Three Sum
Given an integer array nums, return all unique triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, j != k, and nums[i] + nums[j] + nums[k] == 0.
The solution set must not contain duplicate triplets. Output each triplet sorted, one per line, with triplets sorted lexicographically.
Example:
-1,0,1,2,-1,-4
-1 -1 2 -1 0 1
- The input array
numsis first sorted to apply the two-pointer technique: [−4,−1,−1,0,1,2]. - We iterate over the array and for each element, we use two pointers, one starting from the next element and one from the end, to find a pair that sums to −nums[i].
- For nums[0]=−4, no triplet sums to 0, but for nums[1]=−1, we find −1+0+1=0 and for nums[2]=−1, we find −1+−1+2=0.
- These two unique triplets [−1,0,1] and [−1,−1,2] are then sorted and output, resulting in the given sample output.
Constraints:
- 3 <= len(nums) <= 3000
- -10^5 <= nums[i] <= 10^5
Background Knowledge
The Two Pointers technique is a fundamental concept in solving array and string problems. It involves using two pointers, typically starting from the beginning and end of the array, to traverse the data structure. This technique is useful for finding pairs or triplets that satisfy certain conditions. In the context of the "Three Sum" problem, we need to find triplets that sum up to zero.
The key concept here is to understand how to avoid duplicate triplets and ensure that the output is sorted lexicographically. This requires a good understanding of sorting and hashing techniques. We can use sorting to arrange the triplets in lexicographical order, and hashing can help us keep track of unique triplets. Additionally, we need to consider the constraint that i != j, i != k, and j != k, which means we cannot use the same element more than once in a triplet.
To tackle this problem, it's essential to have a solid grasp of array manipulation and iteration techniques. We need to iterate over the array, fix one element, and then use the two pointers technique to find the other two elements that sum up to the remaining target value. This requires a good understanding of looping and conditional statements.
Algorithm/Approach
The general approach to solving this type of problem involves using a combination of sorting, hashing, and the two pointers technique. We can start by sorting the input array, which allows us to apply the two pointers technique efficiently. Then, we can iterate over the array, fixing one element at a time, and use the two pointers to find the other two elements that satisfy the condition.
Step-by-Step Strategy
To implement the solution, follow these steps:
- Sort the input array nums in ascending order.
- Iterate over the array, fixing one element nums[i] at a time.
- Use the two pointers technique to find the other two elements nums[j] and nums[k] that sum up to the remaining target value -nums[i].
- Apply the constraints i != j, i != k, and j != k to ensure unique triplets.
- Store the unique triplets in a result set, ensuring that each triplet is sorted in ascending order.
- Finally, sort the result set lexicographically to obtain the final output.
Common Pitfalls
When implementing the solution, watch out for the following common pitfalls:
- Failing to sort the input array, which can lead to incorrect results.
- Not applying the constraints correctly, resulting in duplicate triplets.
- Not handling edge cases, such as an empty input array or an array with fewer than three elements.
Time & Space Complexity
The expected time complexity for this problem is O(n2), where n is the length of the input array. This is because we are using a nested loop structure to iterate over the array. The space complexity is O(n), as we need to store the result set, which can contain up to n unique triplets. However, the actual space complexity may be lower if we use a hashing technique to store the unique triplets.