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Tested Python solution for LeetCode 503 with 21 pytest cases. Generate a practice environment with lcpy.
lcpy gen -n 503 # by problem number
lcpy gen -s next_greater_element_ii # by problem name
nums (i.e., the next element of nums[nums.length - 1] is nums[0]), return the next greater number for every element in nums.
The next greater number of a number x is the first greater number to its traversing-order next in the array, which means you could search circularly to find its next greater number. If it doesn’t exist, return -1 for this number.
Input: nums = [1,2,1]
Output: [2,-1,2]
Explanation: The first 1's next greater number is 2;
The number 2 can't find next greater number.
The second 1's next greater number needs to search circularly, which is also 2.
Input: nums = [1,2,3,4,3]
Output: [2,3,4,-1,4]
1 <= nums.length <= 10^4-10^9 <= nums[i] <= 10^9class Solution:
# Time: O(n)
# Space: O(n)
def next_greater_elements(self, nums: list[int]) -> list[int]:
n = len(nums)
result = [-1] * n
stack: list[int] = []
for i in range(2 * n):
idx = i % n
while stack and nums[stack[-1]] < nums[idx]:
result[stack.pop()] = nums[idx]
if i < n:
stack.append(idx)
return result
| Time | Space |
|---|---|
| O(n) | O(n) |