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

The optimal solution has a time complexity of O(n).

What data structures are used to solve Number of Subarrays with Bounded Maximum?

This problem uses Array, Two Pointers. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Number of Subarrays with Bounded Maximum?

Insight 1: The core idea is to reframe the problem. Instead of directly counting valid subarrays, we can count the total number of subarrays with a maximum value no greater than `right` and subtract the number of subarrays with a maximum value strictly less than `left` (i.e., no greater than `left - 1`). Insight 2: The problem can be solved using a sliding window approach implicitly. We iterate through the array, tracking the length of the current valid subarray (where all elements are within the bound). Whenever we encounter an element outside the bound, we reset the length of the current valid subarray to 0.

Which companies ask LeetCode 795. Number of Subarrays with Bounded Maximum in interviews?

This problem is frequently asked at Quora.

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Number of Subarrays with Bounded Maximum - LeetCode 795 Solution

Number of Subarrays with Bounded Maximum - Complete Solution Guide

Number of Subarrays with Bounded Maximum is LeetCode problem 795, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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. Example 1: 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]. Example 2: Input: nums = [2,9,2,5,6], left = 2, right = 8 O

Detailed Explanation

The problem asks us to find the number of contiguous subarrays within a given integer array `nums` such that the maximum value within each subarray falls within the inclusive range `[left, right]`. We need to return the count of such subarrays. The constraints ensure the answer will fit within a 32-bit integer.

Solution Approach

The solution calculates the number of subarrays with a maximum value less than or equal to `right` and subtracts the number of subarrays with a maximum value less than `left`. This is done by implementing a helper function `count(bound)` which iterates through the array and counts the number of subarrays where the maximum element is less than or equal to `bound`. The function maintains a running count `cur` of the length of the current valid subarray. If an element `x` is less than or equal to `bound`, `cur` is incremented. Otherwise, `cur` is reset to 0. In each iteration, `cur` is added to the total count `ans`.

Step-by-Step Algorithm

Step 1: Define a helper function `count(bound)` that takes an integer `bound` as input.
Step 2: Initialize `ans` (the total number of subarrays) and `cur` (the length of the current valid subarray) to 0.
Step 3: Iterate through the input array `nums`.
Step 4: For each element `x` in `nums`, check if `x` is less than or equal to `bound`.
Step 5: If `x <= bound`, increment `cur` by 1, indicating that the current valid subarray can be extended.
Step 6: If `x > bound`, reset `cur` to 0, because the current valid subarray has ended.
Step 7: Add the value of `cur` to `ans`. This effectively counts all subarrays ending at the current position with maximum value less than or equal to `bound`.
Step 8: After iterating through the entire array, return `ans`.
Step 9: In the main function `numSubarrayBoundedMax`, call `count(right)` to calculate the number of subarrays with a maximum value no greater than `right`.
Step 10: Call `count(left - 1)` to calculate the number of subarrays with a maximum value no greater than `left - 1`.
Step 11: Return the difference between `count(right)` and `count(left - 1)`. This difference represents the number of subarrays whose maximum value falls within the range `[left, right]`.

Key Insights

Insight 1: The core idea is to reframe the problem. Instead of directly counting valid subarrays, we can count the total number of subarrays with a maximum value no greater than `right` and subtract the number of subarrays with a maximum value strictly less than `left` (i.e., no greater than `left - 1`).
Insight 2: The problem can be solved using a sliding window approach implicitly. We iterate through the array, tracking the length of the current valid subarray (where all elements are within the bound). Whenever we encounter an element outside the bound, we reset the length of the current valid subarray to 0.
Insight 3: The function `count(bound)` calculates the number of subarrays with a maximum value less than or equal to `bound` efficiently in O(n) time. Subtracting `count(left - 1)` from `count(right)` gives the number of subarrays whose maximum value falls within the range `[left, right]`.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Two Pointers.

Companies

Asked at: Quora.

Number of Subarrays with Bounded Maximum | LeetCode 795

Medium

Array Two Pointers

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public int numSubarrayBoundedMax(int[] nums, int left, int right) {
        return count(nums, right) - count(nums, left - 1);
    }

    private int count(int[] nums, int bound) {
        int ans = 0;
        int cur = 0;
        for (int x : nums) {
            if (x <= bound) {
                cur += 1;
            } else {
                cur = 0;
            }
            ans += cur;
        }
        return ans;
    }
}

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Problem Info

Description

Problem 795: Number of Subarrays with Bounded Maximum - Detailed explanation and solution approaches.

