Next Greater Element I
<p>The <strong>next greater element</strong> of some element <code>x</code> in an array is the <strong>first greater</strong> element that is <strong>to the right</strong> of <code>x</code> in the same array.</p>

<p>You are given two <strong>distinct 0-indexed</strong> integer arrays <code>nums1</code> and <code>nums2</code>, where <code>nums1</code> is a subset of <code>nums2</code>.</p>

<p>For each <code>0 &lt;= i &lt; nums1.length</code>, find the index <code>j</code> such that <code>nums1[i] == nums2[j]</code> and determine the <strong>next greater element</strong> of <code>nums2[j]</code> in <code>nums2</code>. If there is no next greater element, then the answer for this query is <code>-1</code>.</p>

<p>Return <em>an array </em><code>ans</code><em> of length </em><code>nums1.length</code><em> such that </em><code>ans[i]</code><em> is the <strong>next greater element</strong> as described above.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums1 = [4,1,2], nums2 = [1,3,4,2]
<strong>Output:</strong> [-1,3,-1]
<strong>Explanation:</strong> The next greater element for each value of nums1 is as follows:
- 4 is underlined in nums2 = [1,3,<u>4</u>,2]. There is no next greater element, so the answer is -1.
- 1 is underlined in nums2 = [<u>1</u>,3,4,2]. The next greater element is 3.
- 2 is underlined in nums2 = [1,3,4,<u>2</u>]. There is no next greater element, so the answer is -1.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums1 = [2,4], nums2 = [1,2,3,4]
<strong>Output:</strong> [3,-1]
<strong>Explanation:</strong> The next greater element for each value of nums1 is as follows:
- 2 is underlined in nums2 = [1,<u>2</u>,3,4]. The next greater element is 3.
- 4 is underlined in nums2 = [1,2,3,<u>4</u>]. There is no next greater element, so the answer is -1.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums1.length &lt;= nums2.length &lt;= 1000</code></li>
	<li><code>0 &lt;= nums1[i], nums2[i] &lt;= 10<sup>4</sup></code></li>
	<li>All integers in <code>nums1</code> and <code>nums2</code> are <strong>unique</strong>.</li>
	<li>All the integers of <code>nums1</code> also appear in <code>nums2</code>.</li>
</ul>

<p>&nbsp;</p>
<strong>Follow up:</strong> Could you find an <code>O(nums1.length + nums2.length)</code> solution?
Stack; Array; Hash Table; Monotonic Stack
class Solution {
    public int[] nextGreaterElement(int[] nums1, int[] nums2) {
        Deque<Integer> stk = new ArrayDeque<>();
        int m = nums1.length, n = nums2.length;
        Map<Integer, Integer> d = new HashMap(n);
        for (int i = n - 1; i >= 0; --i) {
            int x = nums2[i];
            while (!stk.isEmpty() && stk.peek() < x) {
                stk.pop();
            }
            if (!stk.isEmpty()) {
                d.put(x, stk.peek());
            }
            stk.push(x);
        }
        int[] ans = new int[m];
        for (int i = 0; i < m; ++i) {
            ans[i] = d.getOrDefault(nums1[i], -1);
        }
        return ans;
    }
}
