This code implements the "Three Sum" problem using a two-pointer approach. The problem requires finding all unique triplets in the array that sum up to zero.
Let's go through the code step by step:
l = len(nums)
: This line calculates the length of the input listnums
.nums.sort()
: This line sorts the input listnums
in non-decreasing order. Sorting the array allows us to use a two-pointer approach efficiently.out = set()
: This line initializes an empty set calledout
. We are going to use a set to store unique triplets.The code then iterates through the array using a
for
loop:for i in range(l):
: It iterates over each element in the array.
Inside the loop, it sets up two pointers
bam
anddan
:bam = i + 1
:bam
starts from the element after the current elementi
.dan = l - 1
:dan
starts from the last element in the array.
It uses a while loop to find triplets that sum up to zero:
while(bam < dan):
This loop runs as long as the pointerbam
is less thandan
.
It calculates the total sum of the current triplet:
total = nums[i] + nums[bam] + nums[dan]
It checks the sum:
- If
total
is equal to zero, which means we found a triplet that sums up to zero. It adds this triplet to the setout
. - If
total
is less than zero, it means we need to increase the sum, so it incrementsbam
. - If
total
is greater than zero, it means we need to decrease the sum, so it decreasesdan
.
- If
- Once all the elements are processed, the function returns the set
out
containing unique triplets that sum up to zero.
This approach ensures that no duplicate triplets are added to the final output set, thanks to a set data structure. Additionally, sorting the array enables efficient traversal and ensures that identical triplets won't be repeated.