What is the time complexity of Extra Characters in a String?

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

What data structures are used to solve Extra Characters in a String?

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

What are the key insights for solving Extra Characters in a String?

Insight 1: Dynamic Programming is the most effective way to solve this optimization problem, allowing us to build solutions for prefixes of `s` based on solutions for smaller prefixes. Insight 2: Representing the dictionary as a `set` allows for efficient O(1) lookups when checking if a substring is a valid word.

Which companies ask LeetCode 2707. Extra Characters in a String in interviews?

This problem is frequently asked at PornHub.

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Extra Characters in a String - LeetCode 2707 Solution

Extra Characters in a String - Complete Solution Guide

Extra Characters in a String is LeetCode problem 2707, 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 a 0-indexed string s and a dictionary of words dictionary . You have to break s into one or more non-overlapping substrings such that each substring is present in dictionary . There may be some extra characters in s which are not present in any of the substrings. Return the minimum number of extra characters left over if you break up s optimally. Example 1: Input: s = "leetscode", dictionary = ["leet","code","leetcode"] Output: 1 Explanation: We can break s in two substrings: "leet

Detailed Explanation

The problem asks us to find the minimum number of extra characters when breaking down a given string `s` into substrings, where each substring must be present in a given dictionary of words. We are allowed to have some characters in `s` that are not part of any substring from the dictionary. The goal is to minimize the count of these "extra" characters. For example, if `s` is "leetscode" and the dictionary is ["leet", "code", "leetcode"], we can break `s` into "leet" and "code", leaving one extra character 's'. Hence the answer is 1.

Solution Approach

The solution uses dynamic programming to solve this problem. `dp[i]` represents the minimum number of extra characters for the prefix of `s` of length `i`. The base case is `dp[0] = 0` (an empty string has no extra characters). For each index `i` from 1 to `n`, we consider two possibilities: either the character `s[i-1]` is an extra character, or it is part of a word in the dictionary that ends at index `i-1`. If `s[i-1]` is an extra character, then `dp[i] = dp[i-1] + 1`. Otherwise, we iterate through all possible starting positions `j` (from 0 to i-1) and check if the substring `s[j:i]` is a word in the dictionary. If it is, it means s[j:i] doesn't introduce extra character, then `dp[i] = min(dp[i], dp[j])`. After iterating over all the positions, dp[i] holds the optimal number of extra characters for the prefix of `s` of length `i`.

Step-by-Step Algorithm

Step 1: Initialize `dp` array of size `n+1` with 0. `dp[i]` represents the minimum extra characters for `s[:i]`.
Step 2: Create a `set` from the dictionary for faster word lookup.
Step 3: Iterate through the string `s` from index 1 to `n` (inclusive).
Step 4: For each index `i`, initialize `dp[i]` to `dp[i-1] + 1`. This assumes the current character is an extra character.
Step 5: Iterate from `j = 0` to `i - 1` to consider substrings ending at index `i - 1`. The substring being tested is `s[j:i]`.
Step 6: Check if `s[j:i]` is in the dictionary `set`. If it is, update `dp[i]` to `min(dp[i], dp[j])`.
Step 7: After iterating through all `j`, `dp[i]` will store the minimum extra characters for the prefix `s[:i]`.
Step 8: Return `dp[n]` which represents the minimum extra characters for the entire string `s`.

Key Insights

Insight 1: Dynamic Programming is the most effective way to solve this optimization problem, allowing us to build solutions for prefixes of `s` based on solutions for smaller prefixes.
Insight 2: Representing the dictionary as a `set` allows for efficient O(1) lookups when checking if a substring is a valid word.
Insight 3: We need to consider all possible partition points within the string `s` to find the optimal breakdown into dictionary words.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(n)

Topics

This problem involves: Array, Hash Table, String, Dynamic Programming, Trie.

Companies

Asked at: PornHub.

Extra Characters in a String | LeetCode 2707

Medium

Array Hash Table String Dynamic Programming Trie

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

Solution Code

Available in:

import java.util.HashSet;
import java.util.Set;

class Solution {
    public int minExtraChar(String s, String[] dictionary) {
        int n = s.length();
        Set<String> dictionarySet = new HashSet<>();
        for (String word : dictionary) {
            dictionarySet.add(word);
        }
        
        // dp[i] will store the minimum number of extra characters for the prefix s[:i].
        int[] dp = new int[n + 1];
        
        for (int i = 1; i <= n; i++) {
            // By default, we can consider the character s[i-1] as an extra character.
            dp[i] = dp[i-1] + 1;
            
            // We then check if any substring ending at i-1 is a word in the dictionary.
            for (int j = 0; j < i; j++) {
                String substring = s.substring(j, i);
                if (dictionarySet.contains(substring)) {
                    // If s[j:i] is a word, it's a potential partition.
                    // The number of extra characters would be dp[j].
                    // We take the minimum over all such partitions.
                    dp[i] = Math.min(dp[i], dp[j]);
                }
            }
        }
        
        return dp[n];
    }
}

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Problem Info

Description

Problem 2707: Extra Characters in a String - Detailed explanation and solution approaches.

