![]() ![]() Now imagine if the given number will be high like 25, then there is no set of three numbers present that will satisfy the condition. If we pick 1 from the first list, seven from the second list, and 2 from the last list.Īlso, we can pick 2 from the last list, 3 from the second, and 5 from the first list. If the given number is 10 and the linked lists are as follows − List 1: 1 -> 2 -> 3 -> 4 -> 5įrom the above given linked list, we can pick the following sets which have a sum equal to 10. If the sum of the numbers that are picked is equal to the given number, then we have to return yes otherwise false. Here in this problem, we are given three linked lists and a number and we have to pick exactly one element from all the linked lists and the sum of the picked number must be equal to the given number. This problem is the variation of the standard problem of three sums where we are given three arrays and we have to find if there is any triplet present in the array with a sum exactly equal to the given number. Let’s see the problem and implement its code along the key points of the problem. This problem is a kind of variation of the standard and famous three-sum problem but in a linked list manner. In this article, we are going to implement a JavaScript program for finding a triplet from three linked lists with a sum equal to a given number. ![]()
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