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1065. Index Pairs of a String πŸ”’

Description

Given a string text and an array of strings words, return an array of all index pairs [i, j] so that the substring text[i...j] is in words.

Return the pairs [i, j] in sorted order (i.e., sort them by their first coordinate, and in case of ties sort them by their second coordinate).

 

Example 1:

Input: text = "thestoryofleetcodeandme", words = ["story","fleet","leetcode"]
Output: [[3,7],[9,13],[10,17]]

Example 2:

Input: text = "ababa", words = ["aba","ab"]
Output: [[0,1],[0,2],[2,3],[2,4]]
Explanation: Notice that matches can overlap, see "aba" is found in [0,2] and [2,4].

 

Constraints:

  • 1 <= text.length <= 100
  • 1 <= words.length <= 20
  • 1 <= words[i].length <= 50
  • text and words[i] consist of lowercase English letters.
  • All the strings of words are unique.

Solutions

Solution 1

Thinking

Every substring of \(text\) can be tested against the word list. Both sides are short, so a set of words plus all slices \([i,j]\) is enough.

The set answers in expected constant time, and enumerating indices already yields sorted pairs.

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class Solution:
    def indexPairs(self, text: str, words: List[str]) -> List[List[int]]:
        words = set(words)
        n = len(text)
        return [
            [i, j] for i in range(n) for j in range(i, n) if text[i : j + 1] in words
        ]

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class Trie {
    Trie[] children = new Trie[26];
    boolean isEnd = false;

    void insert(String word) {
        Trie node = this;
        for (char c : word.toCharArray()) {
            c -= 'a';
            if (node.children[c] == null) {
                node.children[c] = new Trie();
            }
            node = node.children[c];
        }
        node.isEnd = true;
    }
}

class Solution {
    public int[][] indexPairs(String text, String[] words) {
        Trie trie = new Trie();
        for (String w : words) {
            trie.insert(w);
        }
        int n = text.length();
        List<int[]> ans = new ArrayList<>();
        for (int i = 0; i < n; ++i) {
            Trie node = trie;
            for (int j = i; j < n; ++j) {
                int idx = text.charAt(j) - 'a';
                if (node.children[idx] == null) {
                    break;
                }
                node = node.children[idx];
                if (node.isEnd) {
                    ans.add(new int[] {i, j});
                }
            }
        }
        return ans.toArray(new int[ans.size()][2]);
    }
}

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class Trie {
public:
    vector<Trie*> children;
    bool isEnd = false;

    Trie() {
        children.resize(26);
    }

    void insert(string word) {
        Trie* node = this;
        for (char c : word) {
            c -= 'a';
            if (!node->children[c]) node->children[c] = new Trie();
            node = node->children[c];
        }
        node->isEnd = true;
    }
};

class Solution {
public:
    vector<vector<int>> indexPairs(string text, vector<string>& words) {
        Trie* trie = new Trie();
        for (auto w : words) trie->insert(w);
        int n = text.size();
        vector<vector<int>> ans;
        for (int i = 0; i < n; ++i) {
            Trie* node = trie;
            for (int j = i; j < n; ++j) {
                int idx = text[j] - 'a';
                if (!node->children[idx]) break;
                node = node->children[idx];
                if (node->isEnd) ans.push_back({i, j});
            }
        }
        return ans;
    }
};

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type Trie struct {
    children [26]*Trie
    isEnd    bool
}

func newTrie() *Trie {
    return &Trie{}
}

func (this *Trie) insert(word string) {
    node := this
    for _, c := range word {
        idx := int(c - 'a')
        if node.children[idx] == nil {
            node.children[idx] = newTrie()
        }
        node = node.children[idx]
    }
    node.isEnd = true
}

func indexPairs(text string, words []string) [][]int {
    trie := newTrie()
    for _, w := range words {
        trie.insert(w)
    }
    n := len(text)
    var ans [][]int
    for i := range text {
        node := trie
        for j := i; j < n; j++ {
            idx := int(text[j] - 'a')
            if node.children[idx] == nil {
                break
            }
            node = node.children[idx]
            if node.isEnd {
                ans = append(ans, []int{i, j})
            }
        }
    }
    return ans
}

Solution 2

Thinking

Trying every slice wastes failed lookups on long text. Shared prefixes let us walk a trie from each start and stop at the first mismatch.

Insert the words, then from each \(i\) extend the node to the right and record \([i,j]\) at end marks.

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class Trie:
    def __init__(self):
        self.children = [None] * 26
        self.is_end = False

    def insert(self, word):
        node = self
        for c in word:
            idx = ord(c) - ord('a')
            if node.children[idx] is None:
                node.children[idx] = Trie()
            node = node.children[idx]
        node.is_end = True


class Solution:
    def indexPairs(self, text: str, words: List[str]) -> List[List[int]]:
        trie = Trie()
        for w in words:
            trie.insert(w)
        n = len(text)
        ans = []
        for i in range(n):
            node = trie
            for j in range(i, n):
                idx = ord(text[j]) - ord('a')
                if node.children[idx] is None:
                    break
                node = node.children[idx]
                if node.is_end:
                    ans.append([i, j])
        return ans

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