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Task

Solution

Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.

Note: The solution set must not contain duplicate triplets.

For example, given array S = [-1, 0, 1, 2, -1, -4],

A solution set is:
[
  [-1, 0, 1],
  [-1, -1, 2]
]

Presentation

Solutions

  • Time complexity: O(n!). Solution. Shortest solution. Implementation is based on permutations.
  • Time complexity: O(n^4). Implementation is based on permutations. Use additional binary array (size is equal to original array) with only 3 elements equal to 1. Custom permutation implementation can reduce complexity to O(n^3).
  • Time complexity: O(n^2 * log n). Solution. Brute-force through pairs of values, find a third one with binary search. Asymptote is O(n^2 * log^2 n) due to deduplicating returns by storing them into a std::set. Input sequence is left uncompressed.
  • Time complexity: O(n^2 * log n). Solution. Sort the array, consider all pairs (one iterator is set to the begin and the second iterator is set to the end), find the third element for triplet with binary_search between iterators. If iterator is moved to the repeated value, skip it.
  • Time complexity: O(n^2 * log n). Solution. Sort the array and reduce task to 2-sum problem. Move first pointer k from left to right and solve 2-sum problem for other elements: [i] + [j] = -[k]. Move pointers i and j with binary search.
  • Time complexity: O(n^2). Solution.
    1. Sort input array.
    2. Obtain 2 vectors from input one: first contains sorted unique elements, second - counts of corresponding elements in input vector.
    3. Additional vector 'max_corresp_candidat_idx' for storage min checked third triple element index for a second element in triplet (initialized with nums.size() - 1).
    4. Check triplets of elements with 3 indices: idx[0] is increasing from min element index to max negative index, idx[1] is increasing from idx[0] and stops increasing when triplets is found or when idx[1] == idx[2], idx[2] is decreasing from max_corresp_candidat_idx[idx[1]] and stops decreasing when triples is found or when idx[1] == idx[2].
  • Time complexity: O(n^2). Solution, faster std::unordered_map solution. Use std::unorderd_map to store elements and their quantity. Two loops iterate over elements i and j. The third element -[i] -[j] is found in the unordered_map.
  • Time complexity: O(n^2). Solution. Sort the array and reduce task to 2-sum problem. Move first pointer k from left to right and solve 2-sum problem for other elements: [i] + [j] = -[k].