Hi All, This is the first time I am posting on the Leetcode discuss section related to one minor problem I am facing on one of the Leetcode Hard problems so if I am posting this in the wrong place for my problem to be solved, let me know I will delete it and post at the right section.

So from last weeks I just started solving @Leetcode Hard problems, and I am really enjoying it. But I need some expert help now regarding one famous leetcode hard problem.

Problem URL: https://leetcode.com/problems/count-of-smaller-numbers-after-self/
315. Count of Smaller Numbers After Self

Before seeing the best approach for this problem in discuss section, i just tried to solve this problem using some different method or totally different way compare to all Accepted Solution.

And Luckily my solution is passing  63/65 test cases

So two-trick which I have used here is the reversal approach & Hashmap to get the count if and only if already visited and both are consecutive and adding it to the front of the list using the list.add(0, count[i]).

class Solution {
    public List<Integer> countSmaller(int[] nums) {
        HashMap<Integer,Integer> map=new HashMap<>();
        List<Integer> list=new ArrayList<>();
        if(nums.length==0 || nums==null)
        return list;
        int[] count=new int[nums.length];
        for(int i=nums.length-1;i>=0;i--){
            
             if(map.containsKey(nums[i]) && nums[i+1]==nums[i]){
             list.add(0,map.get(nums[i]));
             continue; 
             }
            
            for(int j=i+1;j<=nums.length-1;j++){ 
                if(nums[i]>nums[j])
                count[i]++;
            }
        map.put(nums[i],count[i]);
        list.add(0,count[i]);
        }
        return list;
    }
}