0496. Next Greater Element I

https://leetcode.com/problems/next-greater-element-i

Description

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] <= 104

• All integers in nums1 and nums2 are unique.

• All the integers of nums1 also appear in nums2.

Follow up: Could you find an O(nums1.length + nums2.length) solution?

ac

class Solution {
    public int[] nextGreaterElement(int[] nums1, int[] nums2) {
        // edge cases
        if (nums1 == null || nums1.length == 0 || nums2 == null || nums2.length == 0 || nums1.length > nums2.length)
            return new int[0];

        Stack<Integer> stack = new Stack<Integer>();
        Map<Integer, Integer> map = new HashMap<>();
        stack.push(nums2[0]);
        for (int i = 1; i < nums2.length; i++) {
            while (!stack.isEmpty() && nums2[i] > stack.peek()) {
                map.put(stack.pop(), nums2[i]);
            }
            stack.push(nums2[i]);
        }

        // get result
        int[] res = new int[nums1.length];
        for (int i = 0; i < nums1.length; i++) {
            res[i] = map.getOrDefault(nums1[i], -1);
        }

        return res;
    }
}