496. Next Greater Element I

//🎯DAY 18 PROBLEM 1
//Apply next greater element problem in nums2 array, there will be three condition, when the array is empty.. When the top of stack is smaller than current element......
//When the top of stack is greater than the element
//1)When stack is empty, i.e. there is no greater element on right, so we insert -1
//2)When the stack's top is greater, we insert stack's top in the answer
//3)But when the stack's top is lesser, then we keep on popping untill stack's top is smaller than current element as we will not need them in further calculations also
//as our current element will serve as the next greater element. 
class Solution {
public:
    vector<int> nextGreaterElement(vector<int>& nums1, vector<int>& nums2) {
        stack<int>st;
      map<int,int>result;
        for(int i=nums2.size()-1;i>=0;i--)
        {
            if(st.empty())
                result.insert({nums2[i],-1});
            else if(st.top()>nums2[i])
               result.insert({nums2[i],st.top()});
            else if(st.size()>0&&st.top()<nums2[i])
            {
                while(st.size()>0&&st.top()<nums2[i])
                {
                    st.pop();
                }
                if(st.empty())
                    result.insert({nums2[i],-1});//if stack's top becomes empty, we push -1
                else
                    result.insert({nums2[i],st.top()});//if stack's top becomes greater, we insert that in the map .
            }
            st.push(nums2[i]);
        }
        vector<int>ans;
        for(int i=0;i<nums1.size();i++)
        {
            ans.push_back(result[nums1[i]]);
        }
        return ans;
    }
};