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 != array2)
return array1 - array2;
else
return array1 - array2;
}
});
int length = indexPairsList.size();
int[][] indexPairs = new int[length];
for (int i = 0; i < length; i++) {
int[] indexPair = indexPairsList.get(i);
indexPairs[i] = indexPair;
indexPairs[i] = indexPair;
}
return indexPairs;
}
}

• Todo

• print("Todo!")