> ## Documentation Index
> Fetch the complete documentation index at: https://leetcode-py.wisl.dev/llms.txt
> Use this file to discover all available pages before exploring further.

> ## Agent Instructions
> leetcode-py is a Python LeetCode practice environment generator with one CLI: lcpy. It is not a service or platform.
> Each problem is a directory under leetcode/ with README.md, solution.py, test_solution.py, helpers.py, and playground.ipynb. lcpy gen creates them from JSON templates bundled with the package.
> Examples are backed by tests; copy them verbatim.

# Next Greater Element I Python Solution

> Tested Python solution for LeetCode 496 with 13 pytest cases. Generate a practice environment with lcpy.

LeetCode 496, [Easy](/catalog/easy). Topics: [Array](/catalog/topics/array), [Hash Table](/catalog/topics/hash-table), [Stack](/catalog/topics/stack), [Monotonic Stack](/catalog/topics/monotonic-stack). [View on LeetCode](https://leetcode.com/problems/next-greater-element-i/description/).

Generate this problem as a practice environment: tested reference solution, 13 [parametrized pytest cases](/practice/testing), and a playground notebook:

```bash theme={"theme":{"light":"github-light","dark":"github-dark"}}
lcpy gen -n 496   # by problem number
lcpy gen -s next_greater_element_i   # by problem name
```

## Problem

The **next greater element** of some element `x` in an array is the **first greater** element that is **to the right** of `x` in the same array.

You are given two **distinct 0-indexed** integer arrays `nums1` and `nums2`, where `nums1` is a subset of `nums2`.

For each `0 <= i < nums1.length`, find the index `j` such that `nums1[i] == nums2[j]` and determine the **next greater element** of `nums2[j]` in `nums2`. If there is no next greater element, then the answer for this query is `-1`.

Return *an array* `ans` *of length* `nums1.length` *such that* `ans[i]` *is the **next greater element** as described above.*

### Examples

```
Input: nums1 = [4,1,2], nums2 = [1,3,4,2]
Output: [-1,3,-1]
Explanation: The next greater element for each value of nums1 is as follows:
- 4 is underlined in nums2 = [1,3,4,2]. There is no next greater element, so the answer is -1.
- 1 is underlined in nums2 = [1,3,4,2]. The next greater element is 3.
- 2 is underlined in nums2 = [1,3,4,2]. There is no next greater element, so the answer is -1.
```

```
Input: nums1 = [2,4], nums2 = [1,2,3,4]
Output: [3,-1]
Explanation: The next greater element for each value of nums1 is as follows:
- 2 is underlined in nums2 = [1,2,3,4]. The next greater element is 3.
- 4 is underlined in nums2 = [1,2,3,4]. There is no next greater element, so the answer is -1.
```

### Constraints

* `1 <= nums1.length <= nums2.length <= 1000`
* `0 <= nums1[i], nums2[i] <= 10^4`
* All integers in `nums1` and `nums2` are unique.
* All the integers of `nums1` also appear in `nums2`.

**Follow up:** Could you find an `O(nums1.length + nums2.length)` solution?

## Solution

Reference implementation from [solution.py on GitHub](https://github.com/wislertt/leetcode-py/blob/main/leetcode/next_greater_element_i/solution.py), full suite in [test\_solution.py](https://github.com/wislertt/leetcode-py/blob/main/leetcode/next_greater_element_i/test_solution.py):

```python theme={"theme":{"light":"github-light","dark":"github-dark"}}
class Solution:
    # Time: O(n1 + n2)
    # Space: O(n2)
    def next_greater_element(self, nums1: list[int], nums2: list[int]) -> list[int]:
        next_greater: dict[int, int] = {}
        stack: list[int] = []
        for num in nums2:
            while stack and stack[-1] < num:
                next_greater[stack.pop()] = num
            stack.append(num)
        return [next_greater.get(num, -1) for num in nums1]
```

## Complexity

| Time | Space |
| - | - |
| O(n1 + n2) | O(n2) |

## Tags

[NeetCode All](/catalog/neetcode).


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