Today I have done leetcode's October LeetCoding Challenge with cpp.

October LeetCoding Challenge 19

Description

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.

Follow up: Could you find an O(nums1.length + nums2.length) solution?

Solution

class Solution {
public:
  vector<int> nextGreaterElement(vector<int>& nums1, vector<int>& nums2) {
    unordered_map<int, int> nextGreater;
    vector<int> monoStack;
    for(auto it = nums2.rbegin(); it != nums2.rend(); ++it) {
      while(monoStack.size() && monoStack.back() <= *it) monoStack.pop_back();
      nextGreater[*it] = monoStack.size() ? monoStack.back() : -1;
      monoStack.push_back(*it);
    }
    vector<int> answer(nums1.size());
    for(int i = 0; i < nums1.size(); ++i) {
      answer[i] = nextGreater[nums1[i]];
    }
    return answer;
  }
};

// Accepted
// 15/15 cases passed (10 ms)
// Your runtime beats 33.18 % of cpp submissions
// Your memory usage beats 58.15 % of cpp submissions (8.8 MB)