Word Break LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

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Hello Programmers/Coders, Today we are going to share solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python. At Each Problem with Successful submission with all Test Cases Passed, you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

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 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<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]