Leetcode #2334: Subarray With Elements Greater Than Varying Threshold
In this guide, we solve Leetcode #2334 Subarray With Elements Greater Than Varying Threshold 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 an integer array nums and an integer threshold. Find any subarray of nums of length k such that every element in the subarray is greater than threshold / k.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Stack, Union Find, Array, Monotonic Stack
Intuition
We need the next greater or smaller element efficiently, which is exactly what a monotonic stack offers.
Each element is pushed and popped at most once, yielding a linear-time scan.
Approach
Maintain a stack that is either increasing or decreasing, depending on the query.
When the invariant is broken, pop and resolve answers for those indices.
Steps:
- Scan elements once.
- Pop while the monotonic condition is violated.
- Use stack indices to update answers.
Example
Input: nums = [1,3,4,3,1], threshold = 6
Output: 3
Explanation: The subarray [3,4,3] has a size of 3, and every element is greater than 6 / 3 = 2.
Note that this is the only valid subarray.
Python Solution
class Solution:
def validSubarraySize(self, nums: List[int], threshold: int) -> int:
def find(x):
if p[x] != x:
p[x] = find(p[x])
return p[x]
def merge(a, b):
pa, pb = find(a), find(b)
if pa == pb:
return
p[pa] = pb
size[pb] += size[pa]
n = len(nums)
p = list(range(n))
size = [1] * n
arr = sorted(zip(nums, range(n)), reverse=True)
vis = [False] * n
for v, i in arr:
if i and vis[i - 1]:
merge(i, i - 1)
if i < n - 1 and vis[i + 1]:
merge(i, i + 1)
if v > threshold // size[find(i)]:
return size[find(i)]
vis[i] = True
return -1
Complexity
The time complexity is O(n). The space complexity is O(n).
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.