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:
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Input: 2->1->3->0->5
Output: 3->3->5->5->0Input: 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. As, we found next greater element for current head we add the next greater element to ans array and at last we return ans array.
Java
// Java program for the above approachimport java.util.*;public class linkedList { ListNode head = null; // ListNode class ListNode { int val; ListNode next; public ListNode(int val) { this.val = val; next = null; } } public int[] nextLargerLL(ListNode head) { // get size of LinkedList int count = sizeOfLL(head); // make size of ans array equal to size of LL i.e // number of nodes in LL int[] ans = new int[count]; // k is for index of ans array int k = 0; // j will be one step ahead of head node that will // check the greater node ListNode j; // temp is for temporary storing the greater node int temp = 0; while (head != null) { // if head.next is null it means there will be // no greater node than head afterwards so add 0 // to ans array if (head.next == null) { ans[k] = 0; break; } // j is one step ahead of head j = head.next; // if head.val is smaller than j.val so add // j.val to ans array and increment index (k) if (head.val < j.val) { ans[k] = j.val; k++; } else if (head.val >= j.val) { // if head.val is greater // than or equal to j.val while ( j != null && head.val >= j.val) { // search j.val such // that it is greater // than head.val j = j.next; } /* if j is not equal to null it means we * have got a value in LL which is greater * than head so add j.val to ans array * increment k after adding j.val */ if (j != null) { ans[k] = j.val; k++; } else { // it means we have not found any // value which is greater than head so // add 0 to ans array and increment // index ans[k] = 0; k++; } } else { ans[k] = 0; k++; } head = head.next; } return ans; } public void push(int new_data) { /* allocate node */ ListNode new_node = new ListNode(new_data); /* link the old list off the new node */ new_node.next = head; /* move the head to point to the new node */ head = new_node; } // Utility function to print the linked list public void printList() { System.out.println(Arrays.toString(nextLargerLL(head))); } //driver code public static void main(String[] args) { linkedList ll = new linkedList(); ll.push(5); ll.push(0); ll.push(3); ll.push(1); ll.push(2); // Function Call ll.nextLargerLL(ll.head); ll.printList(); } //to get size of LinkedList public int sizeOfLL(ListNode head) { int count = 0; while (head != null) { count = count + 1; head = head.next; } return count; }} |
Python3
# Java program for the above approachhead = None# ListNodeclass ListNode: def __init__(self, val): self.val = val self.next = None# to get size of LinkedListdef sizeOfLL(head): count = 0 while (head != None): count = count + 1 head = head.next return countdef nextLargerLL(head): # get size of LinkedList count = sizeOfLL(head) # make size of ans array equal to size of LL i.e # number of nodes in LL ans = [None]*count # k is for index of ans array k = 0 # j will be one step ahead of head node that will # check the greater node j = None # temp is for temporary storing the greater node temp = 0 while (head != None): # if head.next is None it means there will be # no greater node than head afterwards so add 0 # to ans array if (head.next == None): ans[k] = 0 break # j is one step ahead of head j = head.next # if head.val is smaller than j.val so add # j.val to ans array and increment index (k) if (head.val < j.val): ans[k] = j.val k += 1 elif (head.val >= j.val): # if head.val is greater # than or equal to j.val while ( j != None and head.val >= j.val): # search j.val such # that it is greater # than head.val j = j.next # if j is not equal to None it means we # have got a value in LL which is greater # than head so add j.val to ans array # increment k after adding j.val if (j != None): ans[k] = j.val k += 1 else: # it means we have not found any # value which is greater than head so # add 0 to ans array and increment # index ans[k] = 0 k += 1 else: ans[k] = 0 k += 1 head = head.next return ansdef push(new_data): global head # allocate node None new_node = ListNode(new_data) # link the old list off the new node None new_node.next = head # move the head to point to the new node None head = new_node# Utility function to print the linked listdef printList(): print(nextLargerLL(head))# driver codeif __name__ == '__main__': push(5) push(0) push(3) push(1) push(2) # Function Call nextLargerLL(head) printList() |
Output
[3, 3, 5, 5, 0]
Time Complexity: O(N2)
Auxiliary Space: O(1)
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:
- Push the first node to stack.
- 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.
- Finally, push the next node in the stack.
