->left));
printf("\n%d\n", temp->data);
printf("\nright %d\n",inorder(temp->right));
}
}
int main()
{
struct node *root=newNode(1);
root->left=newNode(2);
root->right=newNode(3);
root->left->left=newNode(4);
root->left->right=newNode(5);
inorder(root);
getchar();
return 0;
}
This function should be changed to the following (the first and last print in the original code will only get you more confused!):
int inorder(struct node *temp) {
if (temp != NULL) {
inorder(temp->left);
printf("%d\n", temp->data);
inorder(temp->right);
}
}
The recursion starts with the left branch of a specific node (usually the "root") - printing recursively (in-order) all the nodes on that left-branch, then printing the current node, moving on to printing recursively (in-order) all the nodes in the right branch.
By the way, if you want to keep that tree "ordered" (meaning, all the nodes on the left branch are smaller than the node, and all the nodes on the right branch are bigger or equal to the node) you should change:
root->left->left=newNode(4);
root->left->right=newNode(5);
to:
root->right->right=newNode(4);
root->right->right->right=newNode(5);
so I have a program that is running a bunch of different recursive methods, and I cannot get it to compile/run. The error is in this method, according to my computer:
public static int fibo(int n)
// returns the nth Fibonacci number
{
if (n==0)
{
return 0;
}
else if (n==1)
{
return 1;
}
else if (n>1)
{
return fibo(n-1) + fibo(n-2);
}
}
I have this method called correctly in my main method, so the issue is in this bit of code.
I think I can help you in this. Add return n; after your else if. Outside of the code but before the last curlicue.
The code will work as long as n ≥ 0 btw; another poster here is right in that you may want to add something to catch that error.
Make sure all possible paths have a return statement. In your code, if n < 0, there is no return statement, the compiler recognizes this, and throws the error.
public static int fibo(int n)
// returns the nth Fibonacci number
{
if (n<=0)
{
return 0;
}
else if (n==1)
{
return 1;
}
else // All other cases, i.e. n >= 1
{
return fibo(n-1) + fibo(n-2);
}
}