C++ Program To Delete Nodes Which Have A Greater Value On Right Side

Given a singly linked list, remove all the nodes which have a greater value on the right side. 

Examples: 

Input: 12->15->10->11->5->6->2->3->NULL
Output: 15->11->6->3->NULL
Explanation: 12, 10, 5 and 2 have been deleted because there is a 
             greater value on the right side. When we examine 12, 
             we see that after 12 there is one node with a value 
             greater than 12 (i.e. 15), so we delete 12. When we 
             examine 15, we find no node after 15 that has a value 
             greater than 15, so we keep this node. When we go like 
             this, we get 15->6->3

Input: 10->20->30->40->50->60->NULL
Output: 60->NULL
Explanation: 10, 20, 30, 40, and 50 have been deleted because 
             they all have a greater value on the right side.

Input: 60->50->40->30->20->10->NULL
Output: No Change.

Method 1 (Simple):
Use two loops. In the outer loop, pick nodes of the linked list one by one. In the inner loop, check if there exists a node whose value is greater than the picked node. If there exists a node whose value is greater, then delete the picked node. 
Time Complexity: O(n^2)

Method 2 (Use Reverse): 
Thanks to Paras for providing the below algorithm. 
1. Reverse the list. 
2. Traverse the reversed list. Keep max till now. If the next node is less than max, then delete the next node, otherwise max = next node. 
3. Reverse the list again to retain the original order. 
Time Complexity: O(n)
Thanks to R.Srinivasan for providing the code below. 

C++

// C++ program to delete nodes which
// have a greater value on right side
#include <bits/stdc++.h>
using namespace std;

// Structure of a linked list node
struct Node
{
    int data;
    struct Node* next;
};

// Prototype for utility functions
void reverseList(struct Node** headref);
void _delLesserNodes(struct Node* head);

/* Deletes nodes which have a node with
   greater value node on left side */
void delLesserNodes(struct Node** head_ref)
{
    // 1. Reverse the linked list
    reverseList(head_ref);

    /* 2. In the reversed list, delete nodes
          which have a node with greater value
          node on left side. Note that head node
          is never deleted because it is the
          leftmost node.*/
    _delLesserNodes(*head_ref);

    /* 3. Reverse the linked list again to
          retain the original order */
    reverseList(head_ref);
}

/* Deletes nodes which have
   greater value node(s) on left side */
void _delLesserNodes(struct Node* head)
{
    struct Node* current = head;

    // Initialize max
    struct Node* maxnode = head;
    struct Node* temp;

    while (current != NULL &&
           current->next != NULL)
    {
        /* If current is smaller than max,
           then delete current */
        if (current->next->data <
            maxnode->data)
        {
            temp = current->next;
            current->next = temp->next;
            free(temp);
        }

        /* If current is greater than max,
           then update max and move current */
        else
        {
            current = current->next;
            maxnode = current;
        }
    }
}

/* Utility function to insert a node
   at the beginning */
void push(struct Node** head_ref,
          int new_data)
{
    struct Node* new_node =
           (struct Node*)malloc(sizeof(struct Node));
    new_node->data = new_data;
    new_node->next = *head_ref;
    *head_ref = new_node;
}

/* Utility function to reverse a
   linked list */
void reverseList(struct Node** headref)
{
    struct Node* current = *headref;
    struct Node* prev = NULL;
    struct Node* next;
    while (current != NULL)
    {
        next = current->next;
        current->next = prev;
        prev = current;
        current = next;
    }
    *headref = prev;
}

/* Utility function to print a
   linked list */
void printList(struct Node* head)
{
    while (head != NULL)
    {
        cout << " " << head->data ;
        head = head->next;
    }
    cout << "" ;
}

// Driver code
int main()
{
    struct Node* head = NULL;

    /* Create following linked list
       12->15->10->11->5->6->2->3 */
    push(&head, 3);
    push(&head, 2);
    push(&head, 6);
    push(&head, 5);
    push(&head, 11);
    push(&head, 10);
    push(&head, 15);
    push(&head, 12);

    cout << "Given Linked List " ;
    printList(head);

    delLesserNodes(&head);

    cout << "Modified Linked List " ;
    printList(head);

    return 0;
}
// This code is contributed by shivanisinghss2110

Output:

Given Linked List 
12 15 10 11 5 6 2 3
Modified Linked List 
15 11 6 3

Time Complexity: O(n)

Auxiliary Space: O(1)

Source: https://www.geeksforgeeks.org/forum/topic/amazon-interview-question-for-software-engineerdeveloper-about-linked-lists-6

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