LeetCode [544] Output Contest Matches

During the NBA playoffs, we always arrange the rather strong team to play with the rather weak team, like make the rank 1 team play with the rank nth team, which is a good strategy to make the contest more interesting. Now, you're given n teams, you need to output their final contest matches in the form of a string.

The n teams are given in the form of positive integers from 1 to n, which represents their initial rank. (Rank 1 is the strongest team and Rank n is the weakest team.) We'll use parentheses('(', ')') and commas(',') to represent the contest team pairing - parentheses('(' , ')') for pairing and commas(',') for partition. During the pairing process in each round, you always need to follow the strategy of making the rather strong one pair with the rather weak one.

Example 1:

Input: 2
Output: (1,2)
Explanation: 
Initially, we have the team 1 and the team 2, placed like: 1,2.
Then we pair the team (1,2) together with '(', ')' and ',', which is the final answer.

Example 2:

Input: 4
Output: ((1,4),(2,3))
Explanation: 
In the first round, we pair the team 1 and 4, the team 2 and 3 together, as we need to make the strong team and weak team together.
And we got (1,4),(2,3).
In the second round, the winners of (1,4) and (2,3) need to play again to generate the final winner, so you need to add the paratheses outside them.
And we got the final answer ((1,4),(2,3)).

Example 3:

Input: 8
Output: (((1,8),(4,5)),((2,7),(3,6)))
Explanation: 
First round: (1,8),(2,7),(3,6),(4,5)
Second round: ((1,8),(4,5)),((2,7),(3,6))
Third round: (((1,8),(4,5)),((2,7),(3,6)))
Since the third round will generate the final winner, you need to output the answer (((1,8),(4,5)),((2,7),(3,6))).

Note:

1. The n is in range [2, 212].
2. We ensure that the input n can be converted into the form 2k, where k is a positive integer.

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class Solution { public: string findContestMatch(int n) { vector<string> schedules(n); for (int i = 0; i < n; ++i) { schedules[i] = to_string(i + 1); } while (n) { for (int i = 0; i < n/2; i++) { schedules[i] = "(" + schedules[i] + "," + schedules[n - 1 - i] + ")"; } n = n / 2; } return schedules[0]; } };

test

Windbg on a C sharp application

I've been trying to learn Windbg for a long time. Finally found a log I can successfully follow through.

http://www.cnblogs.com/ceachy/p/WinDBG_Demo1.html

Here's the application.
Compiled using : csc.exe /debug /out:Test3.exe Test3.cs

LeetCode [360] Sort Transformed Array

LeetCode [359] Logger Rate Limiter

359. Logger Rate Limiter

Easy

Design a logger system that receive stream of messages along with its timestamps, each message should be printed if and only if it is not printed in the last 10 seconds.

Given a message and a timestamp (in seconds granularity), return true if the message should be printed in the given timestamp, otherwise returns false.

It is possible that several messages arrive roughly at the same time.

Example:

Logger logger = new Logger();

// logging string "foo" at timestamp 1
logger.shouldPrintMessage(1, "foo"); returns true; 

// logging string "bar" at timestamp 2
logger.shouldPrintMessage(2,"bar"); returns true;

// logging string "foo" at timestamp 3
logger.shouldPrintMessage(3,"foo"); returns false;

// logging string "bar" at timestamp 8
logger.shouldPrintMessage(8,"bar"); returns false;

// logging string "foo" at timestamp 10
logger.shouldPrintMessage(10,"foo"); returns false;

// logging string "foo" at timestamp 11 

logger.shouldPrintMessage(11,"foo"); returns true; 

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class Logger { Map<String, Integer> map; /** Initialize your data structure here. */ public Logger() { map = new HashMap<>(); } /** Returns true if the message should be printed in the given timestamp, otherwise returns false. If this method returns false, the message will not be printed. The timestamp is in seconds granularity. */ public boolean shouldPrintMessage(int timestamp, String message) { if(map.isEmpty() || !map.containsKey(message) || timestamp-map.get(message)+1>10){ map.put(message, timestamp); return true; }else{ return false; } } } /** * Your Logger object will be instantiated and called as such: * Logger obj = new Logger(); * boolean param_1 = obj.shouldPrintMessage(timestamp,message); */

LeetCode [358] Rearrange String k Distance Apart

Given a non-empty string s and an integer k, rearrange the string such that the same characters are at least distance k from each other.

All input strings are given in lowercase letters. If it is not possible to rearrange the string, return an empty string "".

Example 1:

Input: s = "aabbcc", k = 3
Output: "abcabc" 
Explanation: The same letters are at least distance 3 from each other.

Example 2:

Input: s = "aaabc", k = 3
Output: "" 
Explanation: It is not possible to rearrange the string.

Example 3:

Input: s = "aaadbbcc", k = 2
Output: "abacabcd"
Explanation: The same letters are at least distance 2 from each other.

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class Solution { public: //find the next valid char can be pus at "index" int getChar(vector<int>& counts, vector<int>& valid, int index) { int m = 0, offset = -1; for(int i=0; i<26; ++i) { if(counts[i]>m && index>=valid[i]) { m = counts[i];//greedy: use the char with max count offset = i; } } return offset; } string rearrangeString(string s, int k) { if(k==0) return s; int n = s.size(); vector<int> counts(26, 0);//counts[i] is count of char('a'+i) vector<int> valid(26, 0);//valid[i] is the next smallest valid position of char('a'+i) for(auto c:s) counts[c-'a']++; string ret; for(int i=0; i<n; ++i) { int offset = getChar(counts, valid, i); if(offset<0) return "";//cannot find a valid char ret += ('a'+offset); counts[offset]--; valid[offset] = i+k; } return ret; } };

LeetCode [357] Count Numbers with Unique Digits