Given LinkedList, replace each node’s value with its surpasser count. That is the count of elements which are greater towards its right.
Input : 10->12->5->40->21->70->1->49->37->NULL
Output : 6->5->5->2->3->0->2->0->0->NULL
Element in the first node is 10 and the number of elements to the right of the node that are greater than 10 is 6. Therefore replace the node with 6.
Element in the first node is 12 and the number of elements to the right of the node that are greater than 12 is 5. Therefore replace the node with 5.
Similarly, replace for all the elements in the list.
Input : 5->4->6->3->2->NULL
Output : 1->1->0->0->0->NULL
- Take two pointers p and x. The pointer p is used to traverse the list and x is used to traverse the right half of the list for every node.
- Initialize a variable count to count the nodes greater than the current nodes.
- Traverse through all the nodes in the list using the pointer p.
- Initialize the count to 0.
- Initialize the pointer x to point the current node p.
- Count the number of nodes that are greater than the current node.
- Replace the current node with the count.
- Repeat step 4 until the list is traversed completely.
Below is the implementation of the above approach:
6 5 5 2 3 0 2 0 0
Time Complexity: O(N2) where N is the number of nodes in the linked list.
Auxiliary Space: O(1)
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- Linked List | Set 3 (Deleting a node)
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