Leetcode #795: Number of Subarrays with Bounded Maximum
In this guide, we solve Leetcode #795 Number of Subarrays with Bounded Maximum 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
Given an integer array nums and two integers left and right, return the number of contiguous non-empty subarrays such that the value of the maximum array element in that subarray is in the range [left, right]. The test cases are generated so that the answer will fit in a 32-bit integer.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Array, Two Pointers
Intuition
The constraints hint that we can reason about two ends of the data at once, which is perfect for a two-pointer scan.
Moving one pointer at a time keeps the invariant intact and avoids nested loops.
Approach
Place pointers at the left and right ends and move them based on the comparison or target condition.
This yields a clean linear pass after any required sorting.
Steps:
- Set left and right pointers.
- Move a pointer based on the condition.
- Update the best answer while scanning.
Example
Input: nums = [2,1,4,3], left = 2, right = 3
Output: 3
Explanation: There are three subarrays that meet the requirements: [2], [2, 1], [3].
Python Solution
class Solution:
def numSubarrayBoundedMax(self, nums: List[int], left: int, right: int) -> int:
def f(x):
cnt = t = 0
for v in nums:
t = 0 if v > x else t + 1
cnt += t
return cnt
return f(right) - f(left - 1)
Complexity
The time complexity is O(n) (after optional sort O(n log n)). The space complexity is O(1).
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.