954. Array of Doubled Pairs

Problem Link : https://leetcode.com/problems/array-of-doubled-pairs/
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Complexity : Time-O(NlogN), Memory-O(N), TimeTaken-30min
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Logic :
We sort the array and use a map to store the freq. Then we start from the lowest num
and see if there is any num that satisfies the cond if yes decrease both elements freq from
the map. And after the loop see if map is empty or not.

class Solution {
    public boolean canReorderDoubled(int[] arr) {
        Arrays.sort(arr);
        Map<Integer, Integer> map=new HashMap();
        for(int ele : arr){
            map.put(ele, map.getOrDefault(ele, 0)+1);
        }
        
        for(int i=0;i<arr.length;i++){
            int ele2=2*arr[i];
            if(map.containsKey(arr[i]) && map.containsKey(ele2)){
                map.put(arr[i], map.get(arr[i])-1);
                map.put(ele2, map.get(ele2)-1);
                
                if(map.get(arr[i])==0)
                    map.remove(arr[i]);
                if(map.containsKey(ele2) && map.get(ele2)==0)
                    map.remove(ele2);
            }
            if(map.isEmpty())return true;
        }
        return map.isEmpty();
    }
}