# Next Greater Element I – Leetcode Solution

In this post, we are going to solve the 496. Next Greater Element I problem of Leetcode. This problem 496. Next Greater Element I is a Leetcode easy level problem. Let’s see code, 496. Next Greater Element I.

## Problem

In this Leetcode Next Greater Element I problem solution,

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.

### 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] <= 104`
• All integers in `nums1` and `nums2` are unique.
• All the integers of `nums1` also appear in `nums2`.

Now, let’s see the code of 496. Next Greater Element I – Leetcode Solution.

# Next Greater Element I – Leetcode Solution

### 496. Next Greater Element I – Solution in Java

```class Solution {
public int[] nextGreaterElement(int[] nums1, int[] nums2) {
int[] ans = new int[nums1.length];
for(int i=0;i<nums1.length;i++){
ans[i] = nextGreaterElement(nums2,nums1[i]);
}
return ans;
}

public int nextGreaterElement(int[] nums, int tar){
Stack<Integer> s = new Stack<>();
for(int e : nums){
}
int ans = -1;
while(!(s.isEmpty())){
int top = s.peek();
if(top == tar) return ans;
s.pop();

if(top > tar){
ans = top;
}
}
return ans;
}
}```

### 496. Next Greater Element I – Solution in C++

```class Solution {
public:
vector<int> nextGreaterElement(vector<int>& nums1, vector<int>& nums2) {
vector<int> ans(nums1.size());
for(int i=0;i<nums1.size();i++){
ans[i] = nextGreaterElement(nums2,nums1[i]);
}
return ans;
}
int nextGreaterElement(vector<int>& nums, int tar){
stack<int> s;
for(int e : nums){
s.push(e);
}
int ans = -1;
while((s.size()>0)){
int top = s.top();
if(top == tar) return ans;
s.pop();

if(top > tar){
ans = top;
}
}
return ans;
}
};```

### 496. Next Greater Element I– Solution in Python

This is a python 3 solution of Leetcode 496. Next Greater Element I.

```class Solution:
def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]:
ans = {}
res = []
stack = []

for n2 in nums2:
while stack and n2 > stack[-1]:
ans[stack.pop()] = n2
stack.append(n2)

for n1 in nums1:
res.append(ans.get(n1, -1))

return res```

Note: This problem 496. Next Greater Element I is generated by Leetcode but the solution is provided by CodingBroz. This tutorial is only for Educational and Learning purpose.