Google Code jam Solutions: Problem A. Store Credit

Problem

You receive a credit C at a local store and would like to buy two items. You first walk through the store and create a list L of all available items. From this list you would like to buy two items that add up to the entire value of the credit. The solution you provide will consist of the two integers indicating the positions of the items in your list (smaller number first).

Input

The first line of input gives the number of cases, N. N test cases follow. For each test case there will be:

  • One line containing the value C, the amount of credit you have at the store.
  • One line containing the value I, the number of items in the store.
  • One line containing a space separated list of I integers. Each integer P indicates the price of an item in the store.
  • Each test case will have exactly one solution.

Output

For each test case, output one line containing “Case #x: ” followed by the indices of the two items whose price adds up to the store credit. The lower index should be output first.

Limits

5 ≤ C ≤ 1000
1 ≤ P ≤ 1000

Small dataset

N = 10
3 ≤ I ≤ 100

Large dataset

N = 50
3 ≤ I ≤ 2000

Sample

Input Output
3
100
3
5 75 25
200
7
150 24 79 50 88 345 3
8
8
2 1 9 4 4 56 90 3
Case #1: 2 3
Case #2: 1 4
Case #3: 4 5

C++ Solution:

/*
Author      :   Sreejith Sreekantan
Description :   https://code.google.com/codejam/contest/351101/dashboard#s=p0

*/

#include 
#include 
#include 
#include 

using namespace std;

int main(int argc, char const *argv[])
{

    int numOfTestInstances;
    cin >> numOfTestInstances;
    std::vector itemWeight;
    for (int testInstanceNum = 0; testInstanceNum < numOfTestInstances; ++testInstanceNum)
    {
        int limit;
        cin >> limit;

        int numOfItems;
        cin >> numOfItems;

        itemWeight.clear();
        itemWeight.reserve(numOfItems);

        for (int itemNum = 0; itemNum < numOfItems; ++itemNum)
        {
            cin >> itemWeight[itemNum];
        }

        cout << "Case #" << testInstanceNum + 1 << ": ";

        for (int i = 0; i < numOfItems; ++i)
        {
            for (int j = i + 1; j < numOfItems; ++j)
            {
                if (itemWeight[i] > limit)
                {
                    break;
                }
                if (itemWeight[j] > limit)
                {
                    continue;
                }
                if (itemWeight[i] + itemWeight[j] == limit)
                {
                    if (i < j)
                    {
                        cout << i + 1 << " " << j + 1 << endl;
                    }
                    else
                    {
                        cout << j + 1 << " " << i + 1 << endl;
                    }
                }
            }
        }

    }
    return 0;
}

Python Solution:

#! /usr/bin/python

numOfInstances = int(raw_input())

for x in xrange(1,numOfInstances+1):
    limit = int(raw_input())
    numOfItems = int(raw_input())
    d = dict()
    items = [int(x) for x in raw_input().split()]
    for y in xrange(1,numOfItems+1):
        if items[y-1] not in d:
            d[items[y-1]] = set() 
        d[items[y-1]].add(y)

     print d

    for x in d.keys():
        if x <= limit:
            index = d[x].pop()
            if (limit-x) in d and len(d[limit-x])>0:
                index2=d[limit-x].pop()
                d[limit-x].add(index2)
                print min(index, index2), max(index, index2)
            d[x].add(index)


Google Code jam Solutions: Problem B. Reverse Words

Problem

Given a list of space separated words, reverse the order of the words. Each line of text contains L letters and W words. A line will only consist of letters and space characters. There will be exactly one space character between each pair of consecutive words.

Input

The first line of input gives the number of cases, N.
N test cases follow. For each test case there will a line of letters and space characters indicating a list of space separated words. Spaces will not appear at the start or end of a line.

Output

For each test case, output one line containing “Case #x: ” followed by the list of words in reverse order.

Limits

Small dataset

N = 5
1 ≤ L ≤ 25

Large dataset

N = 100
1 ≤ L ≤ 1000

Sample

Input Output
3
this is a test
foