# 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) {
}
}
}
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
}