Topics

Array Two Pointers

Companies

Quora

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:795

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

11 steps

1
Step 1: Define a helper function `count(bound)` that takes an integer `bound` as input.
2
Step 2: Initialize `ans` (the total number of subarrays) and `cur` (the length of the current valid subarray) to 0.
3
Step 3: Iterate through the input array `nums`.
4
Step 4: For each element `x` in `nums`, check if `x` is less than or equal to `bound`.
5
Step 5: If `x <= bound`, increment `cur` by 1, indicating that the current valid subarray can be extended.
6
Step 6: If `x > bound`, reset `cur` to 0, because the current valid subarray has ended.
7
Step 7: Add the value of `cur` to `ans`. This effectively counts all subarrays ending at the current position with maximum value less than or equal to `bound`.
8
Step 8: After iterating through the entire array, return `ans`.
9
Step 9: In the main function `numSubarrayBoundedMax`, call `count(right)` to calculate the number of subarrays with a maximum value no greater than `right`.
10
Step 10: Call `count(left - 1)` to calculate the number of subarrays with a maximum value no greater than `left - 1`.
11
Step 11: Return the difference between `count(right)` and `count(left - 1)`. This difference represents the number of subarrays whose maximum value falls within the range `[left, right]`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is linear because we iterate through the input array a fixed number of times (twice). The space complexity is constant because the extra space used by the algorithm does not depend on the size of the input array.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Two Pointers/Sliding Window

A slightly more direct implementation using two pointers to define a sliding window. Use two pointers, start and end. Move the 'end' pointer along the array. If nums[end] is within [left, right], the number of new subarrays is end - start + 1. If nums[end] > right, reset start = end + 1. If nums[end] < left, continue moving 'end' pointer but keep track of last valid position when element was within range [left, right].

Time: O(n)Space: O(1)

Trade-offs: This approach might be more intuitive for some, but the current solution is more concise and easier to understand due to the abstraction of the `count` function.

✨

Key Insights

1Insight 1: The core idea is to reframe the problem. Instead of directly counting valid subarrays, we can count the total number of subarrays with a maximum value no greater than `right` and subtract the number of subarrays with a maximum value strictly less than `left` (i.e., no greater than `left - 1`).
2Insight 2: The problem can be solved using a sliding window approach implicitly. We iterate through the array, tracking the length of the current valid subarray (where all elements are within the bound). Whenever we encounter an element outside the bound, we reset the length of the current valid subarray to 0.
3Insight 3: The function `count(bound)` calculates the number of subarrays with a maximum value less than or equal to `bound` efficiently in O(n) time. Subtracting `count(left - 1)` from `count(right)` gives the number of subarrays whose maximum value falls within the range `[left, right]`.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Two Pointers

Using two indices to traverse data from different positions simultaneously.

Edge Cases

Edge case 1: Empty input array `nums`. The `count` function will return 0 in both calls, so the final result will be 0, which is correct.
Edge case 2: `left` equals `right`. The solution still works correctly, counting subarrays whose maximum value is equal to `left` (or `right`).
Edge case 3: `left` or `right` are outside the range of numbers in `nums`. The algorithm still produces a correct result since the comparison `x <= bound` will still evaluate appropriately.
Edge case 4: All numbers in `nums` are less than left. The `count(right)` will return the count of all possible subarrays. The `count(left-1)` will also return the same value and subtracting it will return 0.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly implementing the `count` function. Forgetting to reset the `cur` variable when an element exceeds the bound will lead to an overcount of subarrays.

✗Common mistake 2: Incorrectly handling the subtraction in the main function. Subtracting `count(left)` instead of `count(left - 1)` will exclude valid subarrays where the maximum value is equal to `left`.

✗Common mistake 3: Not understanding that counting all subarrays with max <= right and subtracting those with max < left is an efficient method to count subarrays where left <= max <= right.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Analyzing stock prices to find periods where the stock price stayed within a certain range. The algorithm can identify consecutive days where the stock price never exceeded a maximum price or fell below a minimum price.

🏦Real-world scenario 2: Network monitoring. The algorithm can be used to find time intervals where network latency or bandwidth usage remained within acceptable bounds. Each value in the array represents the latency or bandwidth at a specific time, and `left` and `right` represent the acceptable minimum and maximum values, respectively.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 15, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 14

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 7

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How do I avoid infinite loops with two pointers?

Always ensure your loop condition guarantees termination (typically left < right or slow != fast). Move at least one pointer in each iteration, and handle edge cases where pointers might not move (like skipping duplicates). Draw out pointer movement on paper first.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Number of Subarrays with Bounded Maximum"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether two pointers techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?