Next Greater Node in Linked List in Java

Given a linked list L of integers, the task is to return an integer linked list that includes the next bigger element for every element in the supplied linked list. In the case that no element has a greater element, enter 0 for it.

Example 1:

Input:

head = [1, 3, 2, 4]

Output:

The next greater element in the Linked List is [3, 4, 4, 0]

Explanation:

• 3 is the first greater element following one and the next greater element for 1.
• As the first greater element following 3, 4 is the next greater element for 3.
• Four is the first greater element following two, and the next greater element for two.
• Since there isn't another bigger element following 4, the next greater element for 4 is 0.

Example 2:

Input:

head = [2, 1, 5]

Output:

The next greater element in the Linked List is [0, 2, 0]

Explanation:

• Zero is the next bigger element for 2 because there isn't one following it.
• After 1 is the initial greater element, 2 is the second greater element for 1.
• 0 is the next bigger element for 5 because there isn't one following it.

Approach: Using the Brute Force Algorithm

The code uses a hierarchical traverse of the linked list using a brute-force methodology; in the worst scenario, this results in a time complexity of O(n²). The results are stored with an O(n) space complexity since it makes use of an array (res[]) of size n. It is inefficient for huge lists since it sequentially tests each node against all subsequent nodes rather than employing an optimal monotonic stack technique. Before traversal, the method sizeOfList() is called, adding an O(n) overhead for determining the list size. Additionally, the list is constructed in opposite order compared to insertion because push() puts elements at the head. To ensure a readable output, the application uses Arrays.toString() to print the result.

Algorithm:

Step 1: Go from right to left through the list since we require the next greater element.

Step 2: To effectively determine each node's next bigger element, keep the stack monotonically lowering.

Step 3: Remove elements from the stack as long as it is not empty, and its top is less than or equal to the node that is currently in use.

Step 4: If the stack isn't empty, the top of the stack is the next bigger element for the current node.

Step 5: Put the current node on the stack so that it may be compared later.

Step 6: Keep an array in order to hold the subsequent larger components.

Step 7: Finally, print the final results.

Implementation:

Output:

 The next greater element in the Linked List is [7, 4, 7, 0, 0] 

Complexity Analysis:

The Time Complexity of the above code is O(N²), where ‘N’ represents the number of elements in the array and the Space complexity is O(1).

Approach: Using Stack and Linkedlist

The stated for each node in a singly linked list, a Java program determines the next bigger element. To process components from right to left, it first flips the linked list. For each node, the next greater element is efficiently found using a monotonic decreasing stack, ensuring an O(n) time complexity. A bigger element is added to the result list if one is discovered; if not, 0 is allocated. To return the result list to its initial order, it is then inverted. New nodes are added to the list using the push method, and the output is shown by printList. The overall strategy makes use of stacks and linked list reversal to optimize the search for the next bigger element.

Algorithm:

Step 1: Reverse the linked list such that the elements are processed from right to left.

Step 2: To store values in decreasing order, initialize a stack.

Step 3: Find the linked list in reverse order.

Step 3.1: 0 should be stored in the result list if the stack is empty.

Step 3.2: If not, use pop elements that are smaller than or equal to the shown value.

Step 3.3: If there is one, the next bigger element is at the top of the stack; if not, store 0.

Step 4: The value should be pushed onto the stack.

Step 5: To restore the original order, flip the result list.

Implementation:

Output:

 The next greater element in the Linked List is 7 4 7 0 0 

Complexity Analysis:

The Time Complexity of the above code is O(N²), where ‘N’ represents the number of elements in the array and the Space complexity is O(1).