503: Solution with step by step explanation
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n1. Initialize a stack to keep track of indices of elements we haven\'t fo
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n1. Initialize a stack to keep track of indices of elements we haven\'t found a next greater element for.\n\n2. Initialize the result list to contain all -1s (assuming we won\'t find any next greater element).\n\n3. Loop through the array twice (to handle circularity) and process each element:\na. For each element i in the array:\ni. Pop elements off the stack if they are less than the current element and update their result to the current element.\nii. Add the current index i to the stack.\n\n4. Return the result list.\n\nThis algorithm works because it processes each element in the array twice, allowing it to handle circularity. During the first pass, it looks for next greater elements to the right of each element. During the second pass, it looks for next greater elements to the left of each element that haven\'t been found yet. By using a stack to keep track of indices of elements we haven\'t found a next greater element for, we can easily update their results when we find a greater element.\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution:\n    def nextGreaterElements(self, nums: List[int]) -> List[int]:\n        # Initialize a stack to keep track of indices of elements we haven\'t found a next greater element for\n        stack = []\n        \n        # Initialize the result list to contain all -1s (assuming we won\'t find any next greater element)\n        result = [-1] * len(nums)\n        \n        # Loop through the array twice (to handle circularity) and process each element\n        for _ in range(2):\n            for i in range(len(nums)):\n                # Pop elements off the stack if they are less than the current element and update their result\n                while stack and nums[stack[-1]] < nums[i]:\n                    result[stack.pop()] = nums[i]\n                \n                # Add the current index to the stack\n                stack.append(i)\n        \n        # Return the result list\n        return result\n\n```
['Array', 'Stack', 'Monotonic Stack', 'Python', 'Python3']