Topics

Array Hash Table String Dynamic Programming Trie

Companies

PornHub

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2707

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 `dp` array of size `n+1` with 0. `dp[i]` represents the minimum extra characters for `s[:i]`.
2
Step 2: Create a `set` from the dictionary for faster word lookup.
3
Step 3: Iterate through the string `s` from index 1 to `n` (inclusive).
4
Step 4: For each index `i`, initialize `dp[i]` to `dp[i-1] + 1`. This assumes the current character is an extra character.
5
Step 5: Iterate from `j = 0` to `i - 1` to consider substrings ending at index `i - 1`. The substring being tested is `s[j:i]`.
6
Step 6: Check if `s[j:i]` is in the dictionary `set`. If it is, update `dp[i]` to `min(dp[i], dp[j])`.
7
Step 7: After iterating through all `j`, `dp[i]` will store the minimum extra characters for the prefix `s[:i]`.
8
Step 8: Return `dp[n]` which represents the minimum extra characters for the entire string `s`.

💡

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)

📦 Linear space - Memory grows with input size.

📝

Why?

The nested loops in the dynamic programming approach are the primary contributors to the O(n^2) time complexity. The outer loop iterates through each character of the string `s`, and the inner loop checks all possible substrings ending at that character. The `dp` array stores the minimum extra characters for each prefix of the string, leading to O(n) space complexity.

📖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

Trie

Using a Trie data structure to store the dictionary words can optimize the substring search process. We can traverse the string `s` and use the Trie to efficiently check if a substring ending at the current position is a valid word.

Time: O(n*m), where n is the length of s, and m is the average length of dictionary words.Space: O(K), where K is the sum of the lengths of all words in the dictionary. This comes from the Trie data structure.

Trade-offs: Trie can be more efficient when there are many common prefixes among the dictionary words. It can potentially improve performance when the dictionary is large and the string `s` needs to be checked frequently. However, the Trie data structure requires extra space.

✨

Key Insights

1Insight 1: Dynamic Programming is the most effective way to solve this optimization problem, allowing us to build solutions for prefixes of `s` based on solutions for smaller prefixes.
2Insight 2: Representing the dictionary as a `set` allows for efficient O(1) lookups when checking if a substring is a valid word.
3Insight 3: We need to consider all possible partition points within the string `s` to find the optimal breakdown into dictionary words.

📚

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!

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

String

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

Edge Cases

Edge case 1: Empty string `s`. The solution correctly handles this because the loop does not run, and `dp[0]` is correctly initialized to 0.
Edge case 2: Empty dictionary. The solution handles this correctly because the inner loop's if statement will always evaluate to false, so the dp values will simply increment by one for each character, correctly counting all characters as extra.
Edge case 3: Dictionary containing only the entire string `s`. The algorithm correctly identifies `s` as a valid partition and `dp[n]` will be 0.
Edge case 4: Very large dictionary. The `HashSet` or `unordered_set` used to store the dictionary provides efficient lookups, mitigating the performance impact of a very large dictionary.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not using a set for dictionary lookups, resulting in O(n) lookup time instead of O(1), thus increasing the overall time complexity to O(n^3).

✗Common mistake 2: Incorrectly initializing the `dp` array. The base case `dp[0] = 0` is crucial for the dynamic programming approach to work correctly.

✗Common mistake 3: Not considering all possible starting positions `j` for substrings ending at index `i-1`. Failing to do so can lead to suboptimal solutions.

✗Common mistake 4: Memory leaks in C implementation because of missing free() calls after allocating string using malloc and strdup

✅

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: Text segmentation and tokenization. This algorithm can be used to break down a text into meaningful tokens based on a predefined dictionary.

🏦Real-world scenario 2: Error correction and spell checking. This algorithm can be adapted to find the minimum number of incorrect or misspelled characters in a sentence by comparing it against a dictionary of valid words.

🎮Real-world scenario 3: Data compression. This algorithm can be used to identify and replace frequently occurring patterns in a string with shorter codes, thereby compressing the data.

💡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: June 8, 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

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

@TechLeadJun 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.

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

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 "Extra Characters in a String"?

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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