What is the time complexity of Longest Increasing Subsequence?

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

What data structures are used to solve Longest Increasing Subsequence?

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

What are the key insights for solving Longest Increasing Subsequence?

Insight 1: We can use dynamic programming or a binary search based approach to solve this problem efficiently. The binary search approach is more efficient in terms of time complexity (O(n log n)). Insight 2: The core idea of the binary search approach is to maintain a 'tails' array. `tails[i]` is the smallest tail of all increasing subsequences with length `i+1`.

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Longest Increasing Subsequence - LeetCode 300 Solution

Q: How to solve Longest Increasing Subsequence?

The solution uses a dynamic programming strategy optimized with binary search. It maintains an array called `tails`, where `tails[i]` represents the smallest tail of all increasing subsequences of length `i+1`. The algorithm iterates through the input array `nums`. For each number, it either extends the current longest increasing subsequence (if the number is larger than all tails) or finds the appropriate position in the `tails` array to update (using binary search). The length of the `tails` array at the end is the length of the longest increasing subsequence.

Q: Which companies ask LeetCode 300. Longest Increasing Subsequence in interviews?

This problem is frequently asked at Accenture, Atlassian, Autodesk, Citadel, Commvault and 14 more companies.

Longest Increasing Subsequence - Complete Solution Guide

Longest Increasing Subsequence is LeetCode problem 300, 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 , return the length of the longest strictly increasing subsequence . Example 1: Input: nums = [10,9,2,5,3,7,101,18] Output: 4 Explanation: The longest increasing subsequence is [2,3,7,101], therefore the length is 4. Example 2: Input: nums = [0,1,0,3,2,3] Output: 4 Example 3: Input: nums = [7,7,7,7,7,7,7] Output: 1 Constraints: 1 <= nums.length <= 2500 -10 4 <= nums[i] <= 10 4 Follow up: Can you come up with an algorithm that runs in O(n log(n)) time complexity?

Detailed Explanation

The problem asks us to find the length of the longest strictly increasing subsequence within a given array of integers. A subsequence is a sequence that can be derived from an array by deleting some or no elements without changing the order of the remaining elements. 'Strictly increasing' means that each element in the subsequence must be greater than the previous one. The input is an array of integers, and the output is an integer representing the length of the longest increasing subsequence.

Solution Approach

The solution uses a dynamic programming strategy optimized with binary search. It maintains an array called `tails`, where `tails[i]` represents the smallest tail of all increasing subsequences of length `i+1`. The algorithm iterates through the input array `nums`. For each number, it either extends the current longest increasing subsequence (if the number is larger than all tails) or finds the appropriate position in the `tails` array to update (using binary search). The length of the `tails` array at the end is the length of the longest increasing subsequence.

Step-by-Step Algorithm

Step 1: Initialize an empty list `tails` to store the smallest tail of all increasing subsequences of different lengths.
Step 2: Iterate through each `num` in the input array `nums`.
Step 3: Perform a binary search on the `tails` array to find the smallest tail that is greater than or equal to `num`.
Step 4: If the binary search returns an index equal to the length of `tails`, it means `num` is greater than all tails in the `tails` array. In this case, append `num` to the `tails` array, extending the longest increasing subsequence.
Step 5: If the binary search returns an index `idx` less than the length of `tails`, it means `num` is smaller than or equal to `tails[idx]`. Replace `tails[idx]` with `num`. This step does not change the length of the longest increasing subsequence found so far, but it potentially allows us to build a longer subsequence later on because we are replacing the existing tail with a smaller number.
Step 6: After processing all numbers in `nums`, the length of the `tails` array is the length of the longest increasing subsequence. Return the length of `tails`.

Key Insights

Insight 1: We can use dynamic programming or a binary search based approach to solve this problem efficiently. The binary search approach is more efficient in terms of time complexity (O(n log n)).
Insight 2: The core idea of the binary search approach is to maintain a 'tails' array. `tails[i]` is the smallest tail of all increasing subsequences with length `i+1`.
Insight 3: If we encounter a number `num` that is greater than all tails in the `tails` array, it means we can extend the longest increasing subsequence by one. If `num` is smaller than some tail, we can potentially replace that tail with `num` to create a smaller tail, potentially enabling a longer increasing subsequence later on.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(n)

Topics

This problem involves: Array, Binary Search, Dynamic Programming.

Companies

Asked at: Accenture, Atlassian, Autodesk, Citadel, Commvault, Druva, Expedia, Flexport, Goldman Sachs, IBM, Intuit, Licious, PayPal, Pure Storage, Samsung, Splunk, Wayfair, Yandex, Zoho.

