What is the time complexity of Longest String Chain?

The optimal solution has a time complexity of O(N*M^2).

What data structures are used to solve Longest String Chain?

This problem uses Array, Hash Table, Two Pointers, String, Dynamic Programming, Sorting. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Longest String Chain?

Insight 1: Sorting words by length allows us to build the chain in a bottom-up manner. We only need to consider words longer than the current word being processed. Insight 2: Dynamic programming is used to efficiently store and reuse the length of the longest chain ending at a particular word. This avoids redundant calculations.

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Longest String Chain - LeetCode 1048 Solution

Q: Which companies ask LeetCode 1048. Longest String Chain in interviews?

This problem is frequently asked at Atlassian, Citadel, Flipkart, MathWorks, Moloco and 4 more companies.

Longest String Chain - Complete Solution Guide

Longest String Chain is LeetCode problem 1048, 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

You are given an array of words where each word consists of lowercase English letters. word A is a predecessor of word B if and only if we can insert exactly one letter anywhere in word A without changing the order of the other characters to make it equal to word B . For example, "abc" is a predecessor of "ab a c" , while "cba" is not a predecessor of "bcad" . A word chain is a sequence of words [word 1 , word 2 , ..., word k ] with k >= 1 , where word 1 is a predecessor of word 2 , word 2 is a

Detailed Explanation

The problem asks us to find the length of the longest possible word chain within a given array of words. A word chain is a sequence of words where each word is a predecessor of the next word. A word A is a predecessor of word B if we can insert exactly one letter into word A to make it equal to word B, without changing the order of other characters. For example, "abc" is a predecessor of "abac". The input is an array of words, and the output is the length of the longest possible word chain that can be formed using words from the input array.

Solution Approach

The solution uses dynamic programming to find the length of the longest word chain. The words are first sorted by their lengths. This makes it possible to build the chain in increasing order of length. A hash map (or dictionary) `dp` is used to store the longest chain ending at a particular word. For each word, it checks all its possible predecessors by removing one character at each possible index. If the predecessor exists in the `dp`, the current word's chain length is updated to be 1 plus the length of the chain ending at the predecessor. The maximum chain length seen so far is tracked and returned.

Step-by-Step Algorithm

Step 1: Sort the input array of words by length in ascending order.
Step 2: Initialize a hash map `dp` to store the length of the longest chain ending at each word. Initialize `max_len` to 0 to keep track of the longest chain found so far.
Step 3: Iterate through each word in the sorted array.
Step 4: For each word, iterate through each possible index `i` from 0 to the length of the word.
Step 5: Create a predecessor by removing the character at index `i` from the current word.
Step 6: Look up the predecessor in the `dp`. If the predecessor exists, update the `current_max` to be the maximum between the existing `current_max` value and the value associated with the predecessor in the `dp`.
Step 7: After checking all possible predecessors, store the length of the longest chain ending at the current word in the `dp` as `1 + current_max`.
Step 8: Update `max_len` to be the maximum of the current `max_len` and the length of the chain ending at the current word.
Step 9: After iterating through all the words, return `max_len`.

Key Insights

Insight 1: Sorting words by length allows us to build the chain in a bottom-up manner. We only need to consider words longer than the current word being processed.
Insight 2: Dynamic programming is used to efficiently store and reuse the length of the longest chain ending at a particular word. This avoids redundant calculations.
Insight 3: Checking if one word is a predecessor of another involves iterating through the longer word and comparing it to the shorter word after skipping one character at a time. This can be optimized by ensuring only one character is skipped.

Complexity Analysis

Time Complexity: O(N*M^2)

Space Complexity: O(N)

Topics

This problem involves: Array, Hash Table, Two Pointers, String, Dynamic Programming, Sorting.

Companies

Asked at: Atlassian, Citadel, Flipkart, MathWorks, Moloco, Snap, Two Sigma, Verily, Wix.

Longest String Chain | LeetCode 1048

Medium

Array Hash Table Two Pointers String Dynamic Programming Sorting

Time: O(N*M^2)Space: O(N)

Solution Code

Available in:

import java.util.Arrays;
import java.util.HashMap;
import java.util.Map;

class Solution {
    public int longestStrChain(String[] words) {
        Arrays.sort(words, (a, b) -> a.length() - b.length());
        Map<String, Integer> dp = new HashMap<>();
        int max_len = 0;

        for (String word : words) {
            int current_max = 0;
            for (int i = 0; i < word.length(); i++) {
                String predecessor = word.substring(0, i) + word.substring(i + 1);
                current_max = Math.max(current_max, dp.getOrDefault(predecessor, 0));
            }
            
            dp.put(word, 1 + current_max);
            max_len = Math.max(max_len, dp.get(word));
        }

        return max_len;
    }
}

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Problem Info

Description

Problem 1048: Longest String Chain - Detailed explanation and solution approaches.

