4 Sum

'''
Use two for loops for 2 elements and then iterate the rest of the array using two pointer, and once the target matches,push it into the ans.second loop should start after the first one ends.

Trick:
Skip the continous duplicates on all the 4 pointers, i,j,l,r
'''
class Solution:
    def fourSum(self, nums: List[int], target: int) -> List[List[int]]:
        # ans=set()
        nums.sort()
        ans=[]
        for i in range(len(nums)):
            if i>0 and nums[i]==nums[i-1]: #skip the continous duplicates as one appearance for each number
                continue
            for j in range(i+1,len(nums)):
                if j>i+1 and nums[j]==nums[j-1]: #skip the continous duplicates as one appearance for each number but it has to skip the previous jth element, not the previous ith element(2,2,2,2) os j>i+1
                    continue
                l=j+1
                r=len(nums)-1
                while l<r:
                    if nums[l]+nums[r]+nums[i]+nums[j]==target:
                        ans.append((nums[i],nums[j],nums[l],nums[r]))
                        l+=1
                        r-=1
                        while l<r and nums[l]==nums[l-1]: #skip the continous duplicates as one appearance for each number
                            l+=1
                        while r>l and nums[r]==nums[r+1]: #skip the continous duplicates as one appearance for each number
                            r-=1
                    elif nums[l]+nums[r]+nums[i]+nums[j]>target:
                            r-=1
                    else:
                        l+=1

        return ans