How do you solve Next Greater Element II on LeetCode?

Next Greater Element II (LeetCode #503) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a stack helps manage the order of elements efficiently.

What is the brute force solution for Next Greater Element II?

The brute force approach for Next Greater Element II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks each element and looks for the next greater element by traversing the array. Since the array is circular, we need to wrap around to the beginning after reaching the end. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Next Greater Element II?

Next Greater Element II uses the following concepts: Array, Stack, Monotonic Stack. The recommended patterns to study are: Monotonic Stack, Circular Array.

What are the interview tips for Next Greater Element II?

Explain your thought process clearly, especially when using data structures like stacks. Consider edge cases, such as arrays of length 1 or all elements being the same. Practice writing clean and efficient code.

What are common mistakes when solving Next Greater Element II?

Not considering the circular nature of the array. Forgetting to initialize the output array correctly.

#503

Next Greater Element II

Medium

ArrayStackMonotonic StackMonotonic StackCircular Array

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Using a monotonic stack allows us to efficiently find the next greater element for each number in a single pass. We can simulate the circular nature by doubling the array.

⚙️

Algorithm

3 steps

1Step 1: Create a stack to keep track of indices and an output array initialized to -1.
2Step 2: Traverse the array twice (to account for circularity) using the modulo operator.
3Step 3: For each element, while the stack is not empty and the current element is greater than the element at the index stored in the stack, pop from the stack and set the output for that index.

solution.py10 lines

1def nextGreaterElements(nums):
2    n = len(nums)
3    output = [-1] * n
4    stack = []
5    for i in range(2 * n):
6        while stack and nums[stack[-1]] < nums[i % n]:
7            output[stack.pop()] = nums[i % n]
8        if i < n:
9            stack.append(i)
10    return output

ℹ

Complexity note: This complexity is efficient because we only pass through the array twice, and each element is pushed and popped from the stack at most once.

1Using a stack helps manage the order of elements efficiently.
2Circular arrays can be handled by simulating a longer array.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.