Topics

Array Hash Table Two Pointers String Dynamic Programming Sorting

Companies

AtlassianCitadelFlipkartMathWorksMolocoSnapTwo SigmaVerily

Stats

Difficulty:Medium

Languages:4

Companies:9

Problem #:1048

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: Sort the input array of words by length in ascending order.
2
Step 2: Initialize a hash map `dp` to store the length of the longest chain ending at each word. Initialize `max_len` to 0 to keep track of the longest chain found so far.
3
Step 3: Iterate through each word in the sorted array.
4
Step 4: For each word, iterate through each possible index `i` from 0 to the length of the word.
5
Step 5: Create a predecessor by removing the character at index `i` from the current word.
6
Step 6: Look up the predecessor in the `dp`. If the predecessor exists, update the `current_max` to be the maximum between the existing `current_max` value and the value associated with the predecessor in the `dp`.
7
Step 7: After checking all possible predecessors, store the length of the longest chain ending at the current word in the `dp` as `1 + current_max`.
8
Step 8: Update `max_len` to be the maximum of the current `max_len` and the length of the chain ending at the current word.
9
Step 9: After iterating through all the words, return `max_len`.

💡

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*M^2)

💾

Space Complexity

O(N)

📝

Why?

The time complexity is derived from the nested loops and string operations. Sorting contributes O(N log N), but is smaller than O(N*M^2). The space complexity is mainly due to the `dp` hash map storing chain lengths for each word in the input.

📖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

Graph Approach

Model the words as nodes in a graph, where an edge exists between two words if one is a predecessor of the other. Then, perform a topological sort and find the longest path in the graph.

Time: O(N^2 * M), where N is the number of words and M is the average word length. Finding predecessor relationships requires comparing each pair of words O(N^2), and checking each pair is O(M). Topological sort is O(V+E) which in this case would still have a time complexity less than O(N^2 * M).Space: O(N^2), due to the adjacency matrix or list used to represent the graph.

Trade-offs: The graph approach is generally less efficient than the dynamic programming approach, especially when the number of words is large. The graph approach requires O(N^2) space to store the adjacency list or matrix, while the dynamic programming approach only requires O(N) space. The graph approach might be more intuitive for some, but the dynamic programming solution provides better performance.

✨

Key Insights

1Insight 1: Sorting words by length allows us to build the chain in a bottom-up manner. We only need to consider words longer than the current word being processed.
2Insight 2: Dynamic programming is used to efficiently store and reuse the length of the longest chain ending at a particular word. This avoids redundant calculations.
3Insight 3: Checking if one word is a predecessor of another involves iterating through the longer word and comparing it to the shorter word after skipping one character at a time. This can be optimized by ensuring only one character is skipped.

📚

Key Terms Explained

For Students

Array

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

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Two Pointers

Using two indices to traverse data from different positions simultaneously.

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

String

A sequence of characters. Like a word or sentence in programming!

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: Empty input array. The code handles this case correctly by returning 0 as `max_len` is initialized to 0 and the loop does not execute.
Edge case 2: All words have the same length. The algorithm still works correctly as it iterates through each word and tries to form a chain based on predecessor relationships. The dynamic programming approach ensures that the longest chain is calculated correctly even if words are of the same length and no predecessors exist.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not sorting the words by length. If the words are not sorted by length, it becomes more difficult to build the word chain bottom-up, and finding the longest chain ending at each word becomes significantly more complex.

✗Common mistake 2: Inefficiently checking predecessor relationships. A naive implementation might involve generating all possible words by inserting a single character into the shorter word and comparing with the longer word. This would be less efficient than generating all possible predecessors from the longer word by removing one character at a time.

✅

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: Spell checker suggestions. When a user misspells a word, a spell checker might suggest similar words based on edit distance (number of insertions, deletions, or substitutions). Finding the longest chain of words with a single insertion can be useful in generating ranked suggestions.

🏦Real-world scenario 2: Bioinformatics - Sequence Alignment. Identifying similar DNA or protein sequences. The concept of predecessor/successor relationship can be extended to sequence alignment algorithms where insertions/deletions play a role in determining the similarity between sequences.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ 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 27, 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 14

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

@TechLeadJul 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.

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

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 String Chain"?

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

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