Next Greater Element II

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. The next greater number of a number x is the first greater number to its traversing-order next in the array, which means you could search circularly to find its next greater number. If it doesn't exist, return -1 for this number.

For example:

nums = [1,2,1] Output: [2,-1,2] Explanation: The first 1's next greater number is 2; The number 2 can't find next greater number. The second 1's next greater number needs to search circularly, which is also 2.

As another example:

nums = [1,2,3,4,3] Output: [2,3,4,-1,4]

## Next Greater Element II

This problem requires us to find the next greater element for each element in a circular array.  We can use a stack to efficiently keep track of potential next greater elements as we iterate through the array (potentially multiple times to simulate circularity).

### 1. Brute Force Solution

A naive approach would be to iterate through the array for each element and search for the next greater element in a circular fashion.

```python
def nextGreaterElements_brute_force(nums):
    n = len(nums)
    result = [-1] * n
    
    for i in range(n):
        for j in range(1, n): # Start from the next element
            if nums[(i + j) % n] > nums[i]:
                result[i] = nums[(i + j) % n]
                break
    
    return result

Explanation:

• For each element at index i, we iterate through the rest of the array (virtually circular) using the modulo operator (i + j) % n.
• If we find an element greater than nums[i], we update result[i] and break the inner loop.
• If we reach the end of the inner loop without finding a greater element, result[i] remains -1.

2. Optimal Solution (Using Stack)

We can improve upon the brute force approach by using a stack to store indices of elements for which we are looking for the next greater element. This allows us to find the next greater element in O(n) time.

def nextGreaterElements(nums):
    n = len(nums)
    result = [-1] * n
    stack = []  # Stack to store indices

    for i in range(2 * n): # Iterate twice to simulate circular array
        num = nums[i % n]  # Actual element from array
        
        while stack and nums[stack[-1]] < num:
            index = stack.pop()
            result[index] = num
            
        if i < n: # Only push indices from the original array
            stack.append(i)
            
    return result

Explanation:

• We iterate through the array twice (2 * n) to simulate the circular nature. The modulo operator (i % n) helps access elements circularly.
• The stack stores indices of elements for which we haven't found the next greater element yet.
• While the stack is not empty and the current element num is greater than the element at the top of the stack (nums[stack[-1]]), it means we've found the next greater element for the element at the top of the stack. We pop the index from the stack and update the result array.
• We only push indices onto the stack during the first iteration (i < n), as we only want to find next greater elements for the original elements.

3. Big(O) Runtime Analysis

• Optimal Solution: O(n)
  • We iterate through the array a maximum of 2n times. While the while loop looks like it might add to the complexity, each element is pushed onto the stack at most once and popped at most once. Thus, the amortized time complexity for the while loop is O(n). Therefore, the total time complexity is O(2n) which simplifies to O(n).
• Brute Force Solution: O(n^2)
  • The outer loop runs n times, and the inner loop runs up to n times in the worst case, resulting in O(n*n) = O(n^2) complexity.

4. Big(O) Space Usage Analysis

• Optimal Solution: O(n)
  • The result array stores n elements. The stack, in the worst-case scenario (decreasing array), can hold up to n indices. Therefore the space complexity is O(n + n) = O(n).
• Brute Force Solution: O(n)
  • The result array stores n elements. Therefore the space complexity is O(n).

5. Edge Cases and Considerations

• Empty Array: If the input array nums is empty, the function should return an empty array.
• All Elements are Same: If all elements in the array are the same, the next greater element for each will be -1.
• Decreasing Array: If the array is strictly decreasing, the next greater element for each will be -1.
• Array with Maximum Element at the Beginning: If the array starts with its maximum element, elements before the maximum in the original array will not have a "circular" next greater element until the second iteration.

Example Edge Cases:

• nums = [] returns []
• nums = [5, 5, 5] returns [-1, -1, -1]
• nums = [5, 4, 3, 2, 1] returns [-1, -1, -1, -1, -1]
• nums = [5, 1, 2, 3, 4] returns [-1, 2, 3, 4, 5]