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Code Circular Linked List

Replace each node of circular doubly linked list by sum of other nodes

In this problem, we are dealing with a circular doubly linked list that contains integer values. The task is to replace each node of the circular doubly linked list with the sum of the other nodes in the list. Specifically, for each node, we need to compute the sum of all the nodes' data except the current node itself and then update the current node's data with that computed sum.

Problem Statement and Description

The problem requires us to modify the circular doubly linked list by replacing each node's data with the sum of all the other nodes' data. For each node, we exclude its own data while calculating the sum of the other nodes' data.

Example

Let's understand the problem with an example. Consider the following circular doubly linked list:

Original Circular Doubly Linked List:
From Front to Tail: 5 -> 2 -> 3 -> 4 -> 1
From Tail to Front: 1 -> 4 -> 3 -> 2 -> 5

After replacing each node with the sum of the other nodes' data, the circular doubly linked list should look like this:

Modified Circular Doubly Linked List:
From Front to Tail: 10 -> 13 -> 12 -> 11 -> 14
From Tail to Front: 14 -> 11 -> 12 -> 13 -> 10

Idea to Solve

To solve this problem, we can follow these steps:

  1. Traverse the circular doubly linked list once to compute the sum of all node data.
  2. Traverse the circular doubly linked list again. For each node, compute the sum of all other node data by subtracting the current node's data from the total sum computed in step 1.
  3. Update the current node's data with the computed sum.
  4. Repeat steps 2 and 3 for each node in the circular doubly linked list.

Pseudocode

changeByOtherSum():
    if front is null or front.next is front:
        return
    initialize temp as front.next
    initialize sum as front.data
    
    // Compute the sum of all node data
    while temp is not front:
        sum += temp.data
        temp = temp.next
    
    // Traverse again to update node data
    initialize temp as front.next
    while temp is not front:
        temp.data = sum - temp.data
        temp = temp.next

Algorithm Explanation

  1. Traverse the circular doubly linked list using the temp pointer to compute the total sum of all node data.
  2. Traverse the circular doubly linked list again using the temp pointer. For each node: a. Compute the sum of all other node data by subtracting the current node's data from the total sum calculated earlier. b. Update the current node's data with the computed sum.
  3. If temp reaches back to the front, terminate the loop.

Code Solution

Time Complexity

  1. In the first traversal of the circular doubly linked list, we calculate the sum of all node data. This requires traversing the entire list once, resulting in a time complexity of O(N), where N is the number of nodes in the linked list.

  2. In the second traversal of the circular doubly linked list, we update each node's data. Again, this operation requires traversing the entire list once, resulting in a time complexity of O(N).

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