Completed 3sum leetcode question

src/leet.cpp

@@ -3,32 +3,39 @@
 
 std::vector<std::vector<int>> Solution::threeSum(std::vector<int>& nums) {
 	std::vector<std::vector<int>> ans;
+	std::unordered_map<int, int> map;
+
+	std::sort(nums.begin(), nums.end());
+
+	if (nums.size() < 3) {                // Base Case 1
+		return {};
+	}
+	if (nums[0] > 0) {                    // Base Case 2
+		return {};
+	}
 
 	//itterate through map O(n^2) and create a map of the sums of each element
 	for (int i = 0; i < nums.size(); i++)
-	{
+		map[nums[i]] = i;
-		for (int j = i+1; j < nums.size(); j++)
-		{
-			sums[nums[i] + nums[j]] = std::vector<int>{i, j};
-		}
-	}
 
 	for (int i = 0; i < nums.size(); i++)
 	{
-		if (sums.count(0-nums[i]))
+		if (nums[i] > 0)
-		{//If required value is in map
+			break;
-			std::vector<int> indexes = sums[0-nums[i]];
+
-			if (i == indexes[0] || i == indexes[1])
+		for (int j = i + 1; j < nums.size(); j++)
-				continue;
+		{
-			ans.push_back({ nums[i], nums[indexes[0]], nums[indexes[1]] });
+			int numNeeded = (nums[i] + nums[j]) * -1;
+			if (map.count(numNeeded) && map.find(numNeeded)->second > j)
+				ans.push_back({ nums[i], nums[j], numNeeded });
+
+			//Update J to avoid duplicates
+			j = map.find(nums[j])->second;
 		}
+
+		//Update i to avoid duplicates
+		i = map.find(nums[i])->second;
 	}
 
 	return ans;
 }
-
-//
-//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.

src/leet.h

@@ -1,11 +1,9 @@
 #pragma once
 #include <vector>
-#include <map>
+#include <algorithm>
+#include <unordered_map>
 
 class Solution {
 public:
     std::vector<std::vector<int>> threeSum(std::vector<int>& nums);
-private:
-    //Map of sum + indexes
-    std::map<int, std::vector<int>> sums;
 };