# 496. Next Greater Element I

## Description

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?

## Solutions

• class Solution {
public int[] nextGreaterElement(int[] nums1, int[] nums2) {
Deque<Integer> stk = new ArrayDeque<>();
Map<Integer, Integer> mp = new HashMap<>();
for (int num : nums2) {
while (!stk.isEmpty() && stk.peek() < num) {
mp.put(stk.pop(), num);
}
stk.push(num);
}
int n = nums1.length;
int[] ans = new int[n];
for (int i = 0; i < n; ++i) {
ans[i] = mp.getOrDefault(nums1[i], -1);
}
return ans;
}
}

• class Solution {
public:
vector<int> nextGreaterElement(vector<int>& nums1, vector<int>& nums2) {
stack<int> stk;
unordered_map<int, int> m;
for (int& v : nums2) {
while (!stk.empty() && stk.top() < v) {
m[stk.top()] = v;
stk.pop();
}
stk.push(v);
}
vector<int> ans;
for (int& v : nums1) ans.push_back(m.count(v) ? m[v] : -1);
return ans;
}
};

• class Solution:
def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]:
m = {}
stk = []
for v in nums2:
while stk and stk[-1] < v:
m[stk.pop()] = v
stk.append(v)
return [m.get(v, -1) for v in nums1]


• func nextGreaterElement(nums1 []int, nums2 []int) []int {
stk := []int{}
m := map[int]int{}
for _, v := range nums2 {
for len(stk) > 0 && stk[len(stk)-1] < v {
m[stk[len(stk)-1]] = v
stk = stk[:len(stk)-1]
}
stk = append(stk, v)
}
var ans []int
for _, v := range nums1 {
val, ok := m[v]
if !ok {
val = -1
}
ans = append(ans, val)
}
return ans
}

• function nextGreaterElement(nums1: number[], nums2: number[]): number[] {
const map = new Map<number, number>();
const stack: number[] = [Infinity];
for (const num of nums2) {
while (num > stack[stack.length - 1]) {
map.set(stack.pop(), num);
}
stack.push(num);
}
return nums1.map(num => map.get(num) || -1);
}


• /**
* @param {number[]} nums1
* @param {number[]} nums2
* @return {number[]}
*/
var nextGreaterElement = function (nums1, nums2) {
let stk = [];
let m = {};
for (let v of nums2) {
while (stk && stk[stk.length - 1] < v) {
m[stk.pop()] = v;
}
stk.push(v);
}
return nums1.map(e => m[e] || -1);
};


• use std::collections::HashMap;

impl Solution {
pub fn next_greater_element(nums1: Vec<i32>, nums2: Vec<i32>) -> Vec<i32> {
let mut map = HashMap::new();
let mut stack = Vec::new();
for num in nums2 {
while num > *stack.last().unwrap_or(&i32::MAX) {
map.insert(stack.pop().unwrap(), num);
}
stack.push(num);
}
nums1
.iter()
.map(|n| *map.get(n).unwrap_or(&-1))
.collect::<Vec<i32>>()
}
}