Longest Increasing Subsequence | LeetCode 300

Medium

Array Binary Search Dynamic Programming

Time: O(n log n)Space: O(n)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.List;

class Solution {
    public int lengthOfLIS(int[] nums) {
        List<Integer> tails = new ArrayList<>();
        for (int num : nums) {
            int idx = binarySearch(tails, num);
            if (idx == tails.size()) {
                tails.add(num);
            } else {
                tails.set(idx, num);
            }
        }
        return tails.size();
    }

    private int binarySearch(List<Integer> tails, int num) {
        int left = 0;
        int right = tails.size();
        while (left < right) {
            int mid = left + (right - left) / 2;
            if (tails.get(mid) < num) {
                left = mid + 1;
            } else {
                right = mid;
            }
        }
        return left;
    }
}

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Problem Info

Description

Problem 300: Longest Increasing Subsequence - Detailed explanation and solution approaches.

Topics

Array Binary Search Dynamic Programming

Companies

AccentureAtlassianAutodeskCitadelCommvaultDruvaExpediaFlexport

Stats

Difficulty:Medium

Languages:4

Companies:19

Problem #:300

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 an empty list `tails` to store the smallest tail of all increasing subsequences of different lengths.
2
Step 2: Iterate through each `num` in the input array `nums`.
3
Step 3: Perform a binary search on the `tails` array to find the smallest tail that is greater than or equal to `num`.
4
Step 4: If the binary search returns an index equal to the length of `tails`, it means `num` is greater than all tails in the `tails` array. In this case, append `num` to the `tails` array, extending the longest increasing subsequence.
5
Step 5: If the binary search returns an index `idx` less than the length of `tails`, it means `num` is smaller than or equal to `tails[idx]`. Replace `tails[idx]` with `num`. This step does not change the length of the longest increasing subsequence found so far, but it potentially allows us to build a longer subsequence later on because we are replacing the existing tail with a smaller number.
6
Step 6: After processing all numbers in `nums`, the length of the `tails` array is the length of the longest increasing subsequence. Return the length of `tails`.

💡

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 log n)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is determined by the number of iterations of the outer loop (O(n)) multiplied by the time complexity of the binary search within the loop (O(log n)). The space complexity is determined by the size of the `tails` array, which can grow up to the size of the input array in the worst case.

📖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

Dynamic Programming (O(n^2) Time Complexity)

A more straightforward dynamic programming approach can be used where `dp[i]` represents the length of the longest increasing subsequence ending at index `i`. We iterate through all previous indices `j < i` and check if `nums[j] < nums[i]`. If it is, then we update `dp[i] = max(dp[i], dp[j] + 1)`. The final answer is the maximum value in the `dp` array.

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

Trade-offs: This approach is easier to understand and implement, but it has a quadratic time complexity, making it less efficient for larger input arrays. Use this when n is relatively small or if code simplicity is favored over speed.

✨

Key Insights

1Insight 1: We can use dynamic programming or a binary search based approach to solve this problem efficiently. The binary search approach is more efficient in terms of time complexity (O(n log n)).
2Insight 2: The core idea of the binary search approach is to maintain a 'tails' array. `tails[i]` is the smallest tail of all increasing subsequences with length `i+1`.
3Insight 3: If we encounter a number `num` that is greater than all tails in the `tails` array, it means we can extend the longest increasing subsequence by one. If `num` is smaller than some tail, we can potentially replace that tail with `num` to create a smaller tail, potentially enabling a longer increasing subsequence later on.

📚

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!

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: Empty input array. The solution correctly handles this case because the loop won't execute, and an empty `tails` array will be returned, resulting in a length of 0.
Edge case 2: Input array with all equal numbers. In this case, the binary search will always return 0, and the first element of tails is always overwritten, meaning that only a subsequence of length 1 is ever achieved. This handles input like [7,7,7,7].
Edge case 3: Input array in descending order. The tails array will always be updated with the new element and will never increase in length. Therefore, the LIS is always 1.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly implementing the binary search. A common mistake is not handling the boundary conditions correctly, leading to infinite loops or incorrect results.

✗Common mistake 2: Confusing subsequence with substring. A subsequence doesn't have to be contiguous, while a substring must be.

✗Common mistake 3: Using a greedy approach. A purely greedy approach will not work. Choosing local optimal choices does not guarantee the global optimal solution (longest increasing subsequence).

✅

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, determining the longest period of increasing stock values.

🏦Real-world scenario 2: Bioinformatics. Aligning DNA sequences. The longest increasing subsequence can be used to find the similarity between two DNA sequences.

🎮Real-world scenario 3: Data compression. Analyzing data streams and identifying patterns for efficient compression.

💡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: September 29, 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

❓

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.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

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 "Longest Increasing Subsequence"?

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

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