3sum

Author: OfficialCodeVoyage

solution.py

@@ -0,0 +1,74 @@
+# Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]]
+# such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0.
+#
+# Notice that the solution set must not contain duplicate triplets.
+#
+#
+# Example 1:
+#
+# Input: nums = [-1,0,1,2,-1,-4]
+# Output: [[-1,-1,2],[-1,0,1]]
+# Explanation:
+# nums[0] + nums[1] + nums[2] = (-1) + 0 + 1 = 0.
+# nums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0.
+# nums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0.
+# The distinct triplets are [-1,0,1] and [-1,-1,2].
+# Notice that the order of the output and the order of the triplets does not matter.
+
+
+nums = [-1,0,1,2,-1,-4]
+
+def threesum(nums):
+    # for i in nums:
+    #     for j in nums:
+    #         for k in nums:
+    #             if i + j + k == 0:
+    #                 print(f"{i} + {j} + {k} = 0")#
+    # brute force for fun
+    #
+    # dic_nums = {}
+    # for i, v in enumerate(nums):
+    #     dic_nums[i] = v
+    #
+    #
+    # po1, po2 = 0, 1
+    #
+    #
+    # while po2 < len(nums) -1:
+    #     # if nums[left] + nums[right] == dic -value == 0
+    #     iter = nums[po1] + nums[po2]
+    #     listofsums = []
+    #
+    #     for k, v in dic_nums.items():
+    #         if iter + v == 0:
+    #             listofsums.append([nums[po1], nums[po2], nums[k]])
+    #
+    #     po1 += 1
+    #     po2 += 1
+    #
+    # return listofsums
+
+    nums.sort()
+    i = 0
+    left, right = i+1, len(nums)-1
+    arr = []
+
+    while i < len(nums) - 2:
+        if nums[i] + nums[left] + nums[right] == 0:
+            res = [nums[i], nums[left], nums[right]]
+            arr.append(res)
+        i += 1
+        left += 1
+        right -= 1
+
+
+
+    return arr
+
+
+
+
+
+
+
+print(threesum(nums))