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).

Examples:

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(n^{2}).

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).

// C++ program to find pair with given sum in a singly // linked list in O(n) time and no extra space. #include<bits/stdc++.h> using namespace std; /* Link list node */ struct node { int data; /* also constains XOR of next and previous node after conversion*/ struct node* next; }; /* Given a reference (pointer to pointer) to the head of a list and an int, push a new node on the front of the list. */ void insert(struct node** head_ref, int new_data) { /* allocate node */ struct node* new_node = (struct node*) malloc(sizeof(struct node)); /* put in the data */ new_node->data = new_data; /* link the old list off the new node */ new_node->next = (*head_ref); /* move the head to point to the new node */ (*head_ref) = new_node; } /* returns XORed value of the node addresses */ struct node* XOR (struct node *a, struct node *b) { return (struct node*) ((uintptr_t) (a) ^ (uintptr_t) (b)); } // Utility function to convert singly linked list // into XOR doubly linked list void convert(struct node *head) { // first we store address of next node in it // then take XOR of next node and previous node // and store it in next pointer struct node *next_node; // prev node stores the address of previously // visited node struct node *prev = NULL; // traverse list and store xor of address of // next_node and prev node in next pointer of node while (head != NULL) { // address of next node next_node = head->next; // xor of next_node and prev node head->next = XOR(next_node, prev); // update previous node prev = head; // move head forward head = next_node; } } // function to Find pair whose sum is equal to // given value x void pairSum(struct node *head, int x) { // initialize first struct node *first = head; // next_node and prev node to calculate xor again // and find next and prev node while moving forward // and backward direction from both the corners struct node *next_node = NULL, *prev = NULL; // traverse list to initialize second pointer // here we need to move in forward direction so to // calculate next address we have to take xor // with prev pointer because (a^b)^b = a struct node *second = head; while (second->next != prev) { struct node *temp = second; second = XOR(second->next, prev); prev = temp; } // now traverse from both the corners next_node = NULL; prev = NULL; // here if we want to move forward then we must // know the prev address to calculate next node // and if we want to move backward then we must // know the next_node address to calculate prev node bool flag = false; while (first != NULL && second != NULL && first != second && first != next_node) { if ((first->data + second->data)==x) { cout << "(" << first->data << "," << second->data << ")" << endl; flag = true; // move first in forward struct node *temp = first; first = XOR(first->next,prev); prev = temp; // move second in backward temp = second; second = XOR(second->next, next_node); next_node = temp; } else { if ((first->data + second->data) < x) { // move first in forward struct node *temp = first; first = XOR(first->next,prev); prev = temp; } else { // move second in backward struct node *temp = second; second = XOR(second->next, next_node); next_node = temp; } } } if (flag == false) cout << "No pair found" << endl; } // Driver program to run the case int main() { /* Start with the empty list */ struct node* head = NULL; int x = 17; /* Use insert() to construct below list 3-->6-->7-->8-->9-->10-->11 */ insert(&head, 11); insert(&head, 10); insert(&head, 9); insert(&head, 8); insert(&head, 7); insert(&head, 6); insert(&head, 3); // convert singly linked list into XOR doubly // linked list convert(head); pairSum(head,x); return 0; }

Output:

(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.

This article is contributed by **Shashank Mishra ( Gullu )**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.