Teaching Kids Programming – Dynamic Programming Algorithm to Break a String using Given Words (Word Break)

Teaching Kids Programming: Videos on Data Structures and Algorithms

Given a string s and a dictionary of strings wordDict, return true if s can be segmented into a space-separated sequence of one or more dictionary words. Note that the same word in the dictionary may be reused multiple times in the segmentation.

Example 1:
Input: s = “leetcode”, wordDict = [“leet”,”code”]
Output: true
Explanation: Return true because “leetcode” can be segmented as “leet code”.

Example 2:
Input: s = “applepenapple”, wordDict = [“apple”,”pen”]
Output: true
Explanation: Return true because “applepenapple” can be segmented as “apple pen apple”.
Note that you are allowed to reuse a dictionary word.

Example 3:
Input: s = “catsandog”, wordDict = [“cats”,”dog”,”sand”,”and”,”cat”]
Output: false

Constraints:
1 <= s.length <= 300
1 <= wordDict.length <= 1000
1 <= wordDict[i].length <= 20
s and wordDict[i] consist of only lowercase English letters.
All the strings of wordDict are unique.

Dynamic Programming Algorithm to Break a String using Given Words (Word Break)

Let dp[i] represent if we can construct the substring s[:i] using the words given. Then if we have dp[j] is true where j is smaller than i and s[j:i] is given in the list, it is obvious that we can set dp[i] to true. This Dynamic Programming Approach is Bottom Up, as compared to Top Down Dynamic Programming with Memoziation.

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class Solution:
    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
        w = set(wordDict)
        n = len(s)
        dp = [False] * (n + 1)
        dp[0] = True
        for i in range(1, n + 1):
            for j in range(i):
                if s[j:i] in w and dp[j]:
                    dp[i] = True
                    break
        return dp[-1]

class Solution:
    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
        w = set(wordDict)
        n = len(s)
        dp = [False] * (n + 1)
        dp[0] = True
        for i in range(1, n + 1):
            for j in range(i):
                if s[j:i] in w and dp[j]:
                    dp[i] = True
                    break
        return dp[-1]

The time complexity is O(N^3) – and we need an array O(N) to store the dp values (intermediate results are stored/cached so they can be re-used later).

Word Break Algorithms

• Teaching Kids Programming – Dynamic Programming Algorithm to Break a String using Given Words (Word Break)
• Teaching Kids Programming – Break a String into Words (Word Break) via Breadth First Search Algorithm
• C/C++ Coding Exercise – Word Break (DP, BFS, DFS)
• Teaching Kids Programming – Break a String into Words (Word Break) via Recursion with Memoziation – Top Down Dynamic Programming Algorithm

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