71 lines
2.6 KiB
Python
71 lines
2.6 KiB
Python
# Time: O(N^2)
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# Space: O(1)
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# NOTE: i, j, k mentioned in the problem refers to indices, not
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# the values in the array! So, duplicates are A-OK!
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class Solution:
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def threeSumMulti(self, arr: List[int], target: int) -> int:
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# To use 2-pointer technique
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arr.sort()
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n = len(arr)
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result = 0
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for i, _ in enumerate(arr):
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diff = target - arr[i]
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j, k = i + 1, n - 1
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while j < k:
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# Inside this, it's pretty much two-sum and
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# we use the two pointers to move from left
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# and right if they don't equal `diff`
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if arr[j] + arr[k] < diff:
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j += 1
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elif arr[j] + arr[k] > diff:
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k -= 1
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# Following condition but we know arr[j] + arr[k] == diff
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elif arr[j] != arr[k]:
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# Count all of arr[j] and arr[k] instances
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count_j = count_k = 1
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while j + 1 < k and arr[j + 1] == arr[j]:
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count_j += 1
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j += 1
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while k - 1 > j and arr[k - 1] == arr[k]:
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count_k += 1
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k -= 1
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result += count_j * count_k
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j += 1
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k -= 1
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else:
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# If both are equal, then we have a range similar to the following:
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#
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# [4, 4, 4, 4]
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# ^ ^
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# | |
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# j k
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#
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# So, they are the same! We have (k - j + 1) elements to pick from and
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# we can use combination to solve this cause we shouldn't consider (3i, 4i)
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# and (4i, 3i) to be the same
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num_dups = k - j + 1
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# num_dups num_dups!
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# C = ----------------------
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# 2 2! · (num_dups - 2)!
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#
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# = (num_dups * (num_dups - 1)) / 2
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result += (num_dups * (num_dups - 1)) // 2
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# Since it's all duplicates from j to k, we break
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break
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return result % (10 ** 9 + 7)
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