496 Next Greater Element I
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.
Solution:
class Solution {
public int[] nextGreaterElement(int[] nums1, int[] nums2) {
int[] result = new int[nums1.length];
int j;
for (int i = 0; i < nums1.length; i++){
result[i] = -1;
}
for (int i = 0; i < nums1.length; i++){
boolean found = false;
for (j = 0; j < nums2.length; j++){
if (found && nums2[j] > nums1[i]){
result[i] = nums2[j];
break;
}
if (nums1[i] == nums2[j]){
found = true;
}
}
}
return result;
}
}
// TC: O(m*n)
// SC: O(n)