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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: 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
andwordDict[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<string>& wordDict) { const int n = s.length(); unordered_set<string> wordSet{begin(wordDict), end(wordDict)}; vector<bool> 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, List<String> wordDict) { final int n = s.length(); Set<String> 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]