How do you solve Next Greater Element IV on LeetCode?

Next Greater Element IV (LeetCode #2454) 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 keep track of indices efficiently.

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

The brute force approach for Next Greater Element IV is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every element to find the second greater integer by iterating through the array for each element. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Next Greater Element IV?

Clarify the problem requirements before jumping to code. Think about edge cases, such as arrays with all elements equal. Practice stack-related problems to get comfortable with this pattern.

What are common mistakes when solving Next Greater Element IV?

Not using a stack properly to track indices. Overlooking the condition of finding exactly one first greater element.

#2454

Next Greater Element IV

Hard

ArrayBinary SearchStackSortingHeap (Priority Queue)Monotonic StackStackArray

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)

The optimal approach uses a stack to efficiently track the first and second greater elements while iterating through the array only once. This reduces the time complexity significantly.

⚙️

Algorithm

3 steps

1Step 1: Initialize an empty stack to keep track of indices and an array for the result.
2Step 2: Iterate through the nums array from right to left, maintaining a stack of indices where we have not yet found a second greater element.
3Step 3: For each element, pop elements from the stack until we find the first greater element, then check the next element in the stack for the second greater element.

solution.py11 lines

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

ℹ

Complexity note: The time complexity is O(n) because we traverse the nums array once and each element is pushed and popped from the stack at most once. The space complexity is O(n) due to the stack used to store indices.

1Using a stack helps keep track of indices efficiently.
2Iterating from the end allows us to find the second greater element without re-scanning the array.

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