LeetCode-in-All

139. Word Break

Medium

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.

Solution

bool wordBreak(char* s, char** wordDict, int wordDictSize) {
    bool dp[strlen(s)+1];
    memset(dp,false,sizeof(dp));
    dp[strlen(s)] = true;

    for (int i = strlen(s)-1; i >= 0; i--) {
        for (int j = 0; j < wordDictSize; j++) {
            if (i+strlen(wordDict[j]) <= strlen(s)) {
                char string[20] = "\0";
                for (int k = 0; k < strlen(wordDict[j]); k++) {
                    string[k] = s[i+k];
                }
                if (strcmp(wordDict[j], string) == 0) {
                    dp[i] = dp[i+strlen(wordDict[j])];
                }
                if (dp[i] == true) {
                    break;
                }
            }
        }
    }
    return dp[0];
}

This site is open source. Improve this page.