111 lines
3 KiB
Markdown
111 lines
3 KiB
Markdown
# Leetcode Next-Greater-Element-I
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#### 2022-07-27 10:55
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> ##### Data structures:
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>
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> #DS #stack #hash_table
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>
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> ##### Difficulty:
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>
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> #coding_problem #difficulty_easy
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>
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> ##### Additional tags:
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>
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> #leetcode
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>
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> ##### Revisions:
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>
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> N/A
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##### Related topics:
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##### Links:
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- [Link to problem](https://leetcode.com/problems/next-greater-element-i/)
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---
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### Problem
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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.
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You are given two **distinct 0-indexed** integer arrays `nums1` and `nums2`, where `nums1` is a subset of `nums2`.
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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`.
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Return _an array_ `ans` _of length_ `nums1.length` _such that_ `ans[i]` _is the **next greater element** as described above._
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#### Examples
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**Example 1:**
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**Input:** nums1 = [4,1,2], nums2 = [1,3,4,2]
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**Output:** [-1,3,-1]
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**Explanation:** The next greater element for each value of nums1 is as follows:
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- 4 is underlined in nums2 = [1,3,4,2]. There is no next greater element, so the answer is -1.
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- 1 is underlined in nums2 = [1,3,4,2]. The next greater element is 3.
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- 2 is underlined in nums2 = [1,3,4,2]. There is no next greater element, so the answer is -1.
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**Example 2:**
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**Input:** nums1 = [2,4], nums2 = [1,2,3,4]
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**Output:** [3,-1]
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**Explanation:** The next greater element for each value of nums1 is as follows:
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- 2 is underlined in nums2 = [1,2,3,4]. The next greater element is 3.
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- 4 is underlined in nums2 = [1,2,3,4]. There is no next greater element, so the answer is -1.
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#### Constraints
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### Thoughts
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> [!summary]
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> This is a problem using the traits of #stack.
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This problem can be divided into two sub-problems:
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- given a subset number, return its location in the parent set -> hash table
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- given a array, return the next greater element in the array after it -> stack
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#### How is the stack used?
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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.
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### Solution
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```cpp
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class Solution {
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public:
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vector<int> nextGreaterElement(vector<int> &nums1, vector<int> &nums2) {
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// Hash table to link the subsets.
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unordered_map<int, int> greater;
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stack<int> st;
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for (int i = nums2.size() - 1; i >= 0; i--) {
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while (!st.empty() && st.top() < nums2[i]) {
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st.pop();
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}
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if (!st.empty()) {
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greater[nums2[i]] = st.top();
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}
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st.push(nums2[i]);
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}
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// double pointers can't be used, because nums doesn't have a order.
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vector<int> ans(nums1.size());
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for (int i = 0, size1 = nums1.size(); i < size1; i++) {
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if (greater.find(nums1[i]) != greater.end()) {
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ans[i] = greater[nums1[i]];
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} else {
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ans[i] = -1;
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}
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}
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return ans;
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}
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};
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```
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