What is the time complexity of Arithmetic Slices II - Subsequence?

The optimal solution has a time complexity of O(n^2).

What data structures are used to solve Arithmetic Slices II - Subsequence?

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

What are the key insights for solving Arithmetic Slices II - Subsequence?

Insight 1: Dynamic Programming: The problem can be efficiently solved using dynamic programming. The key is to keep track of the number of arithmetic subsequences ending at each index with a specific common difference. Insight 2: Hashing common differences: A hash map is used to store the counts of arithmetic subsequences ending at each index, keyed by the common difference. This allows efficient lookup and update of the counts.

Which companies ask LeetCode 446. Arithmetic Slices II - Subsequence in interviews?

This problem is frequently asked at Baidu, Dunzo.

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Arithmetic Slices II - Subsequence - LeetCode 446 Solution

Q: What is the space complexity of Arithmetic Slices II - Subsequence?

The space complexity is O(n^2).

Arithmetic Slices II - Subsequence - Complete Solution Guide

Arithmetic Slices II - Subsequence is LeetCode problem 446, 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

Given an integer array nums , return the number of all the arithmetic subsequences of nums . A sequence of numbers 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. For example, [1, 1, 2, 5, 7] is not an arithmetic sequence. A subsequence of an array is a sequence that can be formed by removing some elements (possibly non

Detailed Explanation

The problem requires us to find the number of arithmetic subsequences present in a given integer array `nums`. An arithmetic subsequence is a sequence of at least three numbers such that the difference between consecutive elements is constant. A subsequence can be formed by removing zero or more elements from the original array without changing the order of the remaining elements. The task is to count the total number of such arithmetic subsequences.

Solution Approach

The solution uses dynamic programming to build up the count of arithmetic subsequences. For each number `nums[i]` in the array, we iterate through all the previous numbers `nums[j]` (where j < i). We calculate the difference between `nums[i]` and `nums[j]`. Then, we check how many arithmetic subsequences end at `nums[j]` with that same difference. We add this count to the total count of arithmetic subsequences and also update the count of arithmetic subsequences ending at `nums[i]` with that same difference.

Step-by-Step Algorithm

Step 1: Initialize a dynamic programming table `dp` where `dp[i]` is a hash map storing the number of arithmetic subsequences ending at index `i` with each possible common difference.
Step 2: Initialize `total_count` to 0. This variable will store the total number of arithmetic subsequences.
Step 3: Iterate through the `nums` array from `i = 0` to `n - 1`.
Step 4: For each `i`, iterate through the `nums` array from `j = 0` to `i - 1`.
Step 5: Calculate the difference `diff = nums[i] - nums[j]`.
Step 6: Retrieve the count `count_at_j` of arithmetic subsequences ending at index `j` with the common difference `diff` from `dp[j][diff]`.
Step 7: Add `count_at_j` to `total_count`. This is because each arithmetic subsequence ending at `j` with difference `diff`, when extended with `nums[i]`, becomes a new valid arithmetic subsequence.
Step 8: Update the count of arithmetic subsequences ending at index `i` with the common difference `diff` in `dp[i][diff]`. We increment it by `count_at_j + 1`. The `+ 1` accounts for the arithmetic subsequence `[nums[j], nums[i]]` of length 2, which can be extended to form a subsequence of length 3 or more.
Step 9: After iterating through all pairs `(i, j)`, return `total_count`.

Key Insights

Insight 1: Dynamic Programming: The problem can be efficiently solved using dynamic programming. The key is to keep track of the number of arithmetic subsequences ending at each index with a specific common difference.
Insight 2: Hashing common differences: A hash map is used to store the counts of arithmetic subsequences ending at each index, keyed by the common difference. This allows efficient lookup and update of the counts.
Insight 3: Handling Integer Overflow: Since the difference between array elements can be large, using `long` data type can prevent potential integer overflow during the difference calculation, particularly in languages like C++ and C.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(n^2)

Topics

This problem involves: Array, Dynamic Programming.

Companies

Asked at: Baidu, Dunzo.

Arithmetic Slices II - Subsequence | LeetCode 446

Hard

Array Dynamic Programming

Time: O(n^2)Space: O(n^2)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

class Solution {
    public int numberOfArithmeticSlices(int[] nums) {
        int n = nums.length;
        if (n < 3) {
            return 0;
        }

        int total_count = 0;
        List<Map<Integer, Integer>> dp = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            dp.add(new HashMap<>());
        }

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < i; j++) {
                int diff = nums[i] - nums[j];

                int count_at_j = dp.get(j).getOrDefault(diff, 0);

                total_count += count_at_j;

                dp.get(i).put(diff, dp.get(i).getOrDefault(diff, 0) + count_at_j + 1);
            }
        }

        return total_count;
    }
}

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Problem Info

Description

Problem 446: Arithmetic Slices II - Subsequence - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming

