[Dynamic Programming] Word Break - Medium

Word Break Problem Explained

The Word Break problem is a common challenge in algorithms. It asks if we can split a string into words from a given dictionary. We can solve this problem well using dynamic programming. We create a table to help us see how we can split the string based on the words we have. By using simple checks and updates, we can find out if we can break the string using the word list.

In this article, we will look into the details of the Word Break problem. We will start by explaining what the problem is and its important parts. We will see different dynamic programming methods and show how to do them in Java, Python, and C++. We will also talk about how to make these solutions use less space. Moreover, we will discuss backtracking methods to solve this problem in these languages. We will answer common questions about Word Break too. Here are the topics we will cover:

• Dynamic Programming Word Break Problem Explained
• Understanding the Problem Statement for Word Break
• Dynamic Programming Approach for Word Break in Java
• Dynamic Programming Approach for Word Break in Python
• Dynamic Programming Approach for Word Break in C++
• Optimizing Space Complexity in Word Break Solutions
• Backtracking Method for Word Break in Java
• Backtracking Method for Word Break in Python
• Backtracking Method for Word Break in C++
• Frequently Asked Questions

For more ideas on dynamic programming techniques, we can check these articles: Dynamic Programming: Fibonacci Number, Dynamic Programming: Coin Change, and Dynamic Programming: Longest Common Subsequence.

Understanding the Problem Statement for Word Break

The Word Break problem is a well-known challenge in dynamic programming. It asks if a given string can be split into words from a dictionary. We have two main parts for the input:

• A string s that we need to check.
• A list of words wordDict which is our dictionary of valid words.

Problem Statement

We need to check if the string s can be split into a sequence of one or more words from wordDict. If it can, we return true. If not, we return false.

Example

Input: - s = "leetcode" - wordDict = ["leet", "code"]

Output: - true

Input: - s = "applepenapple" - wordDict = ["apple", "pen"]

Output: - true

Input: - s = "catsandog" - wordDict = ["cats", "dog", "sand", "and", "cat"]

Output: - false

Constraints

1. The string s has a length of n where 1 ≤ n ≤ 300.
2. The list wordDict can have at most 1000 different words.
3. Each word in wordDict can be up to 20 letters long.

We want to create a good algorithm using dynamic programming. This will help us solve the Word Break problem in a smart way. We need to keep the time we use as short as possible while following the rules we have.

Dynamic Programming Approach for Word Break in Java

The Word Break problem is about finding if a string can be split into words from a dictionary. We can use a dynamic programming approach to solve this problem. This method uses a boolean array to check which parts of the string can be made with the words from the dictionary.

Java Implementation

Here is a simple Java code for the Word Break problem using dynamic programming:

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

public class WordBreak {
    public boolean wordBreak(String s, Set<String> wordDict) {
        int n = s.length();
        boolean[] dp = new boolean[n + 1];
        dp[0] = true; // Base case: empty string can be split

        for (int i = 1; i <= n; i++) {
            for (int j = 0; j < i; j++) {
                if (dp[j] && wordDict.contains(s.substring(j, i))) {
                    dp[i] = true;
                    break;
                }
            }
        }
        return dp[n];
    }

    public static void main(String[] args) {
        WordBreak wb = new WordBreak();
        Set<String> dict = new HashSet<>();
        dict.add("leet");
        dict.add("code");
        String s = "leetcode";
        System.out.println(wb.wordBreak(s, dict)); // Output: true
    }
}

Explanation of the Code

• Input: The method wordBreak takes a string s and a set of words wordDict.
• DP Array: We create a boolean array dp of size n + 1, where n is the length of the string. The dp[i] shows if the part s[0..i-1] can be split into words from the dictionary.
• Base Case: We set dp[0] to true for the empty substring.
• Nested Loops: The outer loop goes through the string length. The inner loop checks all possible splits. If we find a valid split, we set dp[i] to true.
• Return Value: The method returns dp[n], which tells if the full string can be split.

Time Complexity

• The time complexity is O(n^2 * m), where n is the length of the string and m is the average length of the words in the dictionary. The space complexity is O(n) because of the DP array.

This method solves the Word Break problem well using dynamic programming in Java. It gives an efficient way to handle this task. If we want to learn more about dynamic programming, we can check related topics like Dynamic Programming - Coin Change and Dynamic Programming - Unique Paths II.

Dynamic Programming Approach for Word Break in Python

The Word Break problem is about finding if we can split a given string into words from a dictionary. The dynamic programming way helps us check if we can make the string using the words we have.

