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In this post, you will find the solution for the **Word Break** **in C++, Java & Python-LeetCode problem**. We are providing the **correct and tested solutions** to coding problems present on **LeetCode**. If you are not able to solve any problem, then you can take help from our Blog/website.

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** Link for the Problem** – Word Break– LeetCode Problem

Word Break– LeetCode Problem

**Problem:**

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:trueExplanation:Return true because "leetcode" can be segmented as "leet code".

**Example 2:**

Input:s = "applepenapple", wordDict = ["apple","pen"]Output:trueExplanation: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**.

Template – LeetCode Solutions

Word Break Solution in C++:

class Solution { public: bool wordBreak(string s, vector& wordDict) { const int n = s.length(); unordered_set wordSet{begin(wordDict), end(wordDict)}; vector dp(n + 1); // dp[i] := true if s[0..i) can be segmented dp[0] = true; for (int i = 1; i <= n; ++i) for (int j = 0; j < i; ++j) // s[0..j) can be segmented and s[j..i) in wordSet // so s[0..i) can be segmented if (dp[j] && wordSet.count(s.substr(j, i - j))) { dp[i] = true; break; } return dp[n]; } };

Word Break Solution in Java:

class Solution { public boolean wordBreak(String s, ListwordDict) { final int n = s.length(); Set wordSet = new HashSet<>(wordDict); boolean[] dp = new boolean[n + 1]; // dp[i] := true if s[0..i) can be segmented dp[0] = true; for (int i = 1; i <= n; ++i) for (int j = 0; j < i; ++j) // s[0..j) can be segmented and s[j..i) in wordSet // so s[0..i) can be segmented if (dp[j] && wordSet.contains(s.substring(j, i))) { dp[i] = true; break; } return dp[n]; } }

Word Break Solution in Python:

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