---
title: '496. Next Greater Element I'
description: 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
icon: dot
topics:
  - { name: "Array", slug: "array" }
  - { name: "Hash Table", slug: "hash-table" }
  - { name: "Stack", slug: "stack" }
  - { name: "Monotonic Stack", slug: "monotonic-stack" }
issue: "https://github.com/prdlk/leetcode/issues/74"
sidebar:
  label: 'Next Greater Element I'
  badge: 'Easy'
---

::::warning
Could you find an O(nums1.length + nums2.length) solution?
::::

### Example 1:
- 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`.

### Example 2:
- 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.

## Solution

```py
class Solution:
    def nextGreaterElement(self, nums1: list[int], nums2: list[int]) -> list[int]:
        mono_stack = []
        next_greater = {}

        for cur in nums2:
            while mono_stack and cur > mono_stack[-1]:
                val = mono_stack.pop()
                next_greater[val] = cur
            mono_stack.append(cur)

        return [next_greater.get(x, -1) for x in nums1]
```