If the sum is greater than the given value, decrement the right pointer. If the sum is smaller than the given value, increment the count and the left pointer. Then use the two pointers approach, left pointer starts from 0 and right pointer starts from end.įor every iteration, calculate the sum of the current triplet (element pointed by left + element pointed by right + current element). Initialize three variables: left, right, and count. Approachįirst, sort the given array of numbers in ascending order. In the end, we will be returning the count of such triplets with a sum less than the given value. It is a very useful technique to keep in mind for future problems where we need to find the count of some combinations that satisfy a certain condition. This way, we can efficiently solve the problem in O(n^2) time complexity. The number of such triplets will be the count we will be keeping track of.Īdditionally, we will be updating the count and the pointers based on the sum of the triplet being less or equal to the given value. Firstly, we will sort the array in ascending order and then, for each element in the array, we will use two pointers to check for the triplets with a sum less than the given value. This problem can be solved by sorting the array and using two pointers to check for the possible combinations. We will be writing a JavaScript program to count the number of triplets with a sum smaller than a given value.

Javascript for each array