Next greater element in the Linked List

Given a linked list L of integers, the task is to return a linked list of integers such that it contains next greater element for each element in the given linked list. If there doesn’t any greater element for any element then insert 0 for it.

Examples:

Input: 2->1->3->0->5
Output: 3->3->5->5->0

Input: 1->2->3
Output: 2->3->0

Naive Approach: The naive approach is traverse the linked list L, and for each element in the linked list find the next greater element in the list by traversing the whole string from the current element.



Time Complexity: O(N2)

Efficient Approach: The above naive approach can be optimized by maintaining a monotonically decreasing stack of elements traversed. If a greater element is found append it to the resultant linked list L’ else append 0. Below are the steps:

  1. Push the first node to stack.
  2. Pick the rest of the node one by one and follow the following steps in the loop:
    • Mark the current node as next node.
    • If the stack is not empty, compare the top node value of the stack with next node value.
    • If next node value is greater than the top node value then, Pop the top node from the stack and next is the next greater element for the popped node.
    • Keep popping the node from the stack while the popped node value is smaller than next node value. next node will becomes the next greater element for all such popped node.
  3. Finally, push the next node in the stack.
  4. After the loop in step 2 is over, pop all the node from the stack and print 0 as the next element for them.

Below is the implementation of the above approach:

C++

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// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
  
// List Node
struct ListNode {
    int val;
    ListNode* next;
    ListNode(int x)
    {
        val = x;
        next = NULL;
    }
};
  
// Function to reverse the LL
void rev(ListNode** head)
{
    ListNode *pre, *curr, *nex;
  
    pre = NULL;
    curr = *head;
    nex = curr->next;
  
    // Till current is not NULL
    while (curr) {
        curr->next = pre;
        pre = curr;
        curr = nex;
        nex = (curr)
                  ? curr->next
                  : NULL;
    }
    *head = pre;
}
  
// Function to print a LL node
void printList(ListNode* head)
{
    while (head) {
  
        cout << head->val
             << ' ';
        head = head->next;
    }
}
  
// Function to find the next greater
// element in the list
ListNode* nextLargerLL(ListNode* head)
{
    if (head == NULL)
        return NULL;
  
    // Dummy Node
    ListNode* res
        = new ListNode(-1);
    ListNode* temp = res;
  
    // Reverse the LL
    rev(&head);
    stack<int> st;
  
    while (head) {
  
        // Initial Condition
        if (st.empty()) {
            temp->next
                = new ListNode(0);
            st.push(head->val);
        }
        else {
  
            // Maintain Monotonicity
            // Decreasing stack of element
            while (!st.empty()
                   && st.top()
                          <= head->val)
                st.pop();
  
            // Update result LL
            if (st.empty()) {
                temp->next
                    = new ListNode(0);
  
                st.push(head->val);
            }
            else {
                temp->next
                    = new ListNode(st.top());
                st.push(head->val);
            }
        }
        head = head->next;
        temp = temp->next;
    }
  
    // Delete Dummy Node
    temp = res;
    res = res->next;
    delete temp;
  
    // Reverse result LL
    rev(&res);
    return res;
}
  
// Driver Code
int main()
{
    // Given Linked List
    ListNode* head = new ListNode(2);
    ListNode* curr = head;
  
    curr->next = new ListNode(1);
    curr = curr->next;
  
    curr->next = new ListNode(3);
    curr = curr->next;
  
    curr->next = new ListNode(0);
    curr = curr->next;
  
    curr->next = new ListNode(5);
    curr = curr->next;
  
    // Function Call
    printList(nextLargerLL(head));
    return 0;
}

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Output:


3 3 5 5 0





Time Complexity: O(N)

Auxiliary Space: O(N)


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