"""
We traverse the array from right to left because the next greater element must lie on the right side.
A stack is used to keep only useful candidates in decreasing order.
For each element, we pop smaller/equal values until we find a greater one on top (or stack becomes empty).
That top becomes the answer, then we push the current element for future elements on the left.
"""

class Solution:
    def nextLargerElement(self, arr):
        st=[arr[-1]]
        ans=[-1]
        for i in reversed(arr[:-1]):
            if st and st[-1]>i:
                ans.append(st[-1])
                st.append(i)
            else:
                while len(st) and st[-1]<=i:
                    st.pop()
                if len(st):
                    ans.append(st[-1])
                else:
                    ans.append(-1)
                st.append(i)
        ans.reverse()
        return ans