leetcode

commit 28c87283d061f657fe15b08dce06cb96b1b19d98
parent c35c48dc6d12a59e21f2d59d2b366f9803f1d084
Author: AndrewLockVI <andrewlaack1@gmail.com>
Date:   Wed, 24 May 2023 07:23:32 -0500

Completed majority element ii using dart and a hashmap

Diffstat:
A
majority-element-ii/majority-element-ii.dart
 | 
25
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1 file changed, 25 insertions(+), 0 deletions(-)
diff --git a/majority-element-ii/majority-element-ii.dart b/majority-element-ii/majority-element-ii.dart
@@ -0,0 +1,25 @@
+//Given a list of nums return all values that make up at least
+//1/3 of the elements in the list. The time complexity of this code 
+//is O(n) where n is the number of values in the nums list.
+//My solution iterates through the list placing the values into a hashmap
+//along with how many times the value has been found in the list.
+//Then from there it iterates through the map and adds any values
+//that occur more than 1/3 of the time in the list.
+//Time: 290ms Beats: 100%
+//Memory: 144.9MB Beats: 85.71%
+
+class Solution {
+  List<int> majorityElement(List<int> nums) {
+      Map<int, int> usages = {};
+      for(int i = 0 ; i < nums.length ; ++i){
+          usages[nums[i]] = (usages[nums[i]] ?? 0) + 1;
+      }
+      List<int> return_list = [];
+      for(var entry in usages.entries){
+        if(entry.value > nums.length / 3 ){
+          return_list.add(entry.key);
+        }
+      }
+      return return_list;
+  }
+}