# Next Greater Element I

`Stack`, `Hash Map`

You are given two arrays (without duplicates)`nums1`and`nums2`where`nums1`’s elements are subset of`nums2`. Find all the next greater numbers for`nums1`'s elements in the corresponding places of`nums2`.

The Next Greater Number of a numberxin`nums1`is the first greater number to its right in`nums2`. If it does not exist, output -1 for this number.

Example 1:

``````Input: nums1 = [4,1,2], nums2 = [1,3,4,2].
Output: [-1,3,-1]
Explanation:
For number 4 in the first array, you cannot find the next greater number for it in the second array, so output -1.
For number 1 in the first array, the next greater number for it in the second array is 3.
For number 2 in the first array, there is no next greater number for it in the second array, so output -1.
``````

Example 2:

``````Input: nums1 = [2,4], nums2 = [1,2,3,4].
Output: [3,-1]
Explanation:
For number 2 in the first array, the next greater number for it in the second array is 3.
For number 4 in the first array, there is no next greater number for it in the second array, so output -1.
``````

Note:

1. All elements in `nums1` and `nums2` are unique.
2. The length of both `nums1` and `nums2` would not exceed 1000.

## Analysis

Similar idea to Daily Temperatures problem.

Keep a monotonous stack. 经典问题，单调栈可以轻松解决。

## Solution

Monotonous Stack + HashMap (4 ms, faster than 85.00%)

``````class Solution {
public int[] nextGreaterElement(int[] nums1, int[] nums2) {
Map<Integer, Integer> map = new HashMap<>();
Deque<Integer> stack = new ArrayDeque<>();
int[] res = new int[nums1.length];

for (int i = 0; i < nums2.length; i++) {
while (!stack.isEmpty() && nums2[i] > nums2[stack.peek()]) {
int idx = stack.pop();
map.put(nums2[idx], nums2[i]);
}
stack.push(i);
}

for (int j = 0; j < nums1.length; j++) {
res[j] = map.getOrDefault(nums1[j], -1);
}

return res;
}
}
``````