diff --git a/3Sum.cpp b/3Sum.cpp
new file mode 100644
index 0000000..bc5e069
--- /dev/null
+++ b/3Sum.cpp
@@ -0,0 +1,27 @@
+ vector<vector<int>> threeSum(vector<int>& nums) {
+        sort(nums.begin() , nums.end());     //Sorted Array
+        if(nums.size() < 3){                // Base Case 1
+            return {};
+        }
+        if(nums[0] > 0){                    // Base Case 2
+            return {};
+        }
+        unordered_map<int , int> hashMap;
+        for(int i = 0 ; i < nums.size() ; ++i){     //Hashing of Indices
+            hashMap[nums[i]] = i;
+        }
+        vector<vector<int>> answer;
+        for(int i = 0 ; i < nums.size() - 2 ; ++i){     //Traversing the array to fix the number.
+            if(nums[i] > 0){     //If number fixed is +ve, stop there because we can't make it zero by searching after it.
+                break;
+            }
+            for(int j = i + 1 ; j < nums.size() - 1 ; ++j){     //Fixing another number after first number
+                int required = -1*(nums[i] + nums[j]);    //To make sum 0, we would require the -ve sum of both fixed numbers.
+                if(hashMap.count(required) && hashMap.find(required)->second > j){ //If it exists in hashmap and its last occurrence index > 2nd fixed index, we found our triplet.
+                    answer.push_back({nums[i] , nums[j] , required});
+                }
+                j = hashMap.find(nums[j])->second; //Update j to last occurence of 2nd fixed number to avoid duplicate triplets.
+            }
+            i = hashMap.find(nums[i])->second;     //Update i to last occurence of 1st fixed number to avoid duplicate triplets.
+        }
+        return answer;