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1065. Index Pairs of a String

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

  • 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]); } } 
  • 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; } }; 
  • 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 ] 
  • 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 }

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