3Sum

 

  1. Clarify the problem:

    • The problem requires us to find all unique triplets in an array that add up to zero.
    • We need to implement a function that takes an array of integers as input and returns a list of unique triplets that satisfy the given condition.

  2. Analyze the problem:

    • Input: An array of integers.
    • Output: A list of unique triplets (lists) that add up to zero.
    • Constraints:
      • The length of the input array is in the range [0, 3000].
      • Each integer in the array is in the range [-10^5, 10^5].
      • The solution set must not contain duplicate triplets.

  3. Design an algorithm:

    • We can use a two-pointer approach to solve the problem.
    • First, we sort the input array in ascending order.
    • Then, we iterate through the array and for each element nums[i], we set two pointers, left and right, at positions i+1 and n-1, respectively, where n is the length of the array.
    • While left is less than right, we perform the following steps:
      • Calculate the sum total = nums[i] + nums[left] + nums[right].
      • If total is equal to zero, we have found a valid triplet. We add it to the result list.
      • If total is less than zero, we increment left to move towards larger values.
      • If total is greater than zero, we decrement right to move towards smaller values.
      • If total is zero, we continue incrementing left or decrementing right to find the next distinct triplet.
    • We continue this process for each element in the array.
    • Finally, we return the result list containing all the unique triplets.

  4. Explain your approach:

    • We will implement a function called threeSum to solve the problem.
    • The function will take an array of integers, nums, as input.
    • We will first check if the length of nums is less than 3. If so, we will return an empty list.
    • We will sort the nums array in ascending order.
    • We will create an empty list called result to store the unique triplets.
    • We will iterate through the nums array up to the third-to-last element.
    • For each element nums[i], we will set two pointers, left and right, at positions i+1 and n-1, respectively, where n is the length of the array.
    • While left is less than right, we will calculate the sum total = nums[i] + nums[left] + nums[right] and perform the necessary comparisons to find the triplets.
    • We will handle duplicate values by skipping them using conditional checks.
    • We will return the result list containing all the unique triplets.

  5. Write clean and readable code:

    python
  6. def threeSum(nums):
    if len(nums) < 3:
        return []

    nums.sort()
    result = []

    for i in range(len(nums) - 2):
        if i > 0 and nums[i] == nums[i - 1]:
            continue

        left = i + 1
        right = len(nums) - 1

        while left < right:
            total = nums[i] + nums[left] + nums[right]

            if total == 0:
                result.append([nums[i], nums[left], nums[right]])
                while left < right and nums[left] == nums[left + 1]:
                    left += 1
                while left < right and nums[right] == nums[right - 1]:
                    right -= 1
                left += 1
                right -= 1
            elif total < 0:
                left += 1
            else:
                right -= 1

    return result

  7. Test your code:

    python
  8. # Test case 1
nums = [-1, 0, 1, 2, -1, -4]
# The unique triplets that add up to zero are [-1, -1, 2] and [-1, 0, 1]
assert threeSum(nums) == [[-1, -1, 2], [-1, 0, 1]]

# Test case 2
nums = []
# Empty input, so the output should be an empty list
assert threeSum(nums) == []

# Test case 3
nums = [0, 0, 0]
# The unique triplet [0, 0, 0] adds up to zero
assert threeSum(nums) == [[0, 0, 0]]

# Test case 4
nums = [1, 2, -2, -1]
# There are no unique triplets that add up to zero
assert threeSum(nums) == []

# Test case 5
nums = [1, -1, -1, 0]
# The unique triplet [1, -1, 0] adds up to zero
assert threeSum(nums) == [[-1, 0, 1]]

    We have tested the code with multiple test cases, including cases with positive, negative, and zero integers, as well as empty input. The output for each test case matches the expected results.

  9. Optimize if necessary:

    • The current solution has a time complexity of O(N^2), where N is the length of the input array.
    • This is because we have nested loops to iterate through the array and find the triplets.
    • The space complexity is O(N) because we use additional space to store the result list.
    • The code can be further optimized by avoiding duplicate calculations and unnecessary iterations.

  10. Handle error cases:

    • The code handles the case when the input array has less than three elements and returns an empty list.
    • The code assumes that the input is a valid list of integers.

  11. Discuss complexity analysis:

    • Let N be the length of the input array.
    • The time complexity of the solution is O(N^2) because we have nested loops to iterate through the array and find the triplets.
    • The space complexity is O(N) because we use additional space to store the result list.
    • The code can be further optimized to avoid duplicate calculations and unnecessary iterations, which would reduce the constant factors in the time complexity.
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