How do you solve Number of Subarrays with Bounded Maximum on LeetCode?

Number of Subarrays with Bounded Maximum (LeetCode #795) can be solved using Brute Force. The optimal approach is Brute Force with O(n²) time complexity and O(1) space complexity. Key insight: The brute force approach involves checking every possible subarray and determining if its maximum falls within the specified range. This is straightforward but inefficient for larger arrays.

What is the time complexity of Number of Subarrays with Bounded Maximum?

The optimal solution for Number of Subarrays with Bounded Maximum runs in O(n²) time and uses O(1) space. The time complexity is O(n²) because we are generating all subarrays using two nested loops, leading to a quadratic number of checks.

What algorithm or data structure is used to solve Number of Subarrays with Bounded Maximum?

Number of Subarrays with Bounded Maximum uses the following concepts: Array, Two Pointers. The recommended patterns to study are: Two Pointers, Sliding Window.

What are the interview tips for Number of Subarrays with Bounded Maximum?

Always clarify the problem statement and constraints before jumping into coding. Think about edge cases, such as arrays with all elements below or above the bounds. Practice explaining your thought process as you code; it helps in interviews.

What are common mistakes when solving Number of Subarrays with Bounded Maximum?

Not handling cases where the maximum exceeds the right limit correctly. Confusing the conditions for counting valid subarrays, especially when elements are less than the left limit.

Number of Subarrays with Bounded Maximum — LeetCode #795 (Medium)

Tags: Array, Two Pointers

Related patterns: Two Pointers, Sliding Window

Brute Force approach

Time complexity: O(n²). Space complexity: O(1).

The brute force approach involves checking every possible subarray and determining if its maximum falls within the specified range. This is straightforward but inefficient for larger arrays.

The time complexity is O(n²) because we are generating all subarrays using two nested loops, leading to a quadratic number of checks.

Step 1: Initialize a count variable to zero.
Step 2: Use two nested loops to generate all possible subarrays.
Step 3: For each subarray, find the maximum element and check if it lies within the range [left, right]. If it does, increment the count.

For nums = [2,1,4,3], left = 2, right = 3: 1. Subarray [2]: max = 2 (count = 1) 2. Subarray [2, 1]: max = 2 (count = 2) 3. Subarray [2, 1, 4]: max = 4 (break) 4. Subarray [1]: max = 1 (skip) 5. Subarray [4]: max = 4 (break) 6. Subarray [4, 3]: max = 4 (break) 7. Subarray [3]: max = 3 (count = 3)

Common Mistakes

Not handling cases where the maximum exceeds the right limit correctly.
Confusing the conditions for counting valid subarrays, especially when elements are less than the left limit.

Interview Tips

Always clarify the problem statement and constraints before jumping into coding.
Think about edge cases, such as arrays with all elements below or above the bounds.
Practice explaining your thought process as you code; it helps in interviews.

#795

Number of Subarrays with Bounded Maximum

Medium

Array↗Two Pointers↗Two Pointers↗Sliding Window↗

LeetCode ↗

Approaches

💡

Intuition

Time O(n²)Space O(1)

The brute force approach involves checking every possible subarray and determining if its maximum falls within the specified range. This is straightforward but inefficient for larger arrays.

⚙️

Algorithm

3 steps

1Step 1: Initialize a count variable to zero.
2Step 2: Use two nested loops to generate all possible subarrays.
3Step 3: For each subarray, find the maximum element and check if it lies within the range [left, right]. If it does, increment the count.

solution.py14 lines

1# Full working Python code
2
3def numSubarrayBoundedMax(nums, left, right):
4    count = 0
5    n = len(nums)
6    for i in range(n):
7        max_val = nums[i]
8        for j in range(i, n):
9            max_val = max(max_val, nums[j])
10            if left <= max_val <= right:
11                count += 1
12            if max_val > right:
13                break
14    return count

ℹ

Complexity note: The time complexity is O(n²) because we are generating all subarrays using two nested loops, leading to a quadratic number of checks.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.