The problem requires us to find all unique triplets in an array that sum up to zero. Given the constraints and the need to avoid duplicates, a sorted array and a two-pointer approach are suitable. Sorting helps in easily skipping duplicates and efficiently using two pointers to find pairs that sum up to a target.
- Sort the Array: In order to make it easier to avoid duplicates and use the two-pointer technique.
- Iterate Through the Array:
- Use a loop to fix one element of the triplet. This element will be the current element in the loop.
- Skip duplicate elements to ensure the uniqueness of the triplets.
- Two-Pointer Technique:
- For each fixed element, use two pointers (
left
andright
) to find pairs that sum up to the negative of the fixed element. - Adjust the pointers based on the sum of the triplet:
- If the sum is less than zero, move the left pointer to the right to increase the sum.
- If the sum is greater than zero, move the right pointer to the left to decrease the sum.
- If the sum is zero, add the triplet to the result list and move both pointers, skipping duplicates.
- For each fixed element, use two pointers (
- Return the Result: After processing all elements, return the list of unique triplets.
- Time Complexity:
O(N^2)
, whereN
is the number of elements in the array. Sorting the array takesO(NlogN)
, and the two-pointer technique takesO(N^2)
in the worst case. - Space Complexity:
O(1)
orO(N)
depending on the implementation of sorting the input.