TopCoder SRM 571 - Div I Lvl 1

http://community.topcoder.com/stat?c=problem_statement&pm=12436&rd=15491
The following code prints the first 50 lexicographically sorted values of the records names:


class Test:
def __init__(self):
self.res = []
def f(self,n,x):
if x >n:
return
if len(self.res) < 50:
self.res.append(x)
#print x
else:
return

self.f(n,x*10)
rem = x%10
if rem != 9:
self.f(n,x+1)
def getRes(self,n,x):
self.f(n,x)
print self.res

Sample values printed from the program:
Test().getRes(1010,1)


1
10
100
1000
10000<
1001
10010<
1002
1003
:
1009
101
1010

TopCoder SRM 572 - Div I Lvl 1

http://community.topcoder.com/stat?c=round_overview&er=5&rd=15492


def myfunc(s,k):
    res = 0
    incr = len(s)-k
    for i in range(0,incr):
        # find the first set of alphabets which don't match
        j = i
        t = {}
        ctr = 0
        #print "i:",str(i)
        while j<len(s):
            #print "j:",str(j)
            if s[j] in t:
                t[s[j]] += 1
            else:
                t[s[j]] = 1
            j += incr
            ctr += 1
        mx = max(t.values())
        res += ctr - mx
   
    return res

TopCoder SRM 573 - Div I Lvl 1

Div II Lvl 2 problem: easier version solution
http://community.topcoder.com/stat?c=problem_statement&pm=12471&rd=15493


def myfunc(s):
    our_strength = sum(s[0:3])- min(s[0:3])
    rst = s[3:]
    #print our_strength
    if len(rst) == 0:
        return 1
    # find the rank of our team, start with rank 1 and increment it as you find better ranks
    res = 1
    rst.sort()
    #print rst
    found = True
    while found == True:
        # get the max element
        max = rst[-1]
        rst.remove(max)
        # find the min element which is greater than our_rank
        found = False
        for i in rst:
            if i+max > our_strength:
                rst.remove(i)
                found = True
                res += 1
                break
    return res

http://apps.topcoder.com/wiki/display/tc/SRM+573

TopCoder SRM 575 - DIV 1 Level 1

http://community.topcoder.com/stat?c=problem_statement&pm=12496&rd=15495

The solution for the smaller problem:


def myfunc(n):
    f = [0] *(n+1)
    for i in range(2,n+1):
        for j in range(2,i):
            if(i%j==0 and f[i-j]==0):
                f[i] =1
    print [ (i,f[i]) for i in range(0,len(f))]

The solution for the bigger problem involves getting a pattern from the above solution.
http://apps.topcoder.com/wiki/display/tc/SRM+575

Using rsync to analyse hadoop cluster logs at the name node

If you have worked on a hadoop cluster and tried going through the logs on all different cluster nodes you know how painful can it be. This script can be run on the name node and it will copy all the logs for the given hadoop job-id to the current directory of the name node. You will have to change the rsync parameters to suit yourself.

Python script:

import sys
import subprocess
import os

if len(sys.argv) < 2:
        print "Usage: python ",sys.argv[0]," <hadoop-job-id>"
        sys.exit(0)

nodes = ['192.168.112.117', '192.168.156.63', '192.168.152.31', '192.168.112.118', '192.168.156.65', '192.168.156.62' ]

#subprocess.Popen(["ls","-lr"])

for node in nodes:
        subprocess.Popen(["rsync", "-rav", "root@"+node+":/data/hadoop_logs/userlogs/"+sys.argv[1], ".")

#os.system("rsync -rav root@node-0de4bd:/data/hadoop_logs/userlogs/"+sys.argv[1]+" .")

IMP: Rsync requires that you are able to do the password-less access to these nodes from the current node (on which you are running this script.) For help on setting up password-less access check here.