【Weekly】No.184

C_01 String matching in array

Gives you a string array words, each string in the array can be regarded as a word. Please return all words that are substrings of other words in words in any order.

If you can delete the leftmost and / or rightmost characters of words [j] to get word [i], then the string words [i] is a substring of words [j].

输入：words = ["mass","as","hero","superhero"]
输出：["as","hero"]
解释："as" 是 "mass" 的子字符串，"hero" 是 "superhero" 的子字符串。
["hero","as"] 也是有效的答案。

Method 1: Violence

public List<String> stringMatching(String[] words) {
    LinkedList<String> list = new LinkedList<>();
    for (int i = 0; i < words.length; i++) {
        for (int j = 0; j < words.length; j++) {
            if (i == j)
                continue;
            if (words[i].contains(words[j]) && !list.contains(words[j])) 
                list.add(words[j]);
        }
    }
    return list;
}

Complexity analysis

• time complexity: $O()$，
• Space complexity: $O()$，

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B_02 Query the arrangement with key

Give you an array of queries to check, the elements in the array are positive integers between 1 and m. Please process all the items to be queried according to the following rules [i] (from i = 0 to i = queries.length-1):

• At the beginning, arrange P = [1,2,3, ..., m].
• For the current i, please find the position of the query item queues [i] in the arrangement P (the subscript starts from 0), and then move it from the original position to the starting position of the arrangement P (that is, the index is 0 Office). Note that the position of queries [i] in P is the query results of queries [i].

Please return the query results of the array query to be checked in the form of an array.

• 1 <= m <= 10^3
  1 <= queries.length <= m
  1 <= queries[i] <= m

输入：queries = [3,1,2,1], m = 5
输出：[2,1,2,1] 
解释：待查数组 queries 处理如下：
对于 i=0: queries[i]=3, P=[1,2,3,4,5], 3 在 P 中的位置是 2，接着我们把 3 移动到 P 的起始位置，得到 P=[3,1,2,4,5] 。
对于 i=1: queries[i]=1, P=[3,1,2,4,5], 1 在 P 中的位置是 1，接着我们把 1 移动到 P 的起始位置，得到 P=[1,3,2,4,5] 。 
对于 i=2: queries[i]=2, P=[1,3,2,4,5], 2 在 P 中的位置是 2，接着我们把 2 移动到 P 的起始位置，得到 P=[2,1,3,4,5] 。
对于 i=3: queries[i]=1, P=[2,1,3,4,5], 1 在 P 中的位置是 1，接着我们把 1 移动到 P 的起始位置，得到 P=[1,2,3,4,5] 。 
因此，返回的结果数组为 [2,1,2,1] 。  

Method 1: Simulation

public int[] processQueries(int[] queries, int m) {
    ArrayList<Integer> nums = new ArrayList<>();
    for (int i = 1; i <= m; i++) {
        nums.add(i);
    }
    LinkedList<Integer> res = new LinkedList<Integer>();
    for (int q : queries) {
        int idx = nums.indexOf(q);
        res.add(idx);
        nums.remove(idx);
        nums.add(0, q);
    }
    int p = 0;
    int[] arr = new int[queries.length];
    for (int i : res) {
        arr[p++] = i;
    }
    return arr;
}

Complexity analysis

• time complexity: $O (n)$，
• Space complexity: $O (n)$，

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B_03 HTML entity parser

"HTML entity parser" is a special parser that takes HTML code as input and replaces all these special character entities with characters themselves.

These special characters and their corresponding character entities in HTML include:

• Double quotes: the character entity is "", and the corresponding character is "".
  Single quotes: The character entity is 'and the corresponding character is'.
  And symbol: The character entity is &, and the corresponding character is &.
  Greater than sign: The character entity is>, and the corresponding character is>.
  Less than sign: The character entity is <, and the corresponding character is <.
  Slash: The character entity is ⁄ and the corresponding character is /.
  Enter the string text for you, please implement an HTML entity parser, and return the parsing result of the parser.

Method 1: Replace string

public String entityParser(String text) {
    String res = text;
    res = res.replaceAll(""", "\"");
    res = res.replaceAll("'", "'");
    res = res.replaceAll("&", "&");
    res = res.replaceAll(">", ">");
    res = res.replaceAll("&lt;", "<");
    res = res.replaceAll("&frasl;", "/");
    return res;
}  

Complexity analysis

• time complexity: $O (n)$，
• Space complexity: $O (1)$，

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A_04 Number of schemes to color N x 3 grid graph (no AC)

输入：n = 2
输出：54

输入：n = 3
输出：246

示例 4：
输入：n = 7
输出：106494

示例 5：
输入：n = 5000
输出：30228214

prompt:

• n == grid.length
  grid[i].length == 3
  1 <= n <= 5000

method one:

Complexity analysis

• time complexity: $O()$，
• Space complexity: $O()$，

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