#503

Medium

auto_awesomeStack-based state management

LeetCode Problem Workspace

Next Greater Element II

Solve the Next Greater Element II problem by using a stack-based state management approach for circular arrays.

Problem Statement

Given a circular integer array nums (where the next element after nums[nums.length - 1] is nums[0]), your task is to return the next greater number for each element in the array. For any element, its next greater number is the first number in the array that is greater than it when traversing the array in order, and if no such number exists, return -1.

To solve the problem efficiently, you must account for the circular structure of the array while using a stack to keep track of elements whose next greater number hasn’t been found yet. If an element doesn't have a greater element after it, return -1 for that element.

Examples

Example 1

Input: nums = [1,2,1]

Output: [2,-1,2]

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.

Example 2

Input: nums = [1,2,3,4,3]

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

Example details omitted.

Constraints

1 <= nums.length <= 104
-109 <= nums[i] <= 109

Solution Approach

Monotonic Stack Approach

We can use a monotonic stack to efficiently track the next greater element for each number. Start by iterating through the array, using the stack to store indices of the elements that we haven't yet found the next greater element for. As you iterate, pop elements from the stack if the current element is greater than the element at the index stored in the stack, and update their next greater values.

Circular Array Handling

Since the array is circular, we must traverse the array twice to simulate the circular nature. This means that while iterating through the array, you should consider indices beyond the array length and start again from the beginning once you reach the end. This ensures all elements find their next greater number by traversing the array as if it were not linear.

Time and Space Complexity Considerations

The time complexity of this solution is O(n) because each element is pushed and popped from the stack at most once. The space complexity is O(n) due to the stack storing indices of array elements, which in the worst case could contain all the elements of the array.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity is O(n), where n is the length of the array, as each element is processed once when being added or removed from the stack. The space complexity is also O(n), primarily due to the stack used to keep track of elements that haven't yet found their next greater number.

What Interviewers Usually Probe

Candidate should demonstrate an understanding of stack-based state management in the context of circular arrays.
Look for the use of a monotonic stack for efficient traversal and the handling of the circular nature of the array.
The candidate should explain time and space complexities clearly, ensuring they understand both the efficiency of the solution and its trade-offs.

Common Pitfalls or Variants

Common pitfalls

Confusing circular arrays with linear arrays, leading to incorrect results or missed elements.
Not using the stack to manage elements efficiently, leading to unnecessary iterations or excessive time complexity.
Failing to account for array boundaries, resulting in incorrect index references or missed circular behavior.

Follow-up variants

Consider a variant where the array contains negative integers and how that may affect the behavior of the stack.
Adapt the solution to work with arrays of floating-point numbers.
Handle a variant where the array has only one element or all elements are the same.

FAQ

What is the best way to solve the Next Greater Element II problem?

The best way to solve the Next Greater Element II problem is using a monotonic stack while considering the circular nature of the array to ensure you handle all edge cases.

How does the circular nature of the array affect the solution?

The circular nature requires you to traverse the array twice, ensuring that the first element can find its next greater number even if the larger number is in the initial part of the array.

What is the time complexity of the Next Greater Element II problem?

The time complexity is O(n), as each element is added and removed from the stack at most once during the traversal.

What if there is no greater number for an element in the array?

If there is no greater number for an element, the solution will return -1 for that element.

Can I use a brute force approach to solve the Next Greater Element II problem?

While you can use a brute force approach by checking all subsequent elements for each element in the array, it would be inefficient with a time complexity of O(n^2). The stack-based approach is much more efficient with O(n) time complexity.

terminal

Solution

Solution 1: Monotonic Stack

The problem requires us to find the next greater element for each element. Therefore, we can traverse the array from back to front, which effectively turns the problem into finding the previous greater element. Additionally, since the array is circular, we can traverse the array twice.

class Solution:
    def nextGreaterElements(self, nums: List[int]) -> List[int]:
        n = len(nums)
        ans = [-1] * n
        stk = []
        for i in range(n * 2 - 1, -1, -1):
            i %= n
            while stk and stk[-1] <= nums[i]:
                stk.pop()
            if stk:
                ans[i] = stk[-1]
            stk.append(nums[i])
        return ans

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