From 3fed6cd2f93cd9b9817830c1408828308f412bb8 Mon Sep 17 00:00:00 2001
From: Sangeeth Sudheer <sangeeth96@users.noreply.github.com>
Date: Mon, 28 Feb 2022 06:27:17 +0530
Subject: [PATCH] feat(15-3sum): add golang solution

---
 015_3sum/README.md          | 20 +++++++++++++++
 015_3sum/golang/solution.go | 49 +++++++++++++++++++++++++++++++++++++
 2 files changed, 69 insertions(+)
 create mode 100644 015_3sum/README.md
 create mode 100644 015_3sum/golang/solution.go

diff --git a/015_3sum/README.md b/015_3sum/README.md
new file mode 100644
index 0000000..29f3860
--- /dev/null
+++ b/015_3sum/README.md
@@ -0,0 +1,20 @@
+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]] 
+
+Example 2:
+    
+    Input: nums = [] Output: [] 
+
+Example 3:
+    
+    Input: nums = [0] Output: [] 
+
+Constraints:
+
+* `0 <= nums.length <= 3000`
+* `-105 <= nums[i] <= 105`
diff --git a/015_3sum/golang/solution.go b/015_3sum/golang/solution.go
new file mode 100644
index 0000000..4f9a244
--- /dev/null
+++ b/015_3sum/golang/solution.go
@@ -0,0 +1,49 @@
+func threeSum(nums []int) (result [][]int) {
+	target := 0
+	sort.Ints(nums)
+
+	for i := 0; i < len(nums)-2; i++ {
+		// We don't want duplicates of the fixed value
+		// which can cause duplicate triplets to appear.
+		if i > 0 && nums[i] == nums[i-1] {
+			continue
+		}
+
+		startPtr := i + 1
+		endPtr := len(nums) - 1
+
+		for startPtr < endPtr {
+			sum := nums[i] + nums[startPtr] + nums[endPtr]
+
+			if sum == target {
+				result = append(
+					result,
+					[]int{nums[i], nums[startPtr], nums[endPtr]},
+				)
+
+				// Duplicate values can appear while scanning also. Presence of these
+				// duplicate values can also result in duplicate triplets. So, for the
+				// start and end pointers, we need to keep moving them until we are
+				// clear of duplicates.
+				for startPtr < endPtr && nums[startPtr] == nums[startPtr+1] {
+					startPtr++
+				}
+
+				for endPtr > startPtr && nums[endPtr] == nums[endPtr-1] {
+					endPtr--
+				}
+
+				// Since the loops above stop at the last duplicate value, we need to move
+				// them by one step from both directions to continue with the logic.
+				startPtr++
+				endPtr--
+			} else if sum < target {
+				startPtr++
+			} else {
+				endPtr--
+			}
+		}
+	}
+
+	return
+}