Companies

BaiduDunzo

Stats

Difficulty:Hard

Languages:4

Companies:2

Problem #:446

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

9 steps

1
Step 1: Initialize a dynamic programming table `dp` where `dp[i]` is a hash map storing the number of arithmetic subsequences ending at index `i` with each possible common difference.
2
Step 2: Initialize `total_count` to 0. This variable will store the total number of arithmetic subsequences.
3
Step 3: Iterate through the `nums` array from `i = 0` to `n - 1`.
4
Step 4: For each `i`, iterate through the `nums` array from `j = 0` to `i - 1`.
5
Step 5: Calculate the difference `diff = nums[i] - nums[j]`.
6
Step 6: Retrieve the count `count_at_j` of arithmetic subsequences ending at index `j` with the common difference `diff` from `dp[j][diff]`.
7
Step 7: Add `count_at_j` to `total_count`. This is because each arithmetic subsequence ending at `j` with difference `diff`, when extended with `nums[i]`, becomes a new valid arithmetic subsequence.
8
Step 8: Update the count of arithmetic subsequences ending at index `i` with the common difference `diff` in `dp[i][diff]`. We increment it by `count_at_j + 1`. The `+ 1` accounts for the arithmetic subsequence `[nums[j], nums[i]]` of length 2, which can be extended to form a subsequence of length 3 or more.
9
Step 9: After iterating through all pairs `(i, j)`, return `total_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^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(n^2)

📚 Quadratic space - Uses lots of memory!

📝

Why?

The outer loop iterates `n` times, and the inner loop iterates up to `n` times. Inside the inner loop, hash map operations (get and put) take amortized O(1) time. Therefore, the overall time complexity is O(n^2). The dynamic programming table `dp` stores at most `n` hash maps, each containing at most `n` key-value pairs. Thus, the space complexity is O(n^2).

📖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

Recursion with Memoization

This approach uses recursion to explore all possible subsequences and memoization to avoid redundant calculations. However, it can be less efficient than the dynamic programming approach due to the overhead of recursive calls.

Time: O(n^2)Space: O(n^2)

Trade-offs: Generally, DP is preferred due to lower overhead. Recursion with memoization might be easier to conceptualize initially for some, but the DP solution is typically more performant.

✨

Key Insights

1Insight 1: Dynamic Programming: The problem can be efficiently solved using dynamic programming. The key is to keep track of the number of arithmetic subsequences ending at each index with a specific common difference.
2Insight 2: Hashing common differences: A hash map is used to store the counts of arithmetic subsequences ending at each index, keyed by the common difference. This allows efficient lookup and update of the counts.
3Insight 3: Handling Integer Overflow: Since the difference between array elements can be large, using `long` data type can prevent potential integer overflow during the difference calculation, particularly in languages like C++ and C.

📚

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.

Edge Cases

Edge case 1: Empty array or array with fewer than 3 elements. The solution correctly handles this case by returning 0 because an arithmetic subsequence requires at least 3 elements.
Edge case 2: Arithmetic subsequences with same elements. The solution handles subsequences like [7, 7, 7] correctly because the difference will be zero and the counts are correctly accumulated in the hash maps.
Edge case 3: Integer Overflow. The difference `nums[i]-nums[j]` can exceed the integer range. The provided C++ and C solutions use `long long` to handle such cases, while Java and Python can implicitly handle larger integers. The C solution also checks if the `long long` representation of the difference is out of bounds of `int` before casting it to `int`, avoiding undefined behavior.
Edge case 4: Large input size. The constraints specify that the array length can be up to 1000. The O(n^2) solution can handle this within the time limit.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling integer overflow when calculating the difference between elements. This can lead to incorrect results, especially with large input values. Using `long` or similar larger integer types is necessary.

✗Common mistake 2: Counting arithmetic slices instead of subsequences. The problem asks for subsequences, which are more general than slices (contiguous subarrays). The DP approach is essential for correctly handling non-contiguous subsequences.

✗Common mistake 3: Forgetting to add 1 when updating the count in `dp[i][diff] += count_at_j + 1`. This '+1' represents the subsequence `[nums[j], nums[i]]` which now makes a valid arithmetic subsequence of length 2, which can be further extended.

✗Common mistake 4: Incorrectly handling hash map operations in C. Allocating memory and implementing `get`, `insert`, `increment`, and `freeHashTable` functions correctly is crucial to avoid memory leaks and program crashes.

✅

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: Stock market analysis. Identifying trends in stock prices that follow an arithmetic progression could be useful for predicting future prices.

🏦Real-world scenario 2: Signal processing. Detecting patterns in signals that exhibit arithmetic properties can be used in audio or image analysis.

🎮Real-world scenario 3: Data compression. Identifying arithmetic progressions in data can be used for efficient data compression techniques.

💡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

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ 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: August 26, 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

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

@AlgoMasterJul 14

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

@InterviewProJul 7

❓

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 identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

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 "Arithmetic Slices II - Subsequence"?

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

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