# 503. Next Greater Element II

### Description

Given a circular integer array `nums` (i.e., the next element of `nums[nums.length - 1]` is `nums[0]`), return *the **next greater number** for every element in* `nums`.

The **next greater number** of a number `x` is the first greater number to its traversing-order next in the array, which means you could search circularly to find its next greater number. If it doesn't exist, return `-1` for this number.

### Constraints

* `1 <= nums.length <= 104`
* `-109 <= nums[i] <= 109`

### Approach

### Links

* Binarysearch
* GeeksforGeeks
* [Leetcode](https://leetcode.com/problems/next-greater-element-ii/)
* ProgramCreek
* [YouTube](https://youtu.be/ARkl69eBzhY)

### **Examples**

{% tabs %}
{% tab title="Example 1" %}
**Input:** nums = \[1, 2, 1]

**Output:** \[2, -1, 2]

**Explanation:** The first 1's next greater number is 2;

The number 2 can't find next greater number.

The second 1's next greater number needs to search circularly, which is also 2.
{% endtab %}

{% tab title="Example 2" %}
**Input:** nums = \[1, 2, 3, 4, 3]

**Output:** \[2, 3, 4, -1, 4]
{% endtab %}
{% endtabs %}

### **Solutions**

{% tabs %}
{% tab title="Solution 1" %}

```java
/**
 * Time complexity : 
 * Space complexity : 
 */

class Solution {
    public int[] nextGreaterElements(int[] nums) {
        if(nums == null || nums.length == 0) {
            return new int[0];
        }
        
        int n = nums.length;
        int[] result = new int[n];
        Stack<Integer> stack = new Stack<>();
        Arrays.fill(result, -1);
        
        for(int i = 0; i < 2*n; i++) {
            while(!stack.isEmpty() && nums[stack.peek()] < nums[i%n]) {
                result[stack.pop()] = nums[i%n];
            }
            if(i < n) {
                stack.push(i);
            }
        }
        
        return result;
    }
}
```

{% endtab %}
{% endtabs %}

### **Follow up**

*