[main] Added attempt for 4sum implementation, will refactor later

Author: cthacker-udel

pythonProblems/leetcode/4sum.py

@@ -0,0 +1,51 @@
+from typing import List, Union
+
+# e = a + b + c + d
+# a = b + c + d
+# b = a + c + d
+
+
+## target = 4
+## 4 - 1 == 3, so check for pairs for 1 and 3, if both have pairs then we have a four-sum
+## 4 + (-1 + 2) + (2 + 1) == 4 + (1) + (3) == 4
+
+
+class Solution:
+
+    def fourSum(self, nums: List[int], target: int) -> List[List[int]]:
+        pairs = []
+        for i in range(len(nums)):
+            for j in range(len(nums)):
+                for k in range(len(nums)):
+                    elem_i = nums[i]
+                    elem_j = nums[j]
+                    elem_k = nums[k]
+                    diff = target - (elem_i + elem_j + elem_k)
+                    if i != j and j != k and k != i:
+                        altered_arr = nums[:]
+                        indexes = sorted([i, j, k])
+                        del altered_arr[indexes[2]]
+                        del altered_arr[indexes[1]]
+                        del altered_arr[indexes[0]]
+                        if diff in altered_arr:
+                            sorted_pair = sorted(
+                                [diff, elem_i, elem_j, elem_k])
+                            if sorted_pair not in pairs:
+                                pairs.append(sorted_pair)
+        return pairs
+
+
+if __name__ == '__main__':
+    nums = [2, 2, 2, 2, 2]
+    target = 8
+    sol = Solution()
+    print(sol.fourSum(nums, target))
+
+    # def threeSum(self, nums: List[int], target: int) -> Union[List[int], None]:
+    #     for i in range(len(nums)):
+    #         result = target - nums[i]
+    #         if result in nums:
+    #             j = nums.index(result)
+    #             result = Solution.twoSum(self, nums[:j] + nums[j + 1:], result)
+    #             return [target] + result if result != None else []
+    #     return None