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花花酱 LeetCode 650. 2 Keys Keyboard

这道题目还是可以的。每个阶段有两种选择,求最短的决策次数。有好多种不同的做法,我们一个一个来。

方法1: BFS

把问题看成求无权重图中的最短路径,BFS可以找到最优解。

我们的节点有两个状态,当前的长度,剪贴板(上次复制后)的长度。
长度超过n一定无解,所以我们的状态最多只有n2种,每个状态最多拓展2个新的节点。

时间复杂度:O(n2)
空间复杂度:O(n2)

花花酱 LeetCode 2157. Groups of Strings

You are given a 0-indexed array of strings words. Each string consists of lowercase English letters only. No letter occurs more than once in any string of words.

Two strings s1 and s2 are said to be connected if the set of letters of s2 can be obtained from the set of letters of s1 by any one of the following operations:

  • Adding exactly one letter to the set of the letters of s1.
  • Deleting exactly one letter from the set of the letters of s1.
  • Replacing exactly one letter from the set of the letters of s1 with any letter, including itself.

The array words can be divided into one or more non-intersecting groups. A string belongs to a group if any one of the following is true:

  • It is connected to at least one other string of the group.
  • It is the only string present in the group.

Note that the strings in words should be grouped in such a manner that a string belonging to a group cannot be connected to a string present in any other group. It can be proved that such an arrangement is always unique.

Return an array ans of size 2 where:

  • ans[0] is the total number of groups words can be divided into, and
  • ans[1] is the size of the largest group.

Example 1:

Input: words = ["a","b","ab","cde"]
Output: [2,3]
Explanation:
- words[0] can be used to obtain words[1] (by replacing 'a' with 'b'), and words[2] (by adding 'b'). So words[0] is connected to words[1] and words[2].
- words[1] can be used to obtain words[0] (by replacing 'b' with 'a'), and words[2] (by adding 'a'). So words[1] is connected to words[0] and words[2].
- words[2] can be used to obtain words[0] (by deleting 'b'), and words[1] (by deleting 'a'). So words[2] is connected to words[0] and words[1].
- words[3] is not connected to any string in words.
Thus, words can be divided into 2 groups ["a","b","ab"] and ["cde"]. The size of the largest group is 3.  

Example 2:

Input: words = ["a","ab","abc"]
Output: [1,3]
Explanation:
- words[0] is connected to words[1].
- words[1] is connected to words[0] and words[2].
- words[2] is connected to words[1].
Since all strings are connected to each other, they should be grouped together.
Thus, the size of the largest group is 3.

Constraints:

  • 1 <= words.length <= 2 * 104
  • 1 <= words[i].length <= 26
  • words[i] consists of lowercase English letters only.
  • No letter occurs more than once in words[i].

Solution: Bitmask + DFS

Use a bitmask to represent a string. Use dfs to find connected components.

Time complexity: O(n*262)
Space complexity: O(n)

C++

花花酱 LeetCode 2081. Sum of k-Mirror Numbers

k-mirror number is a positive integer without leading zeros that reads the same both forward and backward in base-10 as well as in base-k.

  • For example, 9 is a 2-mirror number. The representation of 9 in base-10 and base-2 are 9 and 1001 respectively, which read the same both forward and backward.
  • On the contrary, 4 is not a 2-mirror number. The representation of 4 in base-2 is 100, which does not read the same both forward and backward.

Given the base k and the number n, return the sum of the n smallest k-mirror numbers.

Example 1:

Input: k = 2, n = 5
Output: 25
Explanation:
The 5 smallest 2-mirror numbers and their representations in base-2 are listed as follows:
  base-10    base-2
    1          1
    3          11
    5          101
    7          111
    9          1001
Their sum = 1 + 3 + 5 + 7 + 9 = 25. 

Example 2:

Input: k = 3, n = 7
Output: 499
Explanation:
The 7 smallest 3-mirror numbers are and their representations in base-3 are listed as follows:
  base-10    base-3
    1          1
    2          2
    4          11
    8          22
    121        11111
    151        12121
    212        21212
Their sum = 1 + 2 + 4 + 8 + 121 + 151 + 212 = 499.

Example 3:

Input: k = 7, n = 17
Output: 20379000
Explanation: The 17 smallest 7-mirror numbers are:
1, 2, 3, 4, 5, 6, 8, 121, 171, 242, 292, 16561, 65656, 2137312, 4602064, 6597956, 6958596

Constraints:

  • 2 <= k <= 9
  • 1 <= n <= 30

Solution: Generate palindromes in base-k.

