What is the time complexity of Count Subarrays With Score Less Than K?

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

What data structures are used to solve Count Subarrays With Score Less Than K?

This problem uses Array, Binary Search, Sliding Window, Prefix Sum. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Count Subarrays With Score Less Than K?

Insight 1: A brute-force approach (checking all possible subarrays) would be inefficient with a time complexity of O(n^2). Therefore, we need a more efficient method. Insight 2: The sliding window technique allows us to maintain a window of contiguous elements and efficiently update the sum and length as we iterate through the array, leading to a linear time solution.

Which companies ask LeetCode 2302. Count Subarrays With Score Less Than K in interviews?

This problem is frequently asked at Pinterest.

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Count Subarrays With Score Less Than K - LeetCode 2302 Solution

Count Subarrays With Score Less Than K - Complete Solution Guide

Count Subarrays With Score Less Than K is LeetCode problem 2302, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

The score of an array is defined as the product of its sum and its length. For example, the score of [1, 2, 3, 4, 5] is (1 + 2 + 3 + 4 + 5) * 5 = 75 . Given a positive integer array nums and an integer k , return the number of non-empty subarrays of nums whose score is strictly less than k . A subarray is a contiguous sequence of elements within an array. Example 1: Input: nums = [2,1,4,3,5], k = 10 Output: 6 Explanation: The 6 subarrays having scores less than 10 are: - [2] with score 2 * 1 = 2

Detailed Explanation

The problem requires us to count the number of non-empty subarrays within a given array `nums` whose score is strictly less than a given integer `k`. The score of a subarray is calculated as the product of the sum of its elements and the length of the subarray. The input consists of an array of positive integers `nums` and a positive integer `k`. The output is the count of subarrays that meet the specified score condition.

Solution Approach

The solution employs a sliding window approach. We maintain a window defined by `left` and `right` pointers. As we iterate through the array using the `right` pointer, we expand the window by including the current element. We calculate the score of the current window. If the score is greater than or equal to `k`, we shrink the window by moving the `left` pointer and updating the current sum until the score is less than `k`. For each valid window, we increment the count of subarrays by the length of the window (right - left + 1).

Step-by-Step Algorithm

Step 1: Initialize `left` to 0, `current_sum` to 0, and `count` to 0.
Step 2: Iterate through the array `nums` using the `right` pointer.
Step 3: Add the current element `nums[right]` to `current_sum`.
Step 4: While the score `current_sum * (right - left + 1)` is greater than or equal to `k`, subtract `nums[left]` from `current_sum` and increment `left`.
Step 5: Add `(right - left + 1)` to `count`. This represents the number of subarrays ending at index `right` that have a score less than `k`.
Step 6: After the loop finishes, return `count`.

Key Insights

Insight 1: A brute-force approach (checking all possible subarrays) would be inefficient with a time complexity of O(n^2). Therefore, we need a more efficient method.
Insight 2: The sliding window technique allows us to maintain a window of contiguous elements and efficiently update the sum and length as we iterate through the array, leading to a linear time solution.
Insight 3: Since the constraints specify `k` can be up to 10^15, using `long` type in Java/C++ is crucial to avoid potential integer overflow when calculating the score.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Binary Search, Sliding Window, Prefix Sum.

Companies

Asked at: Pinterest.

Count Subarrays With Score Less Than K | LeetCode 2302

Hard

Array Binary Search Sliding Window Prefix Sum

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

Solution Code

Available in:

class Solution {
    public long countSubarrays(int[] nums, long k) {
        int left = 0;
        long current_sum = 0;
        long count = 0;
        for (int right = 0; right < nums.length; right++) {
            current_sum += nums[right];
            while (current_sum * (right - left + 1) >= k) {
                current_sum -= nums[left];
                left += 1;
            }
            count += (right - left + 1);
        }
        return count;
    }
}

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Problem Info

Description

Problem 2302: Count Subarrays With Score Less Than K - Detailed explanation and solution approaches.

Topics

Array Binary Search Sliding Window Prefix Sum

Companies

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:2302

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

6 steps

1
Step 1: Initialize `left` to 0, `current_sum` to 0, and `count` to 0.
2
Step 2: Iterate through the array `nums` using the `right` pointer.
3
Step 3: Add the current element `nums[right]` to `current_sum`.
4
Step 4: While the score `current_sum * (right - left + 1)` is greater than or equal to `k`, subtract `nums[left]` from `current_sum` and increment `left`.
5
Step 5: Add `(right - left + 1)` to `count`. This represents the number of subarrays ending at index `right` that have a score less than `k`.
6
Step 6: After the loop finishes, return `count`.

💡

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 array at most twice (once with the right pointer and at most once with the left pointer). The space complexity is constant because we are not using any auxiliary data structures that scale with the input size. The number of variables we use remains the same regardless of 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

Brute Force

Iterate through all possible subarrays and calculate the score for each. Increment the count if the score is less than `k`.

Time: O(n^2), where n is the length of the input array, because we have nested loops to generate all possible subarrays.Space: O(1). We use a constant amount of extra space.

Trade-offs: The brute-force approach is simple to understand but inefficient for larger input arrays. It is suitable only when `n` is small. The sliding window approach is much more efficient for large `n`.

✨

Key Insights

1Insight 1: A brute-force approach (checking all possible subarrays) would be inefficient with a time complexity of O(n^2). Therefore, we need a more efficient method.
2Insight 2: The sliding window technique allows us to maintain a window of contiguous elements and efficiently update the sum and length as we iterate through the array, leading to a linear time solution.
3Insight 3: Since the constraints specify `k` can be up to 10^15, using `long` type in Java/C++ is crucial to avoid potential integer overflow when calculating the score.

📚

Key Terms Explained

For Students

Array

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

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Sliding Window

A moving "window" over data to process subsets efficiently.

Edge Cases

Edge case 1: `nums` contains only one element and its score is less than `k`. The sliding window will be size 1, and the count will be incremented correctly.
Edge case 2: `k` is a very large number such that the score of all subarrays will be less than `k`. The `while` loop will never be executed, and the `count` will accurately reflect the total number of subarrays.
Edge case 3: `nums` contains only large numbers such that no subarray score is less than k. The count will be 0.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Using `int` instead of `long` for `current_sum` and `count` in Java/C++ can lead to integer overflow issues when the sum or product becomes large, especially considering the constraints on `k`.

✗Common mistake 2: Incorrectly updating the `left` pointer, or forgetting to update `current_sum` after moving `left` can lead to incorrect results. The order of operations within the `while` loop is crucial.

✗Common mistake 3: Forgetting the non-empty constraint, and incorrectly counting empty subarrays.

✅

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 network traffic data where `nums` represents the sizes of packets and `k` is a threshold. You might want to count how many consecutive sequences of packets have a combined score (sum * length) below the threshold, indicating a potential issue.

🏦Real-world scenario 2: Monitoring server performance where `nums` represents the load on the server and `k` is a performance threshold. This algorithm can identify periods where the server load remains within acceptable limits for a certain duration.

💡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

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ Watch Out

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

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: October 21, 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

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

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I avoid off-by-one errors in binary search?

Use the pattern: left = 0, right = n-1, while (left <= right), mid = left + (right - left) / 2. Decide if you need to find exact match, first occurrence, or last occurrence, and adjust your conditions accordingly. Test with arrays of size 0, 1, and 2.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Count Subarrays With Score Less Than K"?

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

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