Word Break · Code

Problem

You are provided with a string s and a set of words called wordDict. Write a function to determine whether s can be broken down into a sequence of one or more words from wordDict, where each word can appear more than once and there are no spaces in s. If s can be segmented in such a way, return true; otherwise, return false.

Input:

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

Output:

false

Explanation: There is no valid segmentation of “catsandog” into dictionary words from wordDict.

Input:

s = "hellointerview", wordDict = ["hello","interview"]

Output:

true

Explanation: Return true because “hellointerview” can be segmented as “hello” and “interview”. Note that you are allowed to reuse a dictionary word.

Explanation

This solution uses bottom-up dynamic programming to solve the problem.

We create a boolean array dp of size n + 1 where n is the length of the input string. dp[i] is true if the first i characters of s can be segmented into a valid sequence of dictionary words. dp[0] is True because the empty string is a valid sequence - all other values are false to start.

def wordBreak(s, wordDict):
  wordSet = set(wordDict)
  dp = [False] * (len(s) + 1)
  dp[0] = True # Empty string is a valid break

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

  return dp[len(s)]

public boolean wordBreak(String s, List<String> wordDict) {
    Set<String> wordSet = new HashSet<>(wordDict);
    boolean[] dp = new boolean[s.length() + 1];
    dp[0] = true; // Empty string is a valid break

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

    return dp[s.length()];
}

func wordBreak(s string, wordDict []string) bool {
    wordSet := make(map[string]bool)
    for _, word := range wordDict {
        wordSet[word] = true
    }
    dp := make([]bool, len(s)+1)
    dp[0] = true // Empty string is a valid break

    for i := 1; i <= len(s); i++ {
        for j := 0; j < i; j++ {
            sub := s[j:i]
            if dp[j] && wordSet[sub] {
                dp[i] = true
                break
            }
        }
    }

    return dp[len(s)]
}

function wordBreak(s: string, wordDict: string[]): boolean {
  let wordSet: Set<string> = new Set(wordDict);
  let dp: boolean[] = new Array(s.length + 1).fill(false);
  dp[0] = true; // Empty string is a valid break

  for (let i = 1; i <= s.length; i++) {
    for (let j = 0; j < i; j++) {
      let sub: string = s.substring(j, i);
      if (dp[j] && wordSet.has(sub)) {
        dp[i] = true;
        break;
      }
    }
  }

  return dp[s.length];
}

We then use a for-loop i which goes from 1 to n to iterate through the string. Inside the body of this loop, we determine the correct value for dp[i]. We do this by using another for-loop j, which goes from 0 to i and represents the start of the substring we are considering. If dp[j] is true and the substring from j to i (sub) is in the dictionary, then we have found a valid segmentation, and can therefore set dp[i] to True.

As soon as we find a valid segmentation, we can break out of the inner for-loop and move on to the next character in the string.

Finally, after we finished iterating, we return dp[n] which is the value of the last element in the array. This value will be True if the entire string can be segmented into valid dictionary words, and False otherwise.

Solution

Alternative Solution

An alternative solution follows the same bottom-up approach. We use the same for loop to iterate through the string, but instead of iterating over all substrings ending at i, we instead iterate over all words in the dictionary. If the current word matches the substring ending at i and if dp[i - word.length] is True, then we have found a valid segmentation and can set dp[i] to True.

def wordBreak(s, wordDict):
  dp = [False] * (len(s) + 1)
  dp[0] = True # Empty string is a valid break

  for i in range(1, len(s) + 1):
    for word in wordDict:
      if i >= len(word) and dp[i - len(word)]:
        sub = s[i - len(word):i]
        if sub == word:
          dp[i] = True
          break

  return dp[len(s)]

public boolean wordBreak(String s, List<String> wordDict) {
    boolean[] dp = new boolean[s.length() + 1];
    dp[0] = true; // Empty string is a valid break

    for (int i = 1; i <= s.length(); i++) {
        for (String word : wordDict) {
            if (i >= word.length() && dp[i - word.length()]) {
                String sub = s.substring(i - word.length(), i);
                if (sub.equals(word)) {
                    dp[i] = true;
                    break;
                }
            }
        }
    }

    return dp[s.length()];
}

func wordBreak(s string, wordDict []string) bool {
    dp := make([]bool, len(s)+1)
    dp[0] = true // Empty string is a valid break

    for i := 1; i <= len(s); i++ {
        for _, word := range wordDict {
            if i >= len(word) && dp[i-len(word)] {
                sub := s[i-len(word):i]
                if sub == word {
                    dp[i] = true
                    break
                }
            }
        }
    }

    return dp[len(s)]
}

function wordBreak(s: string, wordDict: string[]): boolean {
  let dp: boolean[] = new Array(s.length + 1).fill(false);
  dp[0] = true; // Empty string is a valid break

  for (let i = 1; i <= s.length; i++) {
    for (let word of wordDict) {
      if (i >= word.length && dp[i - word.length]) {
        let sub: string = s.substring(i - word.length, i);
        if (sub === word) {
          dp[i] = true;
          break;
        }
      }
    }
  }

  return dp[s.length];
}