Problem Definition

We have a string s and a list of words called wordDict. Our job is to return True if we can break s into words from wordDict. If not, we return False.

Dynamic Programming Solution

We can use a list called dp. In this list, dp[i] will be True if the part of the string s[0:i] can be made using words from wordDict. We will go through the string and see if we can find the words.

Python Code Implementation

def wordBreak(s: str, wordDict: list[str]) -> bool:
    word_set = set(wordDict)
    dp = [False] * (len(s) + 1)
    dp[0] = True  # Base case: empty string can be broken
    
    for i in range(1, len(s) + 1):
        for j in range(i):
            if dp[j] and s[j:i] in word_set:
                dp[i] = True
                break
                
    return dp[len(s)]

Explanation of the Code

• Initialization:
  • We change wordDict into a set so we can look for words fast.
  • We create a dp list with size len(s) + 1 and fill it with False. We set dp[0] to True because the empty string case is true.
• Filling the DP Array:
  • We loop through the string with index i.
  • For each i, we check all earlier indices j.
  • If dp[j] is True and the part s[j:i] is in word_set, we set dp[i] to True.

Example Usage

s = "leetcode"
wordDict = ["leet", "code"]
print(wordBreak(s, wordDict))  # Output: True

s = "applepenapple"
wordDict = ["apple", "pen"]
print(wordBreak(s, wordDict))  # Output: True

s = "catsandog"
wordDict = ["cats", "dog", "sand", "and", "cat"]
print(wordBreak(s, wordDict))  # Output: False

This dynamic programming way helps us solve the problem quickly. The time needed is O(n^2) and space needed is O(n). Here n is the length of the string s.

For more problems like this, we can look at Dynamic Programming - Unique Paths II and Dynamic Programming - Coin Change.

Dynamic Programming Approach for Word Break in C++

We can solve the Word Break problem using a dynamic programming approach in C++. The main aim is to find out if a given string can be split into a space-separated sequence of one or more words from a dictionary.

Problem Statement

We have a string s and a dictionary of strings wordDict. We want to return true if s can be split into a space-separated sequence of one or more dictionary words.

Dynamic Programming Solution

1. Define a DP Array: We create a boolean DP array dp with size n+1 where n is the length of the string s. The value of dp[i] will be true if the substring s[0...i-1] can be split into words from the dictionary.

2. Initialization: We set dp[0] to true because an empty string can always be split.

3. Filling the DP Array: We go through each position i in the string. For each i, we check all previous positions j (from 0 to i). We see if s[j...i-1] is in the dictionary and if dp[j] is true.

4. Final Result: We return the value of dp[n].

C++ Code Implementation

#include <iostream>
#include <vector>
#include <unordered_set>
#include <string>

bool wordBreak(std::string s, std::unordered_set<std::string>& wordDict) {
    int n = s.length();
    std::vector<bool> dp(n + 1, false);
    dp[0] = true; // Base case

    for (int i = 1; i <= n; i++) {
        for (int j = 0; j < i; j++) {
            if (dp[j] && wordDict.find(s.substr(j, i - j)) != wordDict.end()) {
                dp[i] = true;
                break;
            }
        }
    }
    return dp[n];
}

int main() {
    std::string s = "leetcode";
    std::unordered_set<std::string> wordDict = {"leet", "code"};

    if (wordBreak(s, wordDict)) {
        std::cout << "String can be segmented." << std::endl;
    } else {
        std::cout << "String cannot be segmented." << std::endl;
    }
    return 0;
}

Complexity Analysis

• Time Complexity: O(n^2). Here n is the length of the string s. The outer loop runs n times. For each time, the inner loop can also run up to n times in the worst case.
• Space Complexity: O(n) for the DP array.

This dynamic programming solution helps us quickly know if the string can be split. It uses checks for substrings and results we calculated before.

Optimizing Space Complexity in Word Break Solutions

In the Word Break problem, we often use dynamic programming. This helps us store results for string splitting. But, this can use a lot of space. This is especially true when we use a big boolean array or a hashmap. Here are some ways to make space usage better in Word Break solutions.

1. Use a Set for Dictionary Storage

Instead of an array or list for words, we can use a Set. It gives us O(1) average time for lookups. This makes it faster.

2. In-Place Updates

When we use dynamic programming, we can reuse the same array for results. We can update it in-place. This way, we use less extra space.

3. Recursive with Memoization

If we use a recursive solution with memoization, we should only store necessary states. This helps us to use less memory compared to saving results for all prefixes.

4. Bottom-Up Approach

Using a bottom-up DP approach helps us avoid using stack space from recursion. This method builds solutions from smaller parts step by step. So we need less extra space.

