496 Next Greater Element I

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.

Solution:

从右往左倒着遍历，栈中记录下一个更大元素的「候选项」。

由于左边更大元素会「挡住」右边更小的元素，所以右边更小的元素是无用信息（不会成为左边元素的下一个更大元素），这会导致栈底（右边）大，栈顶（左边）小。

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

        for (int i = 0; i < n; i++){
            map.put(nums1[i], i);
        }

        int[] result = new int[n];
        Arrays.fill(result, -1);
        Deque<Integer> stack = new ArrayDeque<>();

        for (int i = nums2.length - 1; i >= 0; i--){
            int cur = nums2[i];

            while(!stack.isEmpty() && cur >= stack.peekLast()){
                stack.pollLast();
            }

            if (!stack.isEmpty() && map.containsKey(cur)){
                result[map.get(cur)] = stack.peekLast();
            }

            stack.offerLast(cur);
        }

        return result;
    }
}
// TC: O(n)
// SC: O(n)

class Solution {
    public int[] nextGreaterElement(int[] nums1, int[] nums2) {
        int[] result = new int[nums1.length];

        int j; 

        for (int i = 0; i < nums1.length; i++){
            result[i] = -1;

        }
        for (int i = 0; i < nums1.length; i++){
            boolean found = false;
            for (j = 0; j < nums2.length; j++){
                if (found && nums2[j] > nums1[i]){
                    result[i] = nums2[j];
                    break;
                }

                if (nums1[i] == nums2[j]){
                    found = true;
                }
            }
        }

        return result;

    }
}

// TC: O(m*n)
// SC: O(n)