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3sum problem (finding triplets) in better than o (n^2) asked 13 years, 5 months ago modified 2 years, 7 months ago viewed 6k times I'm having trouble making my algorithm more time efficient. This looks like the standard approach to solving 3sum in quadratic time
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However, i disagree with the other answers concerning space complexity and believe it is quadratic as there can be quadratically many distinct triples summing to 0. I'm reviewing time complexity for some interviews & I'm studying the 3 sum to implement it on my own, and came across the following implementation with the rules
Given an array s of n integers, are there elements a, b, c in s such that a + b + c =.
Given an array a of integers, find any 3 of them that sum to any given t I saw this on some online post, which claims it has a o(nlogn) solution For 2 numbers, i know hashtable could help for o. Solving the leetcode 3sum problem in python asked 7 years ago modified 7 years ago viewed 2k times
Asked3 years, 3 months ago modified 2 years, 7 months ago viewed 619 times 2 in the scenario i present to you, my solution is supposed to represent o (n^2 * log n), and the pointers solution, which i assume is the fastest way to resolve the 3sum problem, represents o (n^2 * 1) Leaving the question of is o (1) faster than o (log n), exampling it with my code. 4 i am working on the 3sum problem (taken from leetcode), which takes a list as input and finds all unique triplets in the lists such that a+b+c=0 4sum and 3sum are very similar
The difference is that we are looking for unique quadruplets instead of triplets
Following a similar logic, we can implement 5sum by wrapping 4sum in another loop But what about the 6sum, 7sum and so on It's better we should go for ksum solution. How do i return all possible triplets in the 3sum problem
Asked 3 years, 5 months ago modified 2 years, 6 months ago viewed 2k times How can i make use of binary search for improving my algorithms time complexity
