From beginning I belive that you need to make use of 2 pointers first and the next and keep on iterating backward and forward, so it will be way more complex to code,

so maybe starting from back is more suitable in this scenario I guess

This problem specifically benefits by going backwards because the last item in the array will have no greatest element. It will always be -1 in the last spot of the result array.

My first hunch was to solve from the beginning too.

class Solution:
    def nextLargerElement(self, arr):
        res = [-1]*len(arr)
        # ToDo: Write Your Code Here.
        stack = []
        for idx, val in enumerate(arr):
            while stac...