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Formatted question description: https://leetcode.ca/all/1065.html

# 1065. Index Pairs of a String

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:

1. All strings contains only lowercase English letters.
2. It’s guaranteed that all strings in words are different.
3. 1 <= text.length <= 100
4. 1 <= words.length <= 20
5. 1 <= words[i].length <= 50
6. 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;
}
}

• 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


• 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;
}
};

• 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
}