Example in Java

import java.util.*;

public class WordBreak {
    public static boolean wordBreak(String s, Set<String> wordDict) {
        boolean[] dp = new boolean[s.length() + 1];
        dp[0] = true;

        for (int i = 1; i <= s.length(); i++) {
            for (int j = 0; j < i; j++) {
                if (dp[j] && wordDict.contains(s.substring(j, i))) {
                    dp[i] = true;
                    break;
                }
            }
        }
        return dp[s.length()];
    }

    public static void main(String[] args) {
        Set<String> wordDict = new HashSet<>(Arrays.asList("leet", "code"));
        System.out.println(wordBreak("leetcode", wordDict)); // Output: true
    }
}

Example in Python

def wordBreak(s: str, wordDict: set) -> bool:
    dp = [False] * (len(s) + 1)
    dp[0] = True

    for i in range(1, len(s) + 1):
        for j in range(i):
            if dp[j] and s[j:i] in wordDict:
                dp[i] = True
                break
    return dp[len(s)]

wordDict = {"leet", "code"}
print(wordBreak("leetcode", wordDict))  # Output: True

Example in C++

#include <iostream>
#include <unordered_set>
#include <vector>
using namespace std;

bool wordBreak(string s, unordered_set<string>& wordDict) {
    vector<bool> dp(s.length() + 1, false);
    dp[0] = true;

    for (int i = 1; i <= s.length(); i++) {
        for (int j = 0; j < i; j++) {
            if (dp[j] && wordDict.count(s.substr(j, i - j))) {
                dp[i] = true;
                break;
            }
        }
    }
    return dp[s.length()];
}

int main() {
    unordered_set<string> wordDict = {"leet", "code"};
    cout << wordBreak("leetcode", wordDict); // Output: 1 (true)
    return 0;
}

By using these tips, we can reduce space complexity a lot. This helps make our solutions for the Word Break problem more efficient and scalable. For more advanced dynamic programming ideas, we can check topics like Dynamic Programming - Coin Change and Dynamic Programming - Unique Paths.

Backtracking Method for Word Break in Java

We can solve the Word Break problem with the backtracking method. This way, we try to break the word into valid words from a dictionary by using recursion.

Implementation

Here is a simple Java code to show how we can use backtracking to solve the Word Break problem:

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

public class WordBreak {
    public boolean wordBreak(String s, Set<String> wordDict) {
        return backtrack(s, wordDict, 0);
    }

    private boolean backtrack(String s, Set<String> wordDict, int start) {
        if (start == s.length()) {
            return true;
        }

        for (int end = start + 1; end <= s.length(); end++) {
            String word = s.substring(start, end);
            if (wordDict.contains(word) && backtrack(s, wordDict, end)) {
                return true;
            }
        }

        return false;
    }

    public static void main(String[] args) {
        WordBreak wb = new WordBreak();
        String s = "leetcode";
        Set<String> wordDict = new HashSet<>();
        wordDict.add("leet");
        wordDict.add("code");

        boolean result = wb.wordBreak(s, wordDict);
        System.out.println("Can the word be segmented? " + result);
    }
}

Explanation

• Functionality: We start the process with the method wordBreak. The helper method backtrack checks for valid word segments one by one.
• Base Case: When the starting index is equal to the length of the string, it means we can segment the whole string.
• Recursion: For each possible end index, we check if the substring is in the dictionary. If it is, we call the method again with the new end index.
• Performance: This backtracking method might not be fast for large inputs. It tries many options without saving results.

This way gives us a simple and clear way to solve the Word Break problem, especially for small inputs. For larger inputs, we should think about using dynamic programming for better speed. For more on dynamic programming methods, you can check articles on Dynamic Programming - Fibonacci Number or Jump Game.

Backtracking Method for Word Break in Python

We can use the backtracking method to solve the Word Break problem. This method looks at all the ways we can split the input string. Then, it checks if the pieces can make valid words from the given dictionary. This method may be slower than dynamic programming. But it is often easier to use and understand for smaller strings.

Here is how we can implement the backtracking method in Python:

def wordBreak(s: str, wordDict: list) -> bool:
    word_set = set(wordDict)
    
    def backtrack(start: int) -> bool:
        if start == len(s):
            return True
        
        for end in range(start + 1, len(s) + 1):
            if s[start:end] in word_set:
                if backtrack(end):
                    return True
        
        return False
    
    return backtrack(0)

# Example usage
s = "leetcode"
wordDict = ["leet", "code"]
print(wordBreak(s, wordDict))  # Output: True

Explanation of the Code:

• wordBreak Function: We start the backtracking here. This function changes the word dictionary into a set. This makes it faster to check for words.
• backtrack Function: This function runs over and over to try to split the string s from the start index. If we reach the end of the string, it gives back True.
• Loop: This part goes through the possible end points. It checks if the piece from start to end is in the word set. If it is, it calls backtrack again from the end index.

