notes/OJ notes/pages/Leetcode Next-Greater-Element-I.md

Leetcode Next-Greater-Element-I

2022-07-27 10:55

Data structures:

#DS #stack #hash_table

Difficulty:

#coding_problem #difficulty-easy

Additional tags:

#leetcode

Revisions:

N/A

Related topics:

tag:#stack

Links:

• Link to problem

---

Problem

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.

Examples

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

Thoughts

[!summary] This is a problem using the traits of #stack.

This problem can be divided into two sub-problems:

• given a subset number, return its location in the parent set -> hash table
• given a array, return the next greater element in the array after it -> stack

How is the stack used?

Stack is FILO, which means, when iterating from the last element, if you push a number greater into it, it will be used last, ideal for finding the next biggest one.

Solution

class Solution {
public:
  vector<int> nextGreaterElement(vector<int> &nums1, vector<int> &nums2) {
    // Hash table to link the subsets.
    unordered_map<int, int> greater;
    stack<int> st;

    for (int i = nums2.size() - 1; i >= 0; i--) {
      while (!st.empty() && st.top() < nums2[i]) {
        st.pop();
      }
      if (!st.empty()) {
        greater[nums2[i]] = st.top();
      }

      st.push(nums2[i]);
    }

    // double pointers can't be used, because nums doesn't have a order.
    vector<int> ans(nums1.size());
    for (int i = 0, size1 = nums1.size(); i < size1; i++) {
      if (greater.find(nums1[i]) != greater.end()) {
        ans[i] = greater[nums1[i]];
      } else {
        ans[i] = -1;
      }
    }

    return ans;
  }
};