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.**
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
andnums2
are unique.All the integers of
nums1
also appear innums2
.
Follow up: Could you find an O(nums1.length + nums2.length)
solution?
Solution
/**
* @param {number[]} nums1
* @param {number[]} nums2
* @return {number[]}
*/
var nextGreaterElement = function(nums1, nums2) {
var map = {};
var stack = [];
for (var i = nums2.length - 1; i >= 0; i--) {
while (stack.length && stack[stack.length - 1] <= nums2[i]) stack.pop();
map[nums2[i]] = stack.length ? stack[stack.length - 1] : -1;
stack.push(nums2[i]);
}
return nums1.map(num => map[num]);
};
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).