# Next Greater Element | LeetCode | GeeksforGeeks

## Next Greater Element

Given an array arr[ ] of size N having distinct elements, the task is to find the next greater element for each element of the array in order of their appearance in the array.
Next greater element of an element in the array is the nearest element on the right which is greater than the current element.
If there does not exist next greater of current element, then next greater element for current element is -1. For example, next greater of the last element is always -1.

Example 1:

```Input:
N = 4, arr[ ] = [1 3 2 4]
Output:
3 4 4 -1
Explanation:
In the array, the next larger element
to 1 is 3 , 3 is 4 , 2 is 4 and for 4 ?
since it doesn't exist, it is -1.
```

Example 2:

```Input:
N = 5, arr[ ] [6 8 0 1 3]
Output:
8 -1 1 3 -1
Explanation:
In the array, the next larger element to
6 is 8, for 8 there is no larger elements
hence it is -1, for 0 it is 1 , for 1 it
is 3 and then for 3 there is no larger
element on right and hence -1.```

This is a function problem. You only need to complete the function nextLargerElement() that takes list of integers arr[ ] and N as input parameters and returns list of integers of length N denoting the next greater elements for all the corresponding elements in the input array.

Expected Time Complexity : O(N)
Expected Auxiliary Space : O(N)

Constraints:
1 ≤ N ≤ 106
1 ≤ Ai ≤ 1018

### Solution:

The brute force solution can be that we will run a for loop and then at every index, we will find its next greater element by another for a loop.

We can optimize the code by using stack. We will check whether the current element is greater that the element at the top of the stack, if yes then we will pop the element and we will store the current arr[i] in the answer array.

See the code below: