What is the time complexity of Arithmetic Slices?

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

What data structures are used to solve Arithmetic Slices?

This problem uses Array, Dynamic Programming, Sliding Window. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Arithmetic Slices?

Insight 1: The core idea is to iterate through the array and identify contiguous sequences where the difference between consecutive elements is constant. Insight 2: Instead of explicitly generating all subarrays, the solution efficiently counts arithmetic slices by keeping track of the current slice length.

Which companies ask LeetCode 413. Arithmetic Slices in interviews?

This problem is frequently asked at Baidu.

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Arithmetic Slices - LeetCode 413 Solution

Arithmetic Slices - Complete Solution Guide

Arithmetic Slices is LeetCode problem 413, 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

An integer array is called arithmetic if it consists of at least three elements and if the difference between any two consecutive elements is the same. For example, [1,3,5,7,9] , [7,7,7,7] , and [3,-1,-5,-9] are arithmetic sequences. Given an integer array nums , return the number of arithmetic subarrays of nums . A subarray is a contiguous subsequence of the array. Example 1: Input: nums = [1,2,3,4] Output: 3 Explanation: We have 3 arithmetic slices in nums: [1, 2, 3], [2, 3, 4] and [1,2,3,4] i

Detailed Explanation

The problem requires us to find the number of arithmetic subarrays within a given integer array `nums`. An arithmetic subarray is defined as a contiguous subsequence of `nums` that contains at least three elements and has a constant difference between consecutive elements (e.g., [1, 2, 3], [7, 7, 7], [3, 1, -1]). The goal is to count all such arithmetic subarrays present in the input array.

Solution Approach

The provided solution uses a dynamic programming approach that iterates through the array and maintains a counter, `current_slices`, representing the number of arithmetic slices ending at the current index. If the current element extends an existing arithmetic sequence, the `current_slices` counter is incremented. The total number of arithmetic slices is then updated by adding the `current_slices` value to `total_slices`. If the sequence breaks, `current_slices` is reset to 0. This approach avoids generating all possible subarrays, leading to an efficient O(n) solution.

Step-by-Step Algorithm

Step 1: Initialize `total_slices` and `current_slices` to 0. These variables will store the total count of arithmetic slices and the count of arithmetic slices ending at the current index, respectively.
Step 2: Iterate through the array starting from the third element (index 2) since an arithmetic slice must have at least three elements.
Step 3: For each element `nums[i]`, check if the difference between `nums[i]` and `nums[i-1]` is equal to the difference between `nums[i-1]` and `nums[i-2]`. This determines if the current element extends an existing arithmetic sequence.
Step 4: If the differences are equal, increment `current_slices` because adding the current element to the sequence forms one additional arithmetic slice. Specifically, we've extended the prior arithmetic slices ending at `i-1`, adding 1 to the number of valid slices ending at `i`.
Step 5: If the differences are not equal, reset `current_slices` to 0, as the current element does not extend the arithmetic sequence.
Step 6: Add `current_slices` to `total_slices` to accumulate the total count of arithmetic slices.
Step 7: After iterating through the entire array, return `total_slices`.
Step 8: Handle the case where array length is less than 3, and return 0 in such case.

Key Insights

Insight 1: The core idea is to iterate through the array and identify contiguous sequences where the difference between consecutive elements is constant.
Insight 2: Instead of explicitly generating all subarrays, the solution efficiently counts arithmetic slices by keeping track of the current slice length.
Insight 3: The number of arithmetic slices of length `k` in a continuous arithmetic sequence is `k - 2`.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Dynamic Programming, Sliding Window.

Companies

Asked at: Baidu.

Arithmetic Slices | LeetCode 413

Medium

Array Dynamic Programming Sliding Window

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

Solution Code

Available in:

class Solution {
    public int numberOfArithmeticSlices(int[] nums) {
        int n = nums.length;
        if (n < 3) {
            return 0;
        }
        
        int total_slices = 0;
        int current_slices = 0;
        
        for (int i = 2; i < n; i++) {
            if (nums[i] - nums[i-1] == nums[i-1] - nums[i-2]) {
                current_slices += 1;
            } else {
                current_slices = 0;
            }
            total_slices += current_slices;
        }
        
        return total_slices;
    }
}

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Problem Info

Description

Problem 413: Arithmetic Slices - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Sliding Window

Companies

Baidu

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:413

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

8 steps

1
Step 1: Initialize `total_slices` and `current_slices` to 0. These variables will store the total count of arithmetic slices and the count of arithmetic slices ending at the current index, respectively.
2
Step 2: Iterate through the array starting from the third element (index 2) since an arithmetic slice must have at least three elements.
3
Step 3: For each element `nums[i]`, check if the difference between `nums[i]` and `nums[i-1]` is equal to the difference between `nums[i-1]` and `nums[i-2]`. This determines if the current element extends an existing arithmetic sequence.
4
Step 4: If the differences are equal, increment `current_slices` because adding the current element to the sequence forms one additional arithmetic slice. Specifically, we've extended the prior arithmetic slices ending at `i-1`, adding 1 to the number of valid slices ending at `i`.
5
Step 5: If the differences are not equal, reset `current_slices` to 0, as the current element does not extend the arithmetic sequence.
6
Step 6: Add `current_slices` to `total_slices` to accumulate the total count of arithmetic slices.
7
Step 7: After iterating through the entire array, return `total_slices`.
8
Step 8: Handle the case where array length is less than 3, and return 0 in such case.

💡

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 O(n) because the solution iterates through the array only once. The space complexity is O(1) because the solution uses a constant amount of extra space, regardless of the size of the input array. The variables `total_slices` and `current_slices` are the only additional memory used.

📖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

Generate all possible subarrays of length at least 3 and check if each subarray is an arithmetic slice. This involves iterating through all possible start and end indices of subarrays.

Time: O(n^3)Space: O(1)

Trade-offs: The brute force approach is simple to understand but highly inefficient due to the nested loops. It's suitable for very small input sizes but quickly becomes impractical for larger arrays. The O(n) dynamic programming solution is significantly more efficient.

✨

Key Insights

1Insight 1: The core idea is to iterate through the array and identify contiguous sequences where the difference between consecutive elements is constant.
2Insight 2: Instead of explicitly generating all subarrays, the solution efficiently counts arithmetic slices by keeping track of the current slice length.
3Insight 3: The number of arithmetic slices of length `k` in a continuous arithmetic sequence is `k - 2`.

📚

Key Terms Explained

For Students

Array

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

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Sliding Window

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

Edge Cases

Edge case 1: `nums` is empty or has less than 3 elements. The solution correctly returns 0 because an arithmetic slice requires at least three elements.
Edge case 2: All elements in `nums` are the same. The solution correctly identifies and counts all arithmetic slices in this scenario (e.g., [2, 2, 2, 2] has two slices [2, 2, 2] and [2, 2, 2, 2]).
Edge case 3: Arithmetic progression exists at the beginning of the array. The algorithm correctly starts counting from the first possible slice.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to check if the array has at least three elements. If not, the solution should return 0 to avoid errors.

✗Common mistake 2: Incorrectly calculating the number of arithmetic slices within a continuous arithmetic sequence. Remember that an arithmetic sequence of length `k` contains `1 + 2 + ... + (k-2)` arithmetic slices, where each slice has a length of at least 3.

✗Common mistake 3: Not resetting `current_slices` to 0 when the arithmetic sequence breaks. This can lead to overcounting.

✅

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 identify trends. Detecting arithmetic slices can indicate a consistent upward or downward trend in stock prices over a period.

🏦Real-world scenario 2: Signal processing. Identifying arithmetic sequences in sensor data can help detect patterns or anomalies, such as a linearly increasing temperature reading or a consistently changing pressure value.

🎮Real-world scenario 3: Data compression. Arithmetic coding is a form of entropy encoding used in lossless data compression. Identifying repeating arithmetic patterns can help in building efficient data compression algorithms.

💡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

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ 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: July 24, 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

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

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

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

@DevExpert92Jun 29

❓

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.

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

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 "Arithmetic Slices"?

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 dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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