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Dynamic Programming

Word Break

medium

DESCRIPTION (inspired by Leetcode.com)

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.

Visualization


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)]

word break

0 / 1

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.

Visualization


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)]

i = 1

0 / 11

Iterating until we set `dp[i] = True` for the first time.

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.

Test Your Knowledge

Complexity Analysis

Time Complexity: O(n²) where n is the length of the input string s. We use nested loops to check all possible substrings.

Space Complexity: O(n + m) where n is the length of the input string s and m is the length of wordDict. This includes the dp array and the additional space used to create wordSet from wordDict.

Solution

Visualization


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)]

word break

0 / 88

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.

Visualization


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)]

word break

0 / 65

Test Your Knowledge

Complexity Analysis

Time Complexity: O(n * m) where n is the length of the input string s and m is the length of wordDict. We iterate through the string and check against each word in the dictionary.

Space Complexity: O(n) where n is the length of the input string s for the dp array.

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Explanation

Solution

Alternative Solution

Dynamic Programming

Word Break

DESCRIPTION (inspired by Leetcode.com)

Explanation

Test Your Knowledge

Complexity Analysis

Solution

Alternative Solution

Test Your Knowledge

Complexity Analysis

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