Next Greater Element II

Problem

Given a circular integer array nums (i.e., the next element of nums[nums.length - 1] is nums[0]), return the next greater number for every element in nums.

Examples

The brute-force approach

For each index i, scan forward (circularly) until you find an element greater than nums[i]. Use modulo to wrap around.

n = len(nums)
answer = [-1] * n

for i in range(n):
    for j in range(1, n):                    # scan up to n-1 steps forward
        next_idx = (i + j) % n
        if nums[next_idx] > nums[i]:
            answer[i] = nums[next_idx]
            break

return answer

O(n²), for each of the n elements you may scan up to n positions. The monotonic stack handles this in O(n) by processing the array twice (simulating the circular nature) and resolving each element exactly once.

Spotting the pattern

This is a Stack problem. The key question to ask yourself:

How do I handle the wrap-around without scanning circularly from every position?

Answering that is where it clicks, and it's exactly what the guided walkthrough below builds with you: the pattern reasoning, a progressive hint ladder that never spoils the answer, a row-by-row dry run, the optimized solution, and an in-browser editor to run your code against real test cases.