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1813. Sentence Similarity III

Description

A sentence is a list of words that are separated by a single space with no leading or trailing spaces. For example, "Hello World", "HELLO", "hello world hello world" are all sentences. Words consist of only uppercase and lowercase English letters.

Two sentences sentence1 and sentence2 are similar if it is possible to insert an arbitrary sentence (possibly empty) inside one of these sentences such that the two sentences become equal. For example, sentence1 = "Hello my name is Jane" and sentence2 = "Hello Jane" can be made equal by inserting "my name is" between "Hello" and "Jane" in sentence2.

Given two sentences sentence1 and sentence2, return true if sentence1 and sentence2 are similar. Otherwise, return false.

 

Example 1:

Input: sentence1 = "My name is Haley", sentence2 = "My Haley"
Output: true
Explanation: sentence2 can be turned to sentence1 by inserting "name is" between "My" and "Haley".

Example 2:

Input: sentence1 = "of", sentence2 = "A lot of words"
Output: false
Explanation: No single sentence can be inserted inside one of the sentences to make it equal to the other.

Example 3:

Input: sentence1 = "Eating right now", sentence2 = "Eating"
Output: true
Explanation: sentence2 can be turned to sentence1 by inserting "right now" at the end of the sentence.

 

Constraints:

  • 1 <= sentence1.length, sentence2.length <= 100
  • sentence1 and sentence2 consist of lowercase and uppercase English letters and spaces.
  • The words in sentence1 and sentence2 are separated by a single space.

Solutions

Solution 1: Two Pointers

We split the two sentences into two word arrays words1 and words2 by spaces. Let the lengths of words1 and words2 be $m$ and $n$, respectively, and assume that $m \ge nn.

We use two pointers $i$ and $j$, initially $i = j = 0$. Next, we loop to check whether words1[i] is equal to words2[i], and if so, pointer $i$ continues to move right; then we loop to check whether words1[m - 1 - j] is equal to words2[n - 1 - j], and if so, pointer $j$ continues to move right.

After the loop, if $i + j \ge n$, it means that the two sentences are similar, and we return true; otherwise, we return false.

The time complexity is $O(L)$, and the space complexity is $O(L)$, where $L$ is the sum of the lengths of the two sentences.

  • class Solution {
    public boolean areSentencesSimilar(String sentence1, String sentence2) {
        var words1 = sentence1.split(" ");
        var words2 = sentence2.split(" ");
        if (words1.length < words2.length) {
            var t = words1;
            words1 = words2;
            words2 = t;
        }
        int m = words1.length, n = words2.length;
        int i = 0, j = 0;
        while (i < n && words1[i].equals(words2[i])) {
            ++i;
        }
        while (j < n && words1[m - 1 - j].equals(words2[n - 1 - j])) {
            ++j;
        }
        return i + j >= n;
    }
}

  • class Solution {
public:
    bool areSentencesSimilar(string sentence1, string sentence2) {
        auto words1 = split(sentence1, ' ');
        auto words2 = split(sentence2, ' ');
        if (words1.size() < words2.size()) {
            swap(words1, words2);
        }
        int m = words1.size(), n = words2.size();
        int i = 0, j = 0;
        while (i < n && words1[i] == words2[i]) {
            ++i;
        }
        while (j < n && words1[m - 1 - j] == words2[n - 1 - j]) {
            ++j;
        }
        return i + j >= n;
    }

    vector<string> split(string& s, char delim) {
        stringstream ss(s);
        string item;
        vector<string> res;
        while (getline(ss, item, delim)) {
            res.emplace_back(item);
        }
        return res;
    }
};

  • class Solution:
    def areSentencesSimilar(self, sentence1: str, sentence2: str) -> bool:
        words1, words2 = sentence1.split(), sentence2.split()
        m, n = len(words1), len(words2)
        if m < n:
            words1, words2 = words2, words1
            m, n = n, m
        i = j = 0
        while i < n and words1[i] == words2[i]:
            i += 1
        while j < n and words1[m - 1 - j] == words2[n - 1 - j]:
            j += 1
        return i + j >= n


  • func areSentencesSimilar(sentence1 string, sentence2 string) bool {
	words1, words2 := strings.Fields(sentence1), strings.Fields(sentence2)
	if len(words1) < len(words2) {
		words1, words2 = words2, words1
	}
	m, n := len(words1), len(words2)
	i, j := 0, 0
	for i < n && words1[i] == words2[i] {
		i++
	}
	for j < n && words1[m-1-j] == words2[n-1-j] {
		j++
	}
	return i+j >= n
}

  • function areSentencesSimilar(sentence1: string, sentence2: string): boolean {
    const words1 = sentence1.split(' ');
    const words2 = sentence2.split(' ');
    if (words1.length < words2.length) {
        return areSentencesSimilar(sentence2, sentence1);
    }
    const [m, n] = [words1.length, words2.length];
    let [i, j] = [0, 0];
    while (i < n && words1[i] === words2[i]) {
        ++i;
    }
    while (j < n && words1[m - 1 - j] === words2[n - 1 - j]) {
        ++j;
    }
    return i + j >= n;
}


  • function areSentencesSimilar(sentence1, sentence2) {
    const [words1, words2] = [sentence1.split(' '), sentence2.split(' ')];
    const [m, n] = [words1.length, words2.length];

    if (m > n) return areSentencesSimilar(sentence2, sentence1);

    let [l, r] = [0, 0];
    for (let i = 0; i < n; i++) {
        if (l === i && words1[i] === words2[i]) l++;
        if (r === i && words2[n - i - 1] === words1[m - r - 1]) r++;
    }

    return l + r >= m;
}

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