README_EN.md

17.13. Re-Space

中文文档

Description

Oh, no! You have accidentally removed all spaces, punctuation, and capitalization in a lengthy document. A sentence like "I reset the computer. It still didn't boot!" became "iresetthecomputeritstilldidntboot''. You'll deal with the punctuation and capi­talization later; right now you need to re-insert the spaces. Most of the words are in a dictionary but a few are not. Given a dictionary (a list of strings) and the document (a string), design an algorithm to unconcatenate the document in a way that minimizes the number of unrecognized characters. Return the number of unrecognized characters.

Note: This problem is slightly different from the original one in the book.

 

Example:

Input: 

dictionary = ["looked","just","like","her","brother"]

sentence = "jesslookedjustliketimherbrother"

Output:  7

Explanation:  After unconcatenating, we got "jess looked just like tim her brother", which containing 7 unrecognized characters.

Note:

• 0 <= len(sentence) <= 1000
• The total number of characters in dictionary is less than or equal to 150000.
• There are only lowercase letters in dictionary and sentence.

Solutions

Python3

class Solution:
    def respace(self, dictionary: List[str], sentence: str) -> int:
        s = set(dictionary)
        n = len(sentence)
        dp = [0] * (n + 1)
        for i in range(1, n + 1):
            dp[i] = dp[i - 1] + 1
            for j in range(i):
                if sentence[j:i] in s:
                    dp[i] = min(dp[i], dp[j])
        return dp[-1]

Java

class Solution {
    public int respace(String[] dictionary, String sentence) {
        Set<String> dict = new HashSet<>(Arrays.asList(dictionary));
        int n = sentence.length();
        int[] dp = new int[n + 1];
        for (int i = 1; i <= n; i++) {
            dp[i] = dp[i - 1] + 1;
            for (int j = 0; j < i; ++j) {
                if (dict.contains(sentence.substring(j, i))) {
                    dp[i] = Math.min(dp[i], dp[j]);
                }
            }
        }
        return dp[n];
    }
}

C++

class Solution {
public:
    int respace(vector<string>& dictionary, string sentence) {
        unordered_set<string> s(dictionary.begin(), dictionary.end());
        int n = sentence.size();
        vector<int> dp(n + 1);
        for (int i = 1; i <= n; ++i) {
            dp[i] = dp[i - 1] + 1;
            for (int j = 0; j < i; ++j) {
                if (s.count(sentence.substr(j, i - j))) {
                    dp[i] = min(dp[i], dp[j]);
                }
            }
        }
        return dp[n];
    }
};

Go

func respace(dictionary []string, sentence string) int {
	s := map[string]bool{}
	for _, v := range dictionary {
		s[v] = true
	}
	n := len(sentence)
	dp := make([]int, n+1)
	for i := 1; i <= n; i++ {
		dp[i] = dp[i-1] + 1
		for j := 0; j < i; j++ {
			if s[sentence[j:i]] {
				dp[i] = min(dp[i], dp[j])
			}
		}
	}
	return dp[n]
}

func min(a, b int) int {
	if a < b {
		return a
	}
	return b
}

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