Sponsored
Sponsored
This approach involves iterating over each element of nums1 and finding the corresponding element in nums2. Once located, search for the next greater element to its right. This straightforward method checks each pair and ensures correctness but may not be optimal for larger arrays.
The time complexity of this brute force approach is O(n * m), where n is the number of elements in nums1, and m is the number of elements in nums2. The space complexity is O(1) apart from the output array.
1def next_greater_element(nums1, nums2):
2 result = []
3 for num in nums1:
4 try:
5 index = nums2.index(num)
6 greater = -1
7 for n in nums2[index + 1:]:
8 if n > num:
9 greater = n
10 break
11 result.append(greater)
12 except ValueError:
13 result.append(-1)
14 return result
15
16nums1 = [4, 1, 2]
17nums2 = [1, 3, 4, 2]
18print(next_greater_element(nums1, nums2))This Python script applies a brute force method, iterating each num1 element through nums2 to locate the element and find a subsequent greater number. It leverages Python's list slicing and exception handling for cleaner code execution.
This approach utilizes a stack and a hashmap to efficiently solve the problem. As we traverse nums2, we use the stack to store elements for which the next greater element hasn't been found yet. Whenever a greater element is found, it's recorded in the hashmap against the elements in the stack. This technique is optimal and runs in linear time.
The time complexity is O(n + m), approaching linear time with respect to input size and space is approximately O(m) for hashmap tracking.
1
The Java solution manages elements using a stack for temporary storage and a hashmap to link elements to their next greater element efficiently. Optimized for linear performance, it quickly maps results for nums1 after preparing the hashmap.