Vidal Sassoon Colorfinity 2 Minute Shadeprecision Treatment for Rich Darks

Given a list of numbers, the goal is to find the three numbers that sum to a target value. This problem is known as the "3Sum" problem and is a popular coding interview question.

The approach to solving this problem involves sorting the input list and then using a two-pointer technique. The algorithm proceeds as follows:

1. Sort the input list in ascending order.
2. Initialize three pointers: i, left, and right. Set i to the first index, left to i+1, and right to the last index.
3. Calculate the sum of the elements at the three pointers: sum = nums[i] + nums[left] + nums[right].
4. If the sum is equal to the target value, add the triplet (nums[i], nums[left], nums[right]) to the result list.
5. If the sum is less than the target value, increment the left pointer to increase the sum.
6. If the sum is greater than the target value, decrement the right pointer to decrease the sum.
7. Repeat steps 3-6 until left is less than right.
8. Increment i and repeat steps 3-7 until i reaches the second-to-last index.

Here's the Python code that implements this approach:

```python def three_sum(nums, target): result = [] nums.sort() n = len(nums)

for i in range(n-2):
    if i > 0 and nums[i] == nums[i-1]:
        continue
    left, right = i+1, n-1
    while left < right:
        sum_value = nums[i] + nums[left] + nums[right]
        if sum_value == target:
            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 sum_value < target:
            left += 1
        else:
            right -= 1
return result

```

The time complexity of this solution is O(n^2), where n is the length of the input list, as we iterate through the list twice (sorting and two-pointer). The space complexity is O(1), as we only use a constant amount of extra space to store the result list.

This approach is efficient and can handle large input lists. It also avoids duplicates by skipping over elements that have already been used in a triplet.

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