Added solution for Leetcode Problem #15 (3Sum) in Dart

- Implemented the 	hreeSum function to find all unique triplets that sum up to zero.
- Included a time complexity of O(N^2) and space complexity of O(1).
- Added unit tests with edge cases for validation.

array/three_sum.dart

@@ -0,0 +1,46 @@
+//Leetcode problem no. 15: https://leetcode.com/problems/3sum/description/
+//Time Complexity: O(N^2)
+//Space Complexity: O(1)
+
+//Input: A list of integers
+//Output: A list of lists containing all those integers whose sum end up to 0
+
+import 'package:test/test.dart';
+
+import '../sort/heap_Sort.dart';
+
+List<List<int>> threeSum(List<int> l){
+  List<List<int>> ans=[];
+  sort(l);
+  for(int i=0; i<l.length; ++i){
+    if(i!=0 && l[i]==l[i-1]) continue;
+    int j=i+1, k=l.length-1;
+    while(j<k){
+      int sum=l[i]+l[j]+l[k];
+      if(sum<0) j++;
+      else if(sum>0) k--;
+      else{
+        ans.add([l[i], l[j], l[k]]);
+        j++;
+        k--;
+        while(j<k && l[j]==l[j-1]) j++;
+        while(j<k && l[k]==l[k+1]) k--;
+      }
+    }
+  }
+  return ans;
+}
+
+void main(){
+  test('test 1', () {
+    expect(threeSum([-1,0,1,2,-1,-4]), [[-1,-1,2],[-1,0,1]]);
+  });
+
+  test('test 2', () {
+    expect(threeSum([0, 1, 1]), []);
+  });
+
+  test('test 3', () {
+    expect(threeSum([0, 0, 0]), [[0,0,0]]);
+  });
+}