leetcode

majority-element-ii.dart (921B)

---

 //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;
   }
 }