Given three linked lists, say a, b and c, find one node from each list such that the sum of the values of the nodes is equal to a given number.
For example, if the three linked lists are 12->6->29, 23->5->8, and 90->20->59, and the given number is 101, the output should be triple “6 5 90”.
In the following solutions, size of all three linked lists is assumed same for simplicity of analysis. The following solutions work for linked lists of different sizes also.
A simple method to solve this problem is to run three nested loops. The outermost loop picks an element from list a, the middle loop picks an element from b and the innermost loop picks from c. The innermost loop also checks whether the sum of values of current nodes of a, b and c is equal to given number. The time complexity of this method will be O(n^3).
Sorting can be used to reduce the time complexity to O(n*n). Following are the detailed steps.
1) Sort list b in ascending order, and list c in descending order.
2) After the b and c are sorted, one by one pick an element from list a and find the pair by traversing both b and c. See isSumSorted() in the following code. The idea is similar to Quadratic algorithm of 3 sum problem.
Following code implements step 2 only. The solution can be easily modified for unsorted lists by adding the merge sort code discussed here.
// C++ program to find a triplet // from three linked lists with // sum equal to a given number #include <bits/stdc++.h> using namespace std;
/* Link list node */ class Node
{ public :
int data;
Node* next;
}; /* A utility function to insert a node at the beginning of a linked list*/ void push (Node** head_ref, int new_data)
{ /* allocate node */
Node* new_node = new Node();
/* put in the data */
new_node->data = new_data;
/* link the old list of the new node */
new_node->next = (*head_ref);
/* move the head to point to the new node */
(*head_ref) = new_node;
} /* A function to check if there are three elements in a, b and c whose sum is equal to givenNumber. The function assumes that the list b is sorted in ascending order and c is sorted in descending order. */ bool isSumSorted(Node *headA, Node *headB,
Node *headC, int givenNumber)
{ Node *a = headA;
// Traverse through all nodes of a
while (a != NULL)
{
Node *b = headB;
Node *c = headC;
// For every node of list a, prick two nodes
// from lists b abd c
while (b != NULL && c != NULL)
{
// If this a triplet with given sum, print
// it and return true
int sum = a->data + b->data + c->data;
if (sum == givenNumber)
{
cout << "Triplet Found: " << a->data << " " <<
b->data << " " << c->data;
return true ;
}
// If sum of this triplet is smaller, look for
// greater values in b
else if (sum < givenNumber)
b = b->next;
else // If sum is greater, look for smaller values in c
c = c->next;
}
a = a->next; // Move ahead in list a
}
cout << "No such triplet" ;
return false ;
} /* Driver code*/ int main()
{ /* Start with the empty list */
Node* headA = NULL;
Node* headB = NULL;
Node* headC = NULL;
/*create a linked list 'a' 10->15->5->20 */
push (&headA, 20);
push (&headA, 4);
push (&headA, 15);
push (&headA, 10);
/*create a sorted linked list 'b' 2->4->9->10 */
push (&headB, 10);
push (&headB, 9);
push (&headB, 4);
push (&headB, 2);
/*create another sorted
linked list 'c' 8->4->2->1 */
push (&headC, 1);
push (&headC, 2);
push (&headC, 4);
push (&headC, 8);
int givenNumber = 25;
isSumSorted (headA, headB, headC, givenNumber);
return 0;
} // This code is contributed by rathbhupendra |
Output:
Triplet Found: 15 2 8
Time complexity: The linked lists b and c can be sorted in O(nLogn) time using Merge Sort (See this). The step 2 takes O(n*n) time. So the overall time complexity is O(nlogn) + O(nlogn) + O(n*n) = O(n*n).
In this approach, the linked lists b and c are sorted first, so their original order will be lost. If we want to retain the original order of b and c, we can create copy of b and c.
Please refer complete article on Find a triplet from three linked lists with sum equal to a given number for more details!