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Formatted question description: https://leetcode.ca/all/1065.html
1065. Index Pairs of a String
Level
Easy
Description
Given a text
string and words
(a list of strings), return all index pairs [i, j]
so that the substring text[i]...text[j]
is in the list of words.
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].
Note:
- All strings contains only lowercase English letters.
- It’s guaranteed that all strings in
words
are different. 1 <= text.length <= 100
1 <= words.length <= 20
1 <= words[i].length <= 50
- Return the pairs
[i,j]
in sorted order (i.e. sort them by their first coordinate in case of ties sort them by their second coordinate).
Solution
For each word
in words
, find all occurrences of word
in text
, and add the index pairs to the result list. Sort the result list and return.
-
class Solution { public int[][] indexPairs(String text, String[] words) { List<int[]> indexPairsList = new ArrayList<int[]>(); for (String word : words) { int wordLength = word.length(); int curIndex = 0; while (curIndex >= 0) { curIndex = text.indexOf(word, curIndex); if (curIndex >= 0) { indexPairsList.add(new int[]{curIndex, curIndex + wordLength - 1}); curIndex++; } } } Collections.sort(indexPairsList, new Comparator<int[]>() { public int compare(int[] array1, int[] array2) { if (array1[0] != array2[0]) return array1[0] - array2[0]; else return array1[1] - array2[1]; } }); int length = indexPairsList.size(); int[][] indexPairs = new int[length][2]; for (int i = 0; i < length; i++) { int[] indexPair = indexPairsList.get(i); indexPairs[i][0] = indexPair[0]; indexPairs[i][1] = indexPair[1]; } return indexPairs; } }
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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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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 }