3sum

Link: https://leetcode.com/problems/3sum/

Solution:

Topics: two pointer, Two sum II, two sum

Intuition
This is an interesting problem, I had not done Two sum II previously so I initially solved it with the hash map solution which is very memory intensive and difficult to avoid duplicates. The two pointer solution is the perfect algorithm here. Note that the brute force solution is to generate all possible triplets which is n**3, so n**2 should be the goal…I spent a bit of time debating if there is a linear solution without first considering if n**2 is the best we can achieve.

The key here is to use the same technique as Two sum II, but for each element as the possible starting point (hence n**2). We can eliminate duplicates by either eliminating all the duplicates in the array beforehand, or shifting pointers in our algorithm until a new value has been reached (the array must first be sorted of course).

Implementation

def three_sum(nums):
	nums.sort()
	res = []
	start = 0
	while start < len(nums)-2:
		l = start+1
		r = len(nums)-1
		while l < r:
			total = nums[start] + nums[l] + nums[r]
			if total == 0:
				res.append((nums[start], nums[l], nums[r]))
				r -= 1
				l += 1
				while l < r and nums[l] == nums[l-1]:
					l += 1 
			elif total > 0:
				r -= 1
			else:
				l += 1
				
		start += 1
		while start < len(nums)-2 and nums[start] == nums[start-1]:
			start += 1
	return res
		
#time: o(n**2)
#memory: o(1)

Mnemonic
Same as Two sum II but imagine that we have a third person and the target is the average of all 3 laser lines.

Review 1
Great problem. Remember two things:

1. Use while loops
2. while nums[i] == nums[i+1] increment i. Remember to increment once more after the loop!

review
hard