2.9 KiB
Leetcode Next-Greater-Element-I
2022-07-27 10:55
Data structures:
#DS #stack #hash_table
Difficulty:
#coding_problems #difficulty_easy
Additional tags:
#leetcode
Revisions:
N/A
Related topics:
Links:
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;
}
};