Of course. Here is a summary of the video providing a solution to the "3Sum" LeetCode problem.

The problem asks to find all unique triplets in a given array of integers nums that add up to zero.

• Initial Approach (Brute-Force): The most straightforward way is to use three nested loops to check every possible combination of three numbers. However, this is inefficient with a time complexity of O(n³) and makes it difficult to handle duplicate triplets.

• Optimized Approach: A more efficient solution involves sorting the array first. This allows for a two-pointer technique to solve the problem in O(n²) time.

1. Sort the Array: Sorting the input array is crucial. It helps in easily skipping duplicate elements and using the two-pointer method effectively.

   nums.sort()

2. Iterate and Reduce to Two Sum: Iterate through the sorted array with a single loop. For each element a, the problem is reduced to finding two numbers in the rest of the array that sum up to -a. This is a variation of the "Two Sum" problem.

   for i, a in enumerate(nums):
       # Skip positive integers that cannot sum to zero with other positive integers
       if a > 0:
           break
       # Skip duplicate values for 'a'
       if i > 0 and a == nums[i - 1]:
           continue

3. Two-Pointer Technique: For each element a, use two pointers, l (left) and r (right), for the remainder of the array. The l pointer starts at i + 1, and the r pointer starts at the end of the array.

       l, r = i + 1, len(nums) - 1
       while l < r:
           threeSum = a + nums[l] + nums[r]

4. Adjust Pointers:

   • If the threeSum is greater than 0, decrement the r pointer to get a smaller sum.
   • If the threeSum is less than 0, increment the l pointer to get a larger sum.
   • If the threeSum is exactly 0, a triplet has been found. Add it to the result list. Then, increment l and continue skipping any subsequent duplicate values to avoid redundant triplets.

           if threeSum > 0:
               r -= 1
           elif threeSum < 0:
               l += 1
           else:
               res.append([a, nums[l], nums[r]])
               l += 1
               r -= 1
               while nums[l] == nums[l - 1] and l < r:
                   l += 1

class Solution:
  def threeSum(self, nums: list[int]) -> list[list[int]]:
    res = []
    nums.sort()

    for i, a in enumerate(nums):
      # Skip positive integers
      if a > 0:
        break

      if i > 0 and a == nums[i - 1]:
        continue

      l, r = i + 1, len(nums) - 1
      while l < r:
        threeSum = a + nums[l] + nums[r]
        if threeSum > 0:
          r -= 1
        elif threeSum < 0:
          l += 1
        else:
          res.append([a, nums[l], nums[r]])
          l += 1
          r -= 1
          while l < r and nums[l] == nums[l - 1]:
            l += 1
            
    return res