people-whose-list-of-favorite-companies-is-not-a-subset-of-another-list

java

python

python3

mysql

mssql

oraclesql

csharp

javascript

typescript

bash

swift

kotlin

dart

golang

ruby

scala

rust

racket

erlang

elixir

pythondata

postgresql

cangjie

Algorithms

Given the array <code>favoriteCompanies</code> where <code>favoriteCompanies[i]</code> is the list of favorites companies for the <code>ith</code> person (indexed from 0).

Return the indices of people whose list of favorite companies is not a subset of any other list of favorites companies. You must return the indices in increasing order.

&nbsp;
Example 1:

<pre>
Input: favoriteCompanies = [[&quot;leetcode&quot;,&quot;google&quot;,&quot;facebook&quot;],[&quot;google&quot;,&quot;microsoft&quot;],[&quot;google&quot;,&quot;facebook&quot;],[&quot;google&quot;],[&quot;amazon&quot;]]
Output: [0,1,4] 
Explanation: 
Person with index=2 has favoriteCompanies[2]=[&quot;google&quot;,&quot;facebook&quot;] which is a subset of favoriteCompanies[0]=[&quot;leetcode&quot;,&quot;google&quot;,&quot;facebook&quot;] corresponding to the person with index 0. 
Person with index=3 has favoriteCompanies[3]=[&quot;google&quot;] which is a subset of favoriteCompanies[0]=[&quot;leetcode&quot;,&quot;google&quot;,&quot;facebook&quot;] and favoriteCompanies[1]=[&quot;google&quot;,&quot;microsoft&quot;]. 
Other lists of favorite companies are not a subset of another list, therefore, the answer is [0,1,4].
</pre>

Example 2:

<pre>
Input: favoriteCompanies = [[&quot;leetcode&quot;,&quot;google&quot;,&quot;facebook&quot;],[&quot;leetcode&quot;,&quot;amazon&quot;],[&quot;facebook&quot;,&quot;google&quot;]]
Output: [0,1] 
Explanation: In this case favoriteCompanies[2]=[&quot;facebook&quot;,&quot;google&quot;] is a subset of favoriteCompanies[0]=[&quot;leetcode&quot;,&quot;google&quot;,&quot;facebook&quot;], therefore, the answer is [0,1].
</pre>

Example 3:

<pre>
Input: favoriteCompanies = [[&quot;leetcode&quot;],[&quot;google&quot;],[&quot;facebook&quot;],[&quot;amazon&quot;]]
Output: [0,1,2,3]
</pre>

&nbsp;
Constraints:

<ul>
	<li><code>1 &lt;= favoriteCompanies.length &lt;= 100</code></li>
	<li><code>1 &lt;= favoriteCompanies[i].length &lt;= 500</code></li>
	<li><code>1 &lt;= favoriteCompanies[i][j].length &lt;= 20</code></li>
	<li>All strings in <code>favoriteCompanies[i]</code> are distinct.</li>
	<li>All lists of favorite companies are distinct, that is, If we sort alphabetically each list then <code>favoriteCompanies[i] != favoriteCompanies[j].</code></li>
	<li>All strings consist of lowercase English letters only.</li>
</ul>

peopleIndexes

People Whose List of Favorite Companies Is Not a Subset of Another List

Array

Hash Table

String

给你一个数组 <code>favoriteCompanies</code> ，其中 <code>favoriteCompanies[i]</code> 是第 <code>i</code> 名用户收藏的公司清单（下标从 0 开始）。

请找出不是其他任何人收藏的公司清单的子集的收藏清单，并返回该清单下标。下标需要按升序排列。

&nbsp;

示例 1：

<pre>输入：favoriteCompanies = [[&quot;leetcode&quot;,&quot;google&quot;,&quot;facebook&quot;],[&quot;google&quot;,&quot;microsoft&quot;],[&quot;google&quot;,&quot;facebook&quot;],[&quot;google&quot;],[&quot;amazon&quot;]]
输出：[0,1,4] 
解释：
favoriteCompanies[2]=[&quot;google&quot;,&quot;facebook&quot;] 是 favoriteCompanies[0]=[&quot;leetcode&quot;,&quot;google&quot;,&quot;facebook&quot;] 的子集。
favoriteCompanies[3]=[&quot;google&quot;] 是 favoriteCompanies[0]=[&quot;leetcode&quot;,&quot;google&quot;,&quot;facebook&quot;] 和 favoriteCompanies[1]=[&quot;google&quot;,&quot;microsoft&quot;] 的子集。
其余的收藏清单均不是其他任何人收藏的公司清单的子集，因此，答案为 [0,1,4] 。
</pre>

示例 2：

<pre>输入：favoriteCompanies = [[&quot;leetcode&quot;,&quot;google&quot;,&quot;facebook&quot;],[&quot;leetcode&quot;,&quot;amazon&quot;],[&quot;facebook&quot;,&quot;google&quot;]]
输出：[0,1] 
解释：favoriteCompanies[2]=[&quot;facebook&quot;,&quot;google&quot;] 是 favoriteCompanies[0]=[&quot;leetcode&quot;,&quot;google&quot;,&quot;facebook&quot;] 的子集，因此，答案为 [0,1] 。
</pre>

示例 3：

<pre>输入：favoriteCompanies = [[&quot;leetcode&quot;],[&quot;google&quot;],[&quot;facebook&quot;],[&quot;amazon&quot;]]
输出：[0,1,2,3]
</pre>

&nbsp;

提示：

<ul>
	<li><code>1 &lt;=&nbsp;favoriteCompanies.length &lt;= 100</code></li>
	<li><code>1 &lt;=&nbsp;favoriteCompanies[i].length &lt;= 500</code></li>
	<li><code>1 &lt;=&nbsp;favoriteCompanies[i][j].length &lt;= 20</code></li>
	<li><code>favoriteCompanies[i]</code> 中的所有字符串 各不相同 。</li>
	<li>用户收藏的公司清单也 各不相同 ，也就是说，即便我们按字母顺序排序每个清单， <code>favoriteCompanies[i] != favoriteCompanies[j] </code>仍然成立。</li>
	<li>所有字符串仅包含小写英文字母。</li>
</ul>