add 3sum solution

Author: jongwanra

3sum/jongwanra.py

@@ -0,0 +1,106 @@
+"""
+[Problem]
+https://leetcode.com/problems/3sum/description/
+
+[Brainstorm]
+3 <= nums.length <= 3,000
+Brute Force: O(N^3) => 27,000,000,000 => time limited
+O(N^2)의 풀이는 시간 제한에 걸리지 않을 것으로 보인다. 3,000 * 3,000 => 9,000,000
+
+[Plan]
+1. map을 설정한다. key = elements value: index list
+2. nested-loop을 순회한다.
+    2-1. i == j인 경우 제외
+    2-2. map에서 동일한 인덱스인 경우 제외하고 구한다.
+
+"""
+
+from typing import List
+
+
+class Solution:
+    """
+    Attempt-1 My solution (incorrect)
+    time limited
+    """
+
+    def threeSum1(self, nums: List[int]) -> List[List[int]]:
+        map = {}
+        for index in range(len(nums)):
+            values = map.get(nums[index], [])
+            values.append(index)
+            map[nums[index]] = values
+
+        triplets = set()
+        for i in range(len(nums)):
+            for j in range(len(nums)):
+                if i == j:
+                    continue
+                complement = -(nums[i] + nums[j])
+                # print(f"nums[{i}]={nums[i]}, nums[{j}]={nums[j]} , complement={complement}")
+                if complement in map:
+                    values = map[complement]
+                    for k in values:
+                        if k == i or k == j:
+                            continue
+                        triplets.add(tuple(sorted([nums[i], nums[j], nums[k]])))
+        return list(triplets)
+
+    """
+    Attempt-2 Another solution
+    ref: https://www.algodale.com/problems/3sum/
+
+    """
+
+    def threeSum2(self, nums: List[int]) -> List[List[int]]:
+        triplets = set()
+
+        for index in range(len(nums) - 2):
+            seen = set()
+            for j in range(index + 1, len(nums)):
+                complement = -(nums[index] + nums[j])
+                if complement in seen:
+                    triplet = [nums[index], nums[j], complement]
+                    triplets.add(tuple(sorted(triplet)))
+                seen.add(nums[j])
+
+        return list(triplets)
+
+    """
+    Attempt-3 Another solution
+    ref: https://www.algodale.com/problems/3sum/
+
+    [Plan]
+    two-pointer로 접근한다.
+
+    [Complexity]
+    N: nums.length
+    Time: O(N^2)
+    Space: O(1)
+    """
+
+    def threeSum3(self, nums: List[int]) -> List[List[int]]:
+        triplets = set()
+        nums.sort()
+
+        for index in range(len(nums) - 2):
+            left = index + 1
+            right = len(nums) - 1
+
+            while left < right:
+                three_sum = nums[index] + nums[left] + nums[right]
+                if three_sum < 0:
+                    left += 1
+                    continue
+                if three_sum > 0:
+                    right -= 1
+                    continue
+                triplets.add((nums[index], nums[left], nums[right]))
+                left += 1
+                right -= 1
+        return list(triplets)
+
+
+sol = Solution()
+print(sol.threeSum3([-1, 0, 1, 2, -1, 4]))
+