leetcode

commit 3c667e4b721364a6d908bc7d039d41b64f66d4c8
parent 132647c453f6fdf3ea81d7bbbffc71d62dc14483
Author: AndrewLockVI <andrewlaack1@gmail.com>
Date:   Fri, 12 May 2023 16:56:19 -0500

Completed interesction of two arrays using dart and sets

Diffstat:
A
intersection-of-two-arrays/interesection-of-two-arrays.dart
 | 
27
+++++++++++++++++++++++++++

1 file changed, 27 insertions(+), 0 deletions(-)
diff --git a/intersection-of-two-arrays/interesection-of-two-arrays.dart b/intersection-of-two-arrays/interesection-of-two-arrays.dart
@@ -0,0 +1,27 @@
+//Return all integers that are in both lists. 
+//To do this first I added the first list to a set.
+//I then checked the second list to see if the values existed in the set
+//if they do then they get added to a duplicate set. Finally I converted
+//the set to a list and returned it. The time complexity of this code is
+//O(n) where n is the number of integers in both lists combined.
+//Time: 265ms Beats: 90.91%
+//Memory: 143MB Beats: 84.85%
+
+class Solution {
+  List<int> intersection(List<int> nums1, List<int> nums2) {
+      Set<int> set_vals ={};
+      Set<int> return_vals = {};
+      for(int i = 0 ; i < nums1.length ; ++i){
+          set_vals.add(nums1[i]);
+      }
+      for(int i = 0 ; i < nums2.length ; ++i){
+          if(set_vals.contains(nums2[i])){
+              return_vals.add(nums2[i]);
+          }
+      }
+      List<int> return_list = return_vals.toList();
+      return return_list;
+  }
+}
+
+