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.
Javascript
<script> // Javascript program to find a triplet from three linked lists with // sum equal to a given number /* Linked list Node*/class Node { constructor(d) { this.data = d; this.next = null; } } /* 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. */function isSumSorted(la,lb,lc,givenNumber) { let a = la; // Traverse all nodes of la while (a != null) { let b = lb; let c = lc; // for every node in la pick 2 nodes from lb and lc while (b != null && c!=null) { let sum = a.data + b.data + c.data; if (sum == givenNumber) { document.write("Triplet found " + a.data + " " + b.data + " " + c.data+"<br>"); return true; } // If sum is smaller then look for greater value of b else if (sum < givenNumber) b = b.next; else c = c.next; } a = a.next; } document.write("No Triplet found<br>"); return false; } /* 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. */function push(head_ref,new_data) { /* 1 & 2: Allocate the Node & Put in the data*/ let new_node = new Node(new_data); /* 3. Make next of new Node as head */ new_node.next = (head_ref); (head_ref) = new_node; return head_ref; } let headA =null; headA = push (headA, 20) headA = push (headA, 4) headA = push (headA, 15) headA = push (headA, 10) // create a sorted linked list 'b' 2.4.9.10 let headB = null; headB = push (headB, 10) headB = push (headB, 9) headB = push (headB, 4) headB = push (headB, 2) // create another sorted // linked list 'c' 8.4.2.1 let headC = null; headC = push (headC, 1) headC = push (headC, 2) headC = push (headC, 4) headC = push (headC, 8) let givenNumber = 25 isSumSorted (headA, headB, headC, givenNumber) // This code is contributed by avanitrachhadiya2155 </script> |
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!
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