Next Greater Element I

Description

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?

Code

Monotonic Stack

Time Complexity: $O(N)$, Space Complexity: $O(N)$

From left to right:

class Solution {
public:
    vector<int> nextGreaterElement(vector<int>& nums1, vector<int>& nums2) {
        unordered_map<int, int> mp;
        stack<int> st;
        for(auto num: nums2) {
            while(!st.empty() && st.top() < num) {
                mp[st.top()] = num;
                st.pop();
            }
            st.push(num);
        }
 
        vector<int> answer;
        for(auto num: nums1) {
            if(mp.find(num) != mp.end()) {
                answer.push_back(mp[num]);
            } else {
                answer.push_back(-1);
            }
        }
        return answer;
    }
};

From right to left:

class Solution {
public:
    vector<int> nextGreaterElement(vector<int>& nums1, vector<int>& nums2) {
        stack<int> st;
        int n2 = nums2.size();
        unordered_map<int, int> mp;
        for(int i = n2 - 1; i >= 0; i--) {
            while(!st.empty() && st.top() < nums2[i]) {
                st.pop();
            }
            if(!st.empty()) {
                mp[nums2[i]] = st.top();
            } else {
                mp[nums2[i]] = -1;
            }
            st.push(nums2[i]);
        }    
 
        vector<int> res;
        for(int i = 0; i < nums1.size(); i++) {
            res.push_back(mp[nums1[i]]);
        }
        return res;
 
    }
};

Source

• Next Greater Element I - LeetCode