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.

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.

class Solution {
public:
    vector<int> nextGreaterElement(vector<int>& nums1, vector<int>& nums2) {
        // solve it in O( m+n )
        unordered_map<int,int> mp;
        vector<int> result( nums1.size(), -1 );
        stack<int> st;
        for ( int i = 0; i < nums1.size(); i++ ) {
            mp.insert( { nums1[i], i } );
        }

        for ( int i = 0; i < nums2.size(); i++ ) {
            while ( !st.empty() && nums2[i] > nums2[st.top()] ) {
                if ( mp.find( nums2[st.top()] ) != mp.end() ) {
                    int check_num = nums2[st.top()];
                    int index = mp[check_num];
                    result[index] = nums2[i];
                }

                st.pop();
            }

            st.push( i );
        }   

        return result;     
    }
};