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Leet Code 4Sum Solution: https://leetcode.com/problems/4sum/
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class Solution: | |
""" | |
Solution to Leet Code problem 4Sum: https://leetcode.com/problems/4sum/ | |
Runtime: 100 ms beats 93.4% of python3 submissions | |
""" | |
def fourSum(self, nums, target): | |
""" | |
:type nums: List[int] | |
:type target: int | |
:rtype: List[List[int]] | |
""" | |
# Edge cases: | |
if len(nums) < 4: | |
return [] | |
if 4 * max(nums) < target: | |
return [] | |
nums = sorted(nums) | |
n = len(nums) | |
solutions = set() | |
for _a in range(n - 3): | |
ta = target - nums[_a] | |
for _d in range(_a + 3, n): | |
tad = ta - nums[_d] | |
# Skip the following cases: | |
if nums[_d - 2] + nums[_d - 1] < tad: | |
continue | |
if nums[_a + 1] + nums[_a + 2] > tad: | |
continue | |
_b, _c = _a + 1, _d - 1 | |
while _b < _c: | |
b, c = nums[_b], nums[_c] | |
bc = b + c | |
if bc == tad: | |
a, d = nums[_a], nums[_d] | |
solutions.add((a, b, c, d)) | |
_b += 1 | |
elif bc > tad: # decrease c to make bc closer to tad | |
_c -= 1 | |
else: # increase b to make bc closer to tad | |
_b += 1 | |
return list(solutions) |
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