How do you solve Next Greater Element I on LeetCode?

Next Greater Element I (LeetCode #496) 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 efficiently find the next greater elements in one pass.

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

The brute force approach for Next Greater Element I is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks each element in nums1 against all elements in nums2 to find the next greater element. This is straightforward but inefficient, as it involves nested loops. The optimal Optimal Solution approach improves this to O(n).

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

Next Greater Element I uses the following concepts: Array, Hash Table, Stack, Monotonic Stack. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Next Greater Element I?

Explain your thought process clearly; interviewers value understanding over just getting the right answer. Consider edge cases, such as when nums1 has elements that are the largest in nums2. Practice implementing both brute force and optimal solutions to solidify your understanding.

What are common mistakes when solving Next Greater Element I?

Not using a stack correctly, leading to incorrect next greater elements. Forgetting to handle cases where no greater element exists.

#496

Next Greater Element I

Easy

ArrayHash TableStackMonotonic StackHash MapArray

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 stack allows us to efficiently find the next greater elements in a single pass through nums2. This is because we can keep track of elements for which we haven't found a next greater element yet.

⚙️

Algorithm

4 steps

1Step 1: Create a stack to keep track of elements in nums2 and a map to store the next greater elements.
2Step 2: Iterate through nums2. For each element, pop elements from the stack until the current element is less than or equal to the top of the stack.
3Step 3: If the stack is not empty after popping, the current element is the next greater element for the popped elements. Store this in the map.
4Step 4: For each element in nums1, use the map to get the next greater element and build the result list.

solution.py8 lines

1def nextGreaterElement(nums1, nums2):
2    stack = []
3    next_greater = {}
4    for num in nums2:
5        while stack and stack[-1] < num:
6            next_greater[stack.pop()] = num
7        stack.append(num)
8    return [next_greater.get(num, -1) for num in nums1]

ℹ

Complexity note: The time complexity is O(n) because we traverse nums2 once, and each element is pushed and popped from the stack at most once. The space complexity is O(n) due to the stack and the map used to store next greater elements.

1Using a stack helps efficiently find the next greater elements in one pass.
2Mapping results allows for quick lookups when processing nums1.

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