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花花酱 LeetCode 552. Student Attendance Record II

Problem

Given a positive integer n, return the number of all possible attendance records with length n, which will be regarded as rewardable. The answer may be very large, return it after mod 109 + 7.

A student attendance record is a string that only contains the following three characters:

  1. ‘A’ : Absent.
  2. ‘L’ : Late.
  3. ‘P’ : Present.

A record is regarded as rewardable if it doesn’t contain more than one ‘A’ (absent) or more than two continuous ‘L’ (late).

Example 1:

Input: n = 2
Output: 8 
Explanation:
There are 8 records with length 2 will be regarded as rewardable:
"PP" , "AP", "PA", "LP", "PL", "AL", "LA", "LL"
Only "AA" won't be regarded as rewardable owing to more than one absent times. 

Note: The value of n won’t exceed 100,000.

Solution

C++

DFS w/ memorization (stack overflow)

DP

Time complexity: O(n)

Space complexity: O(1)

花花酱 LeetCode 541. Reverse String II

Problem

题目大意:把字符串分成块,每块长度2k,单独反转每一块的前k的字符。

https://leetcode.com/problems/reverse-string-ii/description/

Given a string and an integer k, you need to reverse the first k characters for every 2k characters counting from the start of the string. If there are less than k characters left, reverse all of them. If there are less than 2k but greater than or equal to k characters, then reverse the first k characters and left the other as original.Example:

Input: s = "abcdefg", k = 2
Output: "bacdfeg"

Restrictions:

  1. The string consists of lower English letters only.
  2. Length of the given string and k will in the range [1, 10000]

Solution

C++

Related Problems

花花酱 LeetCode 539. Minimum Time Difference

Problem

Given a list of 24-hour clock time points in “Hour:Minutes” format, find the minimum minutes difference between any two time points in the list.

Example 1:

Input: ["23:59","00:00"]
Output: 1

Note:

  1. The number of time points in the given list is at least 2 and won’t exceed 20000.
  2. The input time is legal and ranges from 00:00 to 23:59.

Solution

Time complexity: O(nlog1440)

Space complexity: O(n)

C++

 

花花酱 LeetCode 538. Convert BST to Greater Tree

Problem

题目大意:把二叉搜索树的每个节点加上比他大的节点的和。

https://leetcode.com/problems/convert-bst-to-greater-tree/description/

Given a Binary Search Tree (BST), convert it to a Greater Tree such that every key of the original BST is changed to the original key plus sum of all keys greater than the original key in BST.

Example:

Input: The root of a Binary Search Tree like this:
              5
            /   \
           2     13

Output: The root of a Greater Tree like this:
             18
            /   \
          20     13

Solution: reversed inorder traversal

in a BST, we can visit every node in the decreasing order. Using a member sum to track the sum of all visited nodes.

Time complexity: O(n)

Space complexity: O(1)

C++

 

花花酱 LeetCode 563. Binary Tree Tilt

Problem

https://leetcode.com/problems/binary-tree-tilt/description/

题目大意:返回树的总tilt值。对于一个节点的tilt值定义为左右子树元素和的绝对之差。

Given a binary tree, return the tilt of the whole tree.

The tilt of a tree node is defined as the absolute difference between the sum of all left subtree node values and the sum of all right subtree node values. Null node has tilt 0.

The tilt of the whole tree is defined as the sum of all nodes’ tilt.

Example:

Input: 
         1
       /   \
      2     3
Output: 1
Explanation: 
Tilt of node 2 : 0
Tilt of node 3 : 0
Tilt of node 1 : |2-3| = 1
Tilt of binary tree : 0 + 0 + 1 = 1

Note:

  1. The sum of node values in any subtree won’t exceed the range of 32-bit integer.
  2. All the tilt values won’t exceed the range of 32-bit integer.

Solution: Recursion

Time complexity: O(n)

Space complexity: O(h)

C++

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