0496 - Next Greater Element I
Description (leetcode)
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 <= 10000 <= nums1[i], nums2[i] <= 10^4- All integers in
nums1andnums2are unique. - All the integers of
nums1also appear innums2.
Follow up: Could you find an O(nums1.length + nums2.length) solution?
submission
impl Solution {
pub fn next_greater_element(nums1: Vec<i32>, nums2: Vec<i32>) -> Vec<i32> {
let mut ans = vec![-1; nums1.len()];
let indices = nums1
.iter()
.copied()
.enumerate()
.map(|(k, v)| (v, k))
.collect::<std::collections::HashMap<i32, usize>>();
let mut stack = Vec::with_capacity(nums2.len());
for val in nums2.into_iter().rev() {
while stack.pop_if(|&mut _v| _v <= val).is_some() {}
if let Some(&_v) = stack.last()
&& let Some(&idx) = indices.get(&val)
{
ans[idx] = _v;
}
stack.push(val);
}
ans
}
}