496. Next Greater Element I.cpp

/*
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?
*/ 

class Solution {
public:
    vector<int> nextGreaterElement(vector<int>& nums1, vector<int>& nums2) {
        unordered_map<int, int> m;
        stack<int> st;
        st.push(-1);
        
        for (int i = nums2.size()-1; i >= 0; i--)
        {
            while (st.top() != -1 && nums2[i] > st.top())  
                st.pop();
            m[nums2[i]] = st.top();
            st.push(nums2[i]);
        }
        for (int i = 0; i < nums1.size(); i++)
            nums1[i] = m[nums1[i]];
        
        return nums1;
    }
};