- 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++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// List Nodestruct ListNode { int val; ListNode* next; ListNode(int x) { val = x; next = NULL; }};// Function to reverse the LLvoid 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 nodevoid printList(ListNode* head){ while (head) { cout << head->val << ' '; head = head->next; }}// Function to find the next greater// element in the listListNode* 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 Codeint 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;} |
Java
// Java program for the above approachimport java.util.*;public class linkedList{ ListNode head = null; // ListNode class ListNode { int val; ListNode next; public ListNode(int val) { this.val = val; next = null; } } // Function to reverse the Linked List ListNode reverse(ListNode head) { ListNode prev = null, next = null, curr = head; while (curr != null) { next = curr.next; curr.next = prev; prev = curr; curr = next; } return prev; } // Function to find the next greater // element in the list ListNode nextLargerLL(ListNode head) { if (head == null) return head; // Dummy Node ListNode res = new ListNode(-1); ListNode temp = res; // Reverse the Linked List head = reverse(head); Stack<Integer> st = new Stack<>(); while (head != null) { // 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.peek() <= head.val) st.pop(); // Update result Linked List if (st.empty()) { temp.next = new ListNode(0); st.push(head.val); } else { temp.next = new ListNode(st.peek()); st.push(head.val); } } head = head.next; temp = temp.next; } temp = res; res = res.next; // Reverse result Linked List res = reverse(res); return res; } public void push(int new_data) { /* allocate node */ ListNode new_node = new ListNode(new_data); /* link the old list off the new node */ new_node.next = head; /* move the head to point to the new node */ head = new_node; } // Utility function to print the linked list public void printList(ListNode head) { ListNode temp = head; while (temp != null) { System.out.print(temp.val + " "); temp = temp.next; } } // Driver Code public static void main(String[] args) { linkedList ll = new linkedList(); ll.push(5); ll.push(0); ll.push(3); ll.push(1); ll.push(2); // Function Call ll.printList(ll.nextLargerLL(ll.head)); }} |
Python3
# Python3 program for the above approach# List Nodeclass ListNode: def __init__(self, x): self.val = x self.next = None # Function to reverse the LLdef rev(head): pre = None; curr = head; nex = curr.next; # Till current is not None while (curr): curr.next = pre; pre = curr; curr = nex; nex = (curr.next) if curr else None head = pre return head # Function to print a LL nodedef printList(head): while(head): print(str(head.val), end = ' ') head = head.next; # Function to find the next greater# element in the listdef nextLargerLL(head): if (head == None): return None; # Dummy Node res = ListNode(-1); temp = res; # Reverse the LL head = rev(head); st = [] while (head): # Initial Condition if (len(st) == 0): temp.next = ListNode(0); st.append(head.val); else: # Maintain Monotonicity # Decreasing stack of element while (len(st) != 0 and st[-1]<= head.val): st.pop(); # Update result LL if (len(st) == 0): temp.next = ListNode(0); st.append(head.val); else: temp.next = ListNode(st[-1]); st.append(head.val); head = head.next; temp = temp.next; # Delete Dummy Node temp = res; res = res.next; del temp; # Reverse result LL res = rev(res); return res; # Driver Codeif __name__=='__main__': # Given Linked List head = ListNode(2); curr = head; curr.next = ListNode(1); curr = curr.next; curr.next = ListNode(3); curr = curr.next; curr.next = ListNode(0); curr = curr.next; curr.next = ListNode(5); curr = curr.next; # Function Call printList(nextLargerLL(head));# This code is contributed by rutvik_56 |
C#
// C# program for the above approachusing System;using System.Collections.Generic;class linkedList{ ListNode head = null;// ListNodepublic class ListNode{ public int val; public ListNode next; public ListNode(int val) { this.val = val; next = null; }}// Function to reverse the Linked ListListNode reverse(ListNode head){ ListNode prev = null, next = null, curr = head; while (curr != null) { next = curr.next; curr.next = prev; prev = curr; curr = next; } return prev;}// Function to find the next greater// element in the listListNode nextLargerLL(ListNode head){ if (head == null) return head; // Dummy Node ListNode res = new ListNode(-1); ListNode temp = res; // Reverse the Linked List head = reverse(head); Stack<int> st = new Stack<int>(); while (head != null) { // Initial Condition if (st.Count == 0) { temp.next = new ListNode(0); st.Push(head.val); } else { // Maintain Monotonicity // Decreasing stack of element while (st.Count != 0 && st.Peek() <= head.val) st.Pop(); // Update result Linked List if (st.Count == 0) { temp.next = new ListNode(0); st.Push(head.val); } else { temp.next = new ListNode(st.Peek()); st.Push(head.val); } } head = head.next; temp = temp.next; } temp = res; res = res.next; // Reverse result Linked List res = reverse(res); return res;}public void Push(int new_data){ // Allocate node ListNode new_node = new ListNode(new_data); // Link the old list off the new node new_node.next = head; // Move the head to point to the new node head = new_node;}// Utility function to print the linked listpublic void printList(ListNode head){ ListNode temp = head; while (temp != null) { Console.Write(temp.val + " "); temp = temp.next; }}// Driver Codepublic static void Main(String[] args){ linkedList ll = new linkedList(); ll.Push(5); ll.Push(0); ll.Push(3); ll.Push(1); ll.Push(2); // Function Call ll.printList(ll.nextLargerLL(ll.head));}}// This code is contributed by Amit Katiyar |
Javascript
<script>// Javascript program for the above approach// List Nodeclass ListNode { constructor(x) { this.val = x; this.next = null; }};// Function to reverse the LLfunction rev(head){ var 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; return head;}// Function to print a LL nodefunction printList( head){ while (head) { document.write( head.val + ' '); head = head.next; }}// Function to find the next greater// element in the listfunction nextLargerLL(head){ if (head == null) return null; // Dummy Node var res = new ListNode(-1); var temp = res; // Reverse the LL head = rev(head); var st = []; while (head) { // Initial Condition if (st.length==0) { temp.next = new ListNode(0); st.push(head.val); } else { // Maintain Monotonicity // Decreasing stack of element while (st.length != 0 && st[st.length - 1] <= head.val) st.pop(); // Update result LL if (st.length == 0) { temp.next = new ListNode(0); st.push(head.val); } else { temp.next = new ListNode(st[st.length - 1]); st.push(head.val); } } head = head.next; temp = temp.next; } // Delete Dummy Node temp = res; res = res.next; delete temp; // Reverse result LL res = rev(res); return res;}// Driver Code// Given Linked Listvar head = new ListNode(2);var 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 CallprintList(nextLargerLL(head));// This code is contributed by noob2000.</script> |
Output
3 3 5 5 0
Time Complexity: O(N)
Auxiliary Space: O(N)
