3sum

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15. 3Sum (Medium)

Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0.

Notice that the solution set must not contain duplicate triplets.

 

Example 1:

Input: nums = [-1,0,1,2,-1,-4]
Output: [[-1,-1,2],[-1,0,1]]

Example 2:

Input: nums = []
Output: []

Example 3:

Input: nums = [0]
Output: []

 

Constraints:

• 0 <= nums.length <= 3000
• -105 <= nums[i] <= 105

Related Topics

[Array] [Two Pointers] [Sorting]

Similar Questions

1. Two Sum (Easy)
2. 3Sum Closest (Medium)
3. 4Sum (Medium)
4. 3Sum Smaller (Medium)

Hints

Hint 1

So, we essentially need to find three numbers x, y, and z such that they add up to the given value. If we fix one of the numbers say x, we are left with the two-sum problem at hand!

Hint 2

For the two-sum problem, if we fix one of the numbers, say

x

, we have to scan the entire array to find the next number

y

which is

value - x

where value is the input parameter. Can we change our array somehow so that this search becomes faster?

Hint 3

The second train of thought for two-sum is, without changing the array, can we use additional space somehow? Like maybe a hash map to speed up the search?