42 lines
1.7 KiB
Python
42 lines
1.7 KiB
Python
# NOTE: Far from the most optimal solution to the problem.
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# Time complexity: O(n^3)
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# Space complexity: O(n^2) // NOTE: Really? And how?
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class Solution:
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def fourSum(self, nums: List[int], target: int) -> List[List[int]]:
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arr_len = len(nums)
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# Helps keep out duplicates
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nums.sort()
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# Will be a dict with each key storing a `set` of 2-tuples (pair of nums)
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diff_cache = {}
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quadruplets = set()
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# No point in iterating with the last element since we won't be adding
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# any new quadruplets at that point
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for i in range(1, arr_len - 1):
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for j in range(i + 1, arr_len):
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current_sum = nums[i] + nums[j]
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difference = target - current_sum
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if difference in diff_cache:
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# diff_cache could have multiple pairs that add up to the same
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# `difference`. We need to form quadruplets with all of them
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# and store them in `quadruplets`
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for pair in diff_cache[difference]:
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quadruplets.add(pair + (nums[i], nums[j]))
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# Compute sum with every element to the left of i and store
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# in diff_cache. We do this at this later point to avoid duplicates
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for k in range(0, i):
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current_sum = nums[i] + nums[k]
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if current_sum in diff_cache:
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diff_cache[current_sum].add((nums[i], nums[k]))
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else:
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# This has to be a set to avoid duplicates
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diff_cache[current_sum] = { (nums[i], nums[k]) }
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return quadruplets
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