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
For each num in the array, find two other numbers from the rest of the array whose sum is -num
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class Solution:
def threeSum(self, nums: List[int]) -> List[List[int]]:
res, dups = set(), set()
seen = {}
for i, v1 in enumerate(nums):
# dup case is already coverd by the prev iterations
if v1 not in dups:
dups.add(v1)
for j, v2 in enumerate(nums[i+1:]):
v3 = -(v1 + v2)
if v3 in seen and seen[v3] == i:
# seen in the same outer iteration
res.add(tuple(sorted([v1, v2, v3])))
seen[v2] = i # label the outer iteration idx
return res
'''
Sort first
'''
class Solution:
def threeSum(self, nums: List[int]) -> List[List[int]]:
res = []
nums.sort()
for i in range(len(nums)):
if nums[i] > 0:
break
if i == 0 or nums[i - 1] != nums[i]:
self.find_two_sum(nums, i, res)
return res
def find_two_sum(self, nums: List[int], i: int, res: List[List[int]]):
lo, hi = i + 1, len(nums) - 1
while (lo < hi):
s = nums[i] + nums[lo] + nums[hi]
if s < 0:
lo += 1
elif s > 0:
hi -= 1
else:
res.append([nums[i], nums[lo], nums[hi]])
lo += 1
hi -= 1
while lo < hi and nums[lo] == nums[lo - 1]:
lo += 1
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func threeSum(nums []int) [][]int {
var res [][]int
sort.Ints(nums)
for i := 0; i < len(nums)-2; i++ {
if i > 0 && nums[i] == nums[i-1] {
// case already covered by the previous iterations
continue
}
target, left, right := -nums[i], i+1, len(nums)-1
for left < right {
sum := nums[left] + nums[right]
if sum == target {
res = append(res, []int{nums[i], nums[left], nums[right]})
left++
right--
for left < right && nums[left] == nums[left-1] {
left++
}
for left < right && nums[right] == nums[right+1] {
right--
}
} else if sum > target {
right--
} else if sum < target {
left++
}
}
}
return res
}
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