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# Next Greater Element I – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

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] <= 10`

^{4}- All integers in
`nums1`

and`nums2`

are**unique**. - All the integers of
`nums1`

also appear in`nums2`

.

# C++ Next Greater Element I LeetCode Solution

```
``````
class Solution {
public:
vector<int> nextGreaterElement(vector<int>& findNums, vector<int>& nums) {
stack<int> s;
unordered_map<int, int> m;
for (int n : nums) {
while (s.size() && s.top() < n) {
m[s.top()] = n;
s.pop();
}
s.push(n);
}
vector<int> ans;
for (int n : findNums) ans.push_back(m.count(n) ? m[n] : -1);
return ans;
}
};
```

# Java Next Greater Element I LeetCode Solution

```
``````
public int[] nextGreaterElement(int[] findNums, int[] nums) {
Map<Integer, Integer> map = new HashMap<>(); // map from x to next greater element of x
Stack<Integer> stack = new Stack<>();
for (int num : nums) {
while (!stack.isEmpty() && stack.peek() < num)
map.put(stack.pop(), num);
stack.push(num);
}
for (int i = 0; i < findNums.length; i++)
findNums[i] = map.getOrDefault(findNums[i], -1);
return findNums;
}
```

# Python 3 Next Greater Element I LeetCode Solution

```
``````
def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]:
if not nums2:
return None
mapping = {}
result = []
stack = []
stack.append(nums2[0])
for i in range(1, len(nums2)):
while stack and nums2[i] > stack[-1]: # if stack is not empty, then compare it's last element with nums2[i]
mapping[stack[-1]] = nums2[i] # if the new element is greater than stack's top element, then add this to dictionary
stack.pop() # since we found a pair for the top element, remove it.
stack.append(nums2[i]) # add the element nums2[i] to the stack because we need to find a number greater than this
for element in stack: # if there are elements in the stack for which we didn't find a greater number, map them to -1
mapping[element] = -1
for i in range(len(nums1)):
result.append(mapping[nums1[i]])
return result
```

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