ronith
7/23/2018 - 3:14 AM

## Find a triplet from three linked lists with sum equal to a given number

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 tripel “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.
``````// https://www.geeksforgeeks.org/find-a-triplet-from-three-linked-lists-with-sum-equal-to-a-given-number/
#include <iostream>
#include <list>
using namespace std;

int data;
};

void print (node* n) {
while(n) {
cout<< n->data<< "->";
n = n->next;
}
cout<< "NULL";
}

void insert(node **headref, int n) {
node* temp = new node;
temp->data = n;
temp->next = NULL;
if (last == NULL) {
return;
}
while (last->next)
last = last->next;
last->next = temp;
return;
}

while (temp1) {
while (temp2&&temp3) {
if (temp1->data + temp2->data + temp3->data == n) {
cout<< "Triplet are :" << temp1->data << " " << temp2->data << " " << temp3->data << "\n";
temp2=temp2->next;
temp3=temp3->next;
}
else if (temp1->data + temp2->data + temp3->data < n)
temp2=temp2->next;
else
temp3=temp3->next;
}
temp1=temp1->next;
}
}

int main() {
int n; //input such that LL1 can be in any order, LL2 has to be in ascending order, LL3 has to be in descending order.
while (true) {
cout<< "Enter the number: ";
cin>>n;
if (n==-9)
break;
else
}
while (true) {
cout<< "Enter the number: ";
cin>>n;
if (n==-9)
break;
else
}