Time Complexity:

The time complexity is O(2^n) in the worst case. Here, n is the length of the string. Every character can either be in a word or not.

Space Complexity:

The space complexity is O(n) because of the recursion stack.

This backtracking method is easy to follow but might not work well with big strings because it can take a lot of time. For bigger data, we should think about using dynamic programming for better speed.

If you want to read more about related topics, look at these resources on dynamic programming: Dynamic Programming: Coin Change and Dynamic Programming: Decode Ways.

Backtracking Method for Word Break in C++

We can use the backtracking method for the Word Break problem. This method checks all ways to split the input string into valid words from the dictionary. It might not be fast for large inputs because it looks at every combination. Below is a C++ code that shows how this method works.

C++ Implementation

#include <iostream>
#include <unordered_set>
#include <string>

using namespace std;

class WordBreak {
public:
    bool wordBreak(string s, unordered_set<string>& dict) {
        return backtrack(s, dict, 0);
    }

private:
    bool backtrack(string& s, unordered_set<string>& dict, int start) {
        if (start == s.length()) {
            return true; // We reach the end of the string
        }

        for (int end = start + 1; end <= s.length(); ++end) {
            string word = s.substr(start, end - start);
            if (dict.find(word) != dict.end() && backtrack(s, dict, end)) {
                return true; // We found a valid word and keep searching
            }
        }

        return false; // No valid way found
    }
};

int main() {
    WordBreak wb;
    unordered_set<string> dict = {"leet", "code"};
    string s = "leetcode";
    
    if (wb.wordBreak(s, dict)) {
        cout << "The string can be split into words from the dictionary." << endl;
    } else {
        cout << "The string cannot be split into words from the dictionary." << endl;
    }

    return 0;
}

Explanation of the Code

• The WordBreak class has a public method wordBreak that takes a string s and a dictionary dict.
• The private method backtrack does the searching:
  • It checks if we reached the end of the string.
  • It looks at possible end points to make substrings.
  • If we find a valid word in the dictionary, it checks the rest of the string again.
• The main function sets up the dictionary and tests if the string “leetcode” can be split.

Complexity Analysis

• Time Complexity: The worst time complexity is O(2^n) because of the many calls in recursion.
• Space Complexity: O(n) because of the recursion stack.

This backtracking method gives a simple way to solve the Word Break problem. But it may not be the best choice for bigger inputs. If we want better speed, we can try dynamic programming or memoization. For more on dynamic programming, we can look at articles like Dynamic Programming - Fibonacci Number or Dynamic Programming - Coin Change.

Frequently Asked Questions

1. What is the Word Break problem in dynamic programming?

The Word Break problem is a well-known challenge in dynamic programming. Here, we need to find out if a string can be split into words from a dictionary. We often solve this problem using memoization or tabulation. These methods help us check the possible ways to break the string. Understanding this problem is very important for making good solutions in languages like Java, Python, and C++.

2. How does dynamic programming solve the Word Break problem?

Dynamic programming helps us solve the Word Break problem by dividing it into smaller parts. We store the results of these parts to avoid doing the same work again. By using a boolean array to track if a substring can be broken down, we can quickly decide if we can segment the whole string. This way is much faster than a simple recursive solution.

3. Is backtracking a good method for solving the Word Break problem?

Yes, backtracking can work for the Word Break problem. But it is not as fast as dynamic programming. This method looks at all possible ways to split the string and checks if they match words in the dictionary. It is easier to implement but can take a lot of time for longer strings. So, it is not the best choice compared to dynamic programming.

4. How can I make space better in the Word Break solutions?

To make space usage better in Word Break solutions, we can use one boolean array instead of a 2D table. By going through the string and changing the array, we can use only O(n) space instead of O(n^2). This is very helpful when we deal with longer strings and still keep good time efficiency.

5. What other dynamic programming problems can I try after Word Break?

After we learn the Word Break problem, we can try other dynamic programming problems. Some good ones are the Fibonacci sequence, Climbing Stairs, or the Coin Change problem. These problems help us practice concepts like memoization and tabulation. For example, the Dynamic Programming: Coin Change problem is a nice way to use our skills in another situation.