496

496. Next Greater Element I (Easy)

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?

Companies:
Facebook, Amazon, Adobe, Bloomberg, Hotstar

Related Topics:
Array, Hash Table, Stack, Monotonic Stack

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Solution 1. Monotonic Stack

// OJ: https://leetcode.com/problems/next-greater-element-i/
// Author: github.com/lzl124631x
// Time: O(A + B)
// Space: O(B) extra space
class Solution {
public:
    vector<int> nextGreaterElement(vector<int>& A, vector<int>& B) {
        unordered_map<int, int> m;
        stack<int> s;
        for (int n : B) {
            while (s.size() && n > s.top()) {
                m[s.top()] = n;
                s.pop();
            }
            s.push(n);
        }
        vector<int> ans;
        for (int n : A) ans.push_back(m.count(n) ? m[n] : -1);
        return ans;
    }
};

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