Find pair for given sum in a sorted singly linked without extra space
Given a sorted singly linked list and a value x, the task is to find pair whose sum is equal to x. We are not allowed to use any extra space and expected time complexity is O(n).
Input : head = 3-->6-->7-->8-->9-->10-->11 , x=17 Output: (6, 11), (7, 10), (8, 9)
Hint : We are allowed to modify original linked list
A simple solution for this problem is to take each element one by one and traverse the remaining list in forward direction to find second element whose sum is equal to given value x. Time complexity for this approach will be O(n2).
An efficient solution for this problem is based on ideas discussed in below articles.
Find pair in doubly linked list : We use the same algorithm that traverses linked list from both ends.
XOR Linked list : In singly linked list, we can traverse list only in forward direction. We use XOR concept to convert a singly linked list to doubly linked list.
Below are steps :
- First we need to convert our singly linked list into doubly linked list. Here we are given singly linked list structure node which have only next pointer not prev pointer, so to convert our singly linked list into doubly linked list we use memory efficient doubly linked list ( XOR linked list ).
- In XOR linked list each next pointer of singly linked list contains XOR of next and prev pointer.
- After converting singly linked list into doubly linked list we initialize two pointers variables to find the candidate elements in the sorted doubly linked list. Initialize first with start of doubly linked list i.e; first = head and initialize second with last node of doubly linked list i.e; second = last_node.
- Here we don’t have random access, so to initialize pointer, we traverse the list till last node and assign last node to second.
- If current sum of first and second is less than x, then we move first in forward direction. If current sum of first and second element is greater than x, then we move second in backward direction.
- Loop termination conditions are also different from arrays. The loop terminates when either of two pointers become NULL, or they cross each other (first=next_node), or they become same (first == second).
(6,11) , (7,10) , (8,9)
Time complexity : O(n)
If linked list is not sorted, then we can sort the list as a first step. But in that case overall time complexity would become O(n Log n). We can use Hashing in such cases if extra space is not a constraint. The hashing based solution is same as method 2 here.
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