The 3Sum Problem is a popular algorithmic challenge that involves finding all unique triplets in an array of integers such that the sum of the three elements is zero. In other words, we need to find all possible combinations of three numbers in the array that add up to zero.

Step-by-Step Approach:

1. We are given an array of integers.
2. The goal is to find all unique triplets (a, b, c) such that a + b + c = 0.
3. To efficiently solve this problem, we sort the array in ascending order.
4. We use a two-pointers technique to iterate through the sorted array and find the triplets that sum up to zero.
5. We fix the first element (a) and then use two pointers, one starting from the element after a (b) and the other from the end of the array (c).
6. If the sum of a + b + c is zero, we found a valid triplet. If the sum is less than zero, we move the left pointer (b) to the right to get a larger sum. If the sum is greater than zero, we move the right pointer (c) to the left to get a smaller sum.
7. We repeat steps 5 and 6 until we have considered all possible combinations of triplets in the array.

Java Solution:

import java.util.*;

public class ThreeSum {

    public static List<List<Integer>> threeSum(int[] nums) {
        Arrays.sort(nums);
        List<List<Integer>> result = new ArrayList<>();

        for (int i = 0; i < nums.length - 2; i++) {
            if (i == 0 || (i > 0 && nums[i] != nums[i - 1])) {
                int left = i + 1, right = nums.length - 1, target = -nums[i];

                while (left < right) {
                    int sum = nums[left] + nums[right];

                    if (sum == target) {
                        result.add(Arrays.asList(nums[i], nums[left], nums[right]));

                        while (left < right && nums[left] == nums[left + 1])
                            left++;
                        while (left < right && nums[right] == nums[right - 1])
                            right--;

                        left++;
                        right--;
                    } else if (sum < target) {
                        left++;
                    } else {
                        right--;
                    }
                }
            }
        }

        return result;
    }

    public static void main(String[] args) {
        int[] nums = {-1, 0, 1, 2, -1, -4};
        List<List<Integer>> triplets = threeSum(nums);

        for (List<Integer> triplet : triplets) {
            System.out.println(triplet);
        }
    }
}

Python Solution:

def three_sum(nums):
    nums.sort()
    result = []

    for i in range(len(nums) - 2):
        if i == 0 or (i > 0 and nums[i] != nums[i - 1]):
            left, right, target = i + 1, len(nums) - 1, -nums[i]

            while left < right:
                sum = nums[left] + nums[right]

                if sum == 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 < target:
                    left += 1
                else:
                    right -= 1

    return result

if __name__ == "__main__":
    nums = [-1, 0, 1, 2, -1, -4]
    triplets = three_sum(nums)
    print(triplets)

Explanation of Java/Python Solution:
The Java and Python solutions implement the 3Sum Problem using a two-pointer technique to find all unique triplets in the given array that sum up to zero.

The threeSum function in Java and three_sum function in Python accept an array nums as input and return a list of lists containing the unique triplets that satisfy the sum condition.

The provided example array nums = [-1, 0, 1, 2, -1, -4] demonstrates how the two-pointer technique efficiently finds the triplets [[-1, -1, 2], [-1, 0, 1]] with the sum of zero. These triplets are the valid solutions to the 3Sum Problem for the given input array.

Explanation of 3Sum Problem:

The 3Sum Problem is a popular algorithmic challenge that involves finding all unique triplets in an array of integers such that the sum of the three elements is zero. In other words, we need to find all possible combinations of three numbers in the array that add up to zero.

Step-by-Step Approach:

1. We are given an array of integers.
2. The goal is to find all unique triplets (a, b, c) such that a + b + c = 0.
3. To efficiently solve this problem, we sort the array in ascending order.
4. We use a two-pointers technique to iterate through the sorted array and find the triplets that sum up to zero.
5. We fix the first element (a) and then use two pointers, one starting from the element after a (b) and the other from the end of the array (c).
6. If the sum of a + b + c is zero, we found a valid triplet. If the sum is less than zero, we move the left pointer (b) to the right to get a larger sum. If the sum is greater than zero, we move the right pointer (c) to the left to get a smaller sum.
7. We repeat steps 5 and 6 until we have considered all possible combinations of triplets in the array.

Java Solution:

import java.util.*;

public class ThreeSum {

    public static List<List<Integer>> threeSum(int[] nums) {
        Arrays.sort(nums);
        List<List<Integer>> result = new ArrayList<>();

        for (int i = 0; i < nums.length - 2; i++) {
            if (i == 0 || (i > 0 && nums[i] != nums[i - 1])) {
                int left = i + 1, right = nums.length - 1, target = -nums[i];

                while (left < right) {
                    int sum = nums[left] + nums[right];

                    if (sum == target) {
                        result.add(Arrays.asList(nums[i], nums[left], nums[right]));

                        while (left < right && nums[left] == nums[left + 1])
                            left++;
                        while (left < right && nums[right] == nums[right - 1])
                            right--;

                        left++;
                        right--;
                    } else if (sum < target) {
                        left++;
                    } else {
                        right--;
                    }
                }
            }
        }

        return result;
    }

    public static void main(String[] args) {
        int[] nums = {-1, 0, 1, 2, -1, -4};
        List<List<Integer>> triplets = threeSum(nums);

        for (List<Integer> triplet : triplets) {
            System.out.println(triplet);
        }
    }
}

Python Solution:

def three_sum(nums):
    nums.sort()
    result = []

    for i in range(len(nums) - 2):
        if i == 0 or (i > 0 and nums[i] != nums[i - 1]):
            left, right, target = i + 1, len(nums) - 1, -nums[i]

            while left < right:
                sum = nums[left] + nums[right]

                if sum == 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 < target:
                    left += 1
                else:
                    right -= 1

    return result

if __name__ == "__main__":
    nums = [-1, 0, 1, 2, -1, -4]
    triplets = three_sum(nums)
    print(triplets)

Explanation of Java/Python Solution:
The Java and Python solutions implement the 3Sum Problem using a two-pointer technique to find all unique triplets in the given array that sum up to zero.

The threeSum function in Java and three_sum function in Python accept an array nums as input and return a list of lists containing the unique triplets that satisfy the sum condition.

The provided example array nums = [-1, 0, 1, 2, -1, -4] demonstrates how the two-pointer technique efficiently finds the triplets [[-1, -1, 2], [-1, 0, 1]] with the sum of zero. These triplets are the valid solutions to the 3Sum Problem for the given input array.