Need help with this question

[hasty pond]

I think I should solve it using a stack but I'm not sure how to implement it

[storm jolt]

You iterate through the list with O(N) time.
Comparing elements on left and right should be O(1)
You just have to set all elements initially to -1 and use 2 if to compare left and right.
Just beware of the list's start and end since it will go out of bounds if list[i-1] when i-1 < 0 and list[i+1] when i+1 > 0
I don't really know how to use python so here's just the main idea.
Do you happen to have a more formal statement or something?

Wait does the closest element must be next to the current element or just find the closest element that is strictly greater?

[hasty pond]

Just find the closest element that is strictly greater I think

[copper canyon]

One way to solve this problem in linear time is to use a stack. We can iterate through the array from left to right and maintain a stack of indices of elements that are strictly greater than the current element. For each element, we pop elements from the stack until we find an element that is greater than the current element, and we record that element as the closest greater element for the current element. If the stack becomes empty, we know that there is no greater element to the left of the current element, so we record -1 as the closest greater element for the current element.

Here's an implementation of this algorithm in Python:

def closest_greater_elements(arr):
    n = len(arr)
    stack = []
    result = [-1] * n
    for i in range(n):
        while stack and arr[stack[-1]] < arr[i]:
            j = stack.pop()
            result[j] = arr[i]
        stack.append(i)
    return result

This function takes an array arr as input and returns an array of the same length containing the closest greater element for each index in arr, or -1 if there is no greater element.

Do you have any other questions about this solution or the problem in general?

[hasty pond]

Yeah I think I had to use a stack but I am having troubles understanding it