# 3Sum in Python

PythonServer Side ProgrammingProgramming

Suppose we have an array of numbers. It stores n integers, there are there elements a, b, c in the array, such that a + b + c = 0. Find all unique triplets in the array which satisfies the situation. So if the array is like [-1,0,1,2,-1,-4], then the result will be [[-1, 1, 0], [-1, -1, 2]]

To solve this, we will follow these steps −

• Sort the array nums, and define an array res
• for i in range 0 to length of nums – 3
• if i > 0 and nums[i] = nums[i - 1], then skip the next part and continue
• l := i + 1 and r := length of nums – 1
• while l < r
• sum := sum of nums[i], nums[l] and nums[r]
• if sum < 0, then l := l + 1, otherwise when sum > 0, then r := r – 1
• otherwise insert nums[i], nums[l], nums[r] into the res array
• while l < length of nums – 1 and nums[l] = nums[l + 1]
• increase l by 1
• while r > 0 and nums[r] = nums[r - 1]
• decrease r by 1
• increase l by 1 and decrease r by 1
• return res

## Example(Python)

Let us see the following implementation to get better understanding −

Live Demo

class Solution(object):
def threeSum(self, nums):
nums.sort()
result = []
for i in range(len(nums)-2):
if i> 0 and nums[i] == nums[i-1]:
continue
l = i+1
r = len(nums)-1
while(l<r):
sum = nums[i] + nums[l] + nums[r]
if sum<0:
l+=1
elif sum >0:
r-=1
else:
result.append([nums[i],nums[l],nums[r]])
while l<len(nums)-1 and nums[l] == nums[l + 1] : l += 1
while r>0 and nums[r] == nums[r - 1]: r -= 1
l+=1
r-=1
return result
ob1 = Solution()
print(ob1.threeSum([-1,0,1,2,-1,-4]))

## Input

[-1,0,1,2,-1,-4]

## Output

[[-1,-1,2],[-1,0,1]]