## Word Break

###### DESCRIPTION (credit 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.

###### EXAMPLES

Input:

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

Output:

`false`

Explanation: There is no valid segmentation of "catsanddog" 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.

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## 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 breakfor 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] = Truebreakreturn dp[len(s)]

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.

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

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.

## Complexity Analysis

**Time Complexity**: O(n2) where n is the length of the input string s.
**Space Complexity**: O(n + m) where n is the length of the input string s and m is the length of wordDict for the dp array and the additional space used to create wordSet from wordDict.

## Solution

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

## 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 breakfor 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] = Truebreakreturn dp[len(s)]

## Complexity Analysis

**Time Complexity**: O(n * m) where n is the length of the input string s and m is the length of wordDict.
**Space Complexity**: O(n) where n is the length of the input string s for the dp array.

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