Leetcode #2454: Next Greater Element IV
In this guide, we solve Leetcode #2454 Next Greater Element IV in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given a 0-indexed array of non-negative integers nums. For each integer in nums, you must find its respective second greater integer.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Stack, Array, Binary Search, Sorting, Monotonic Stack, Heap (Priority Queue)
Intuition
The problem structure suggests a monotonic decision, which makes binary search a natural fit.
By halving the search space each step, we reach the answer efficiently.
Approach
Search either directly on a sorted array or on the answer space using a check function.
Each check is fast, and the logarithmic search keeps the overall runtime low.
Steps:
- Define the search bounds.
- Check the mid point condition.
- Narrow the bounds until convergence.
Example
Input: nums = [2,4,0,9,6]
Output: [9,6,6,-1,-1]
Explanation:
0th index: 4 is the first integer greater than 2, and 9 is the second integer greater than 2, to the right of 2.
1st index: 9 is the first, and 6 is the second integer greater than 4, to the right of 4.
2nd index: 9 is the first, and 6 is the second integer greater than 0, to the right of 0.
3rd index: There is no integer greater than 9 to its right, so the second greater integer is considered to be -1.
4th index: There is no integer greater than 6 to its right, so the second greater integer is considered to be -1.
Thus, we return [9,6,6,-1,-1].
Python Solution
class Solution:
def secondGreaterElement(self, nums: List[int]) -> List[int]:
arr = [(x, i) for i, x in enumerate(nums)]
arr.sort(key=lambda x: -x[0])
sl = SortedList()
n = len(nums)
ans = [-1] * n
for _, i in arr:
j = sl.bisect_right(i)
if j + 1 < len(sl):
ans[i] = nums[sl[j + 1]]
sl.add(i)
return ans
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.