Python

花花酱 LeetCode 2002. Maximum Product of the Length of Two Palindromic Subsequences

Given a string s, find two disjoint palindromic subsequences of s such that the product of their lengths is maximized. The two subsequences are disjoint if they do not both pick a character at the same index.

Return the maximum possible product of the lengths of the two palindromic subsequences.

subsequence is a string that can be derived from another string by deleting some or no characters without changing the order of the remaining characters. A string is palindromic if it reads the same forward and backward.

Example 1:

example-1
Input: s = "leetcodecom"
Output: 9
Explanation: An optimal solution is to choose "ete" for the 1st subsequence and "cdc" for the 2nd subsequence.
The product of their lengths is: 3 * 3 = 9.

Example 2:

Input: s = "bb"
Output: 1
Explanation: An optimal solution is to choose "b" (the first character) for the 1st subsequence and "b" (the second character) for the 2nd subsequence.
The product of their lengths is: 1 * 1 = 1.

Example 3:

Input: s = "accbcaxxcxx"
Output: 25
Explanation: An optimal solution is to choose "accca" for the 1st subsequence and "xxcxx" for the 2nd subsequence.
The product of their lengths is: 5 * 5 = 25.

Constraints:

  • 2 <= s.length <= 12
  • s consists of lowercase English letters only.

Solution 1: DFS

Time complexity: O(3n*n)
Space complexity: O(n)

C++

Solution: Subsets + Bitmask + All Pairs

Time complexity: O(22n)
Space complexity: O(2n)

C++

花花酱 LeetCode 2065. Maximum Path Quality of a Graph

There is an undirected graph with n nodes numbered from 0 to n - 1 (inclusive). You are given a 0-indexed integer array values where values[i] is the value of the ith node. You are also given a 0-indexed 2D integer array edges, where each edges[j] = [uj, vj, timej] indicates that there is an undirected edge between the nodes uj and vj,and it takes timej seconds to travel between the two nodes. Finally, you are given an integer maxTime.

valid path in the graph is any path that starts at node 0, ends at node 0, and takes at most maxTime seconds to complete. You may visit the same node multiple times. The quality of a valid path is the sum of the values of the unique nodes visited in the path (each node’s value is added at most once to the sum).

Return the maximum quality of a valid path.

Note: There are at most four edges connected to each node.

Example 1:

Input: values = [0,32,10,43], edges = [[0,1,10],[1,2,15],[0,3,10]], maxTime = 49
Output: 75
Explanation:
One possible path is 0 -> 1 -> 0 -> 3 -> 0. The total time taken is 10 + 10 + 10 + 10 = 40 <= 49.
The nodes visited are 0, 1, and 3, giving a maximal path quality of 0 + 32 + 43 = 75.

Example 2:

Input: values = [5,10,15,20], edges = [[0,1,10],[1,2,10],[0,3,10]], maxTime = 30
Output: 25
Explanation:
One possible path is 0 -> 3 -> 0. The total time taken is 10 + 10 = 20 <= 30.
The nodes visited are 0 and 3, giving a maximal path quality of 5 + 20 = 25.

Example 3:

Input: values = [1,2,3,4], edges = [[0,1,10],[1,2,11],[2,3,12],[1,3,13]], maxTime = 50
Output: 7
Explanation:
One possible path is 0 -> 1 -> 3 -> 1 -> 0. The total time taken is 10 + 13 + 13 + 10 = 46 <= 50.
The nodes visited are 0, 1, and 3, giving a maximal path quality of 1 + 2 + 4 = 7.

Example 4:

Input: values = [0,1,2], edges = [[1,2,10]], maxTime = 10
Output: 0
Explanation: 
The only path is 0. The total time taken is 0.
The only node visited is 0, giving a maximal path quality of 0.

Constraints:

  • n == values.length
  • 1 <= n <= 1000
  • 0 <= values[i] <= 108
  • 0 <= edges.length <= 2000
  • edges[j].length == 3
  • 0 <= u< vj <= n - 1
  • 10 <= timej, maxTime <= 100
  • All the pairs [uj, vj] are unique.
  • There are at most four edges connected to each node.
  • The graph may not be connected.

Solution: DFS

Given time >= 10 and maxTime <= 100, the path length is at most 10, given at most four edges connected to each node.
Time complexity: O(410)
Space complexity: O(n)

C++