LIKE CODING

Wednesday, August 15, 2018

test

Friday, December 29, 2017

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

Thursday, August 11, 2016

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;

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.

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;
    }
};