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 <= 3000-10^5 <= nums[i] <= 10^5func threeSum(nums []int) [][]int { h := map[int]int{} for _, num := range nums { h[num]++ } // fmt.Println("H", h) uniq := []int{} for k := range h { uniq = append(uniq, k) } // fmt.Println("Uniq", uniq) result := [][]int{} for i := 0; i < len(uniq); i++ { a := uniq[i] for j := i; j < len(uniq); j++ { b := uniq[j] c := -a - b if c >= a && c >= b { counts := map[int]int{} counts[a]++ counts[b]++ counts[c]++ canAdd := true for k, v := range counts { if v > h[k] { canAdd = false break } } if !canAdd { continue } result = append(result, []int{a, b, c}) } } } return result }