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Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0.Notice that the solution set must not contain duplicate triplets. Example 1:Input: nums = [-1,0,1,2,-1,-4]Output: [[-1,-1,2],[-1,0,1]]Explanation: nums[0] + nums[1] + nums[2] = (-1) + 0 + 1 = 0.nums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0.nums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0.The distinct triplets are [-1,0,1] and [-1,-1,2].Notice that the order of the output and the order of the triplets does not matter.Example 2:Input: nums = [0,1,1]Output: []Explanation: The only possible triplet does not sum up to 0.Example 3:Input: nums = [0,0,0]Output: [[0,0,0]]Explanation: The only possible triplet sums up to 0. Constraints:3 <= nums.length <= 3000-105 <= nums[i] <= 105

Question

Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0.Notice that the solution set must not contain duplicate triplets. Example 1:Input: nums = [-1,0,1,2,-1,-4]Output: [[-1,-1,2],[-1,0,1]]Explanation: nums[0] + nums[1] + nums[2] = (-1) + 0 + 1 = 0.nums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0.nums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0.The distinct triplets are [-1,0,1] and [-1,-1,2].Notice that the order of the output and the order of the triplets does not matter.Example 2:Input: nums = [0,1,1]Output: []Explanation: The only possible triplet does not sum up to 0.Example 3:Input: nums = [0,0,0]Output: [[0,0,0]]Explanation: The only possible triplet sums up to 0. Constraints:3 <= nums.length <= 3000-105 <= nums[i] <= 105

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Solution 1

Para resolver este problema, podemos usar un enfoque de dos punteros después de ordenar el array. Aquí están los pasos detallados:

  1. Ordenar el array nums.
  2. Iterar a través del array con un índice i, considerando cada elemento como un posible primer elemento del triplete.
  3. Para cada i, usar dos punteros, uno comenzando justo después de i (left = i + 1) y otro comenzando desde el final del array (right = nums.length - 1).
  4. Calcular la suma de los elementos en las posiciones i, left y right.
    • Si la suma es 0, añadir el triplete a la lista de resultados y mover ambos punteros (left y right) para buscar otros posibles tripletes.
    • Si la suma es menor que 0, mover el puntero left hacia la derecha para aumentar la suma.
    • Si la suma es mayor que 0, mover el puntero right hacia la izquierda para disminuir la suma.
  5. Evitar duplicados moviendo los punteros left y right mientras los elementos sean iguales a los anteriores.
  6. Continuar hasta que left sea mayor o igual a right.

Aquí está el código que implementa este enfoque:

def threeSum(nums):
    nums.sort()
    result = []
    for i in range(len(nums) - 2):
        if i > 0 and nums[i] == nums[i - 1]:
            continue
        left, right = i + 1, len(nums) - 1
        while left < right:
            total = nums[i] + nums

This problem has been solved

Solution 2

This problem is asking to find all unique triplets in the array which gives the sum of zero. A triplet is a set of three numbers.

Here is a step by step solution:

  1. First, sort the input array in ascending order.

  2. Then, iterate through the array and for each element, use two pointers (one at the start of the array and one at the end) to find the pair that makes the sum zero.

  3. If the sum of the numbers at the two pointers is less than zero, move the start pointer to the right (increase it). If the sum is greater than zero, move the end pointer to the left (de

This problem has been solved

Solution 3

This problem can be solved using a two-pointer technique.

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