496. Next Greater Element I

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496. Next Greater Element I

Description

Difficulty: Easy

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

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:

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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:

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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?

Hints/Notes

  • Monotonic Stack

Solution

Language: C++

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class Solution {
public:
    vector<int> nextGreaterElement(vector<int>& nums1, vector<int>& nums2) {
        map<int, int> mapping;
        vector<int> res;
        vector<int> greater = nextGreater(nums2);
        for (int i = 0; i < nums2.size(); i++) {
            mapping[nums2[i]] = greater[i];
        }
        for (int num : nums1) {
            res.push_back(mapping[num]);
        }
        return res;
    }

    vector<int> nextGreater(vector<int>& nums) {
        vector<int> res(nums.size(), 0);
        stack<int> greater;
        for (int i = nums.size() - 1; i >= 0; i--) {
            while (!greater.empty() && greater.top() <= nums[i]) {
                greater.pop();
            }
            res[i] = greater.empty() ? -1 : greater.top();
            greater.push(nums[i]);
        }
        return res;
    }
};