496. Next Greater Element I

Problem

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] <= 10^4

• 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?

Solution (Java)

class Solution {
    public int[] nextGreaterElement(int[] nums1, int[] nums2) {
        Map<Integer, Integer> indexMap = new HashMap<>();
        for (int i = 0; i < nums2.length; i++) {
            indexMap.put(nums2[i], i);
        }
        for (int i = 0; i < nums1.length; i++) {
            int num = nums1[i];
            int index = indexMap.get(num);
            if (index == nums2.length - 1) {
                nums1[i] = -1;
            } else {
                boolean found = false;
                while (index < nums2.length) {
                    if (nums2[index] > num) {
                        nums1[i] = nums2[index];
                        found = true;
                        break;
                    }
                    index++;
                }
                if (!found) {
                    nums1[i] = -1;
                }
            }
        }
        return nums1;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).