1253. Reconstruct a 2-Row Binary Matrix

Problem Link : https://leetcode.com/problems/reconstruct-a-2-row-binary-matrix/
--------------------------------------------------------------------------------
Complexity : Time-O(N), Memory-O(1), TimeTaken-<6hr
--------------------------------------------------------------------------------
Logic :
So the logic is that at each col[i] where its sum is 2 we place 1 in both row 
if 1 then we see which row sum is greater and act accordingly. 
And for the impossible 
part after the nth col processing we see if upper & lower !=0 it either means value was
not assigned fully or needed more value in that case return empty list.

class Solution {
    public List<List<Integer>> reconstructMatrix(int upper, int lower, int[] colsum) {
        List<List<Integer>> ans=new ArrayList();
        ans.add(new ArrayList());
        ans.add(new ArrayList());
        int n=colsum.length;
        for(int i=0;i<n;i++){
            if(colsum[i]==0){
                ans.get(0).add(0);
                ans.get(1).add(0);
            } else if(colsum[i]==2){
                upper--;
                lower--;
                ans.get(0).add(1);
                ans.get(1).add(1);
            } else {
                if(upper>lower){
                    upper--;
                    ans.get(0).add(1);
                    ans.get(1).add(0);
                } else {
                    lower--;
                    ans.get(0).add(0);
                    ans.get(1).add(1);
                }
            }
            if(upper<0 || lower<0)
                return new ArrayList();
        }
        if(upper!=0 || lower!=0)
            return new ArrayList();
        return ans;
    }
}