Point to Greatest Value Node
Posted: 1 Dec, 2020
You are given a singly linked list with every node having an additional “arbitrary” pointer that currently points to NULL. We need to make the “arbitrary” pointer to the greatest value node in a linked list on its right side.
More formally, each Linked List node has three attributes ‘data’, which is the value of the node, the ‘next’ pointer, which points to the next node, and an ‘arbit’ pointer which is initially NULL.
You need to return the head after populating the ‘arbit’ pointer.
Given ‘head’ as 3 -> 5 -> 2 -> 1.
After populating the arbitrary pointer, each node's ‘arbit’ pointer would point to 5 2 1 -1 respectively.
The first line of input contains an integer ‘T’ denoting the number of test cases. The second line of each test case contains space-separated integers, denoting the elements in linked list nodes, and -1 denotes the end of the linked list.
Output Format :
For each test case, print integers denoting the value of the nodes being pointed by the arbitrary pointer of that node, if the arbitrary pointer points to NULL, print -1.
You are supposed to return the head of the linked list, whose arbitrary pointer is populated. You are not required to print the expected output; it has already been taken care of. Just implement the function.
1 <= ‘T’ <= 10 1 <= ‘N’ <= 5000 0 <= ‘data’ <= 10 ^ 4 and ‘data’ != -1 Time Limit: 1sec.
The idea is to loop for every node and find the node with the greatest value with respect to the current node.
The steps are as follows:
- Maintain a ‘tempNode’ which points at the 'head'.
- While ‘tempNode’ is not NULL:
- Maintain a ‘currentNode’ which points to the next of ‘tempNode’, a variable ‘mxVal’ which helps to find the max value towards the right of ‘currentNode’, and ‘maxValNode’, which points to the node with maximum value.
- Point the ‘arbit’ pointer of ‘tempNode’ to ‘maxValNode’.
- Move the ‘tempNode’ ahead.
- Return ‘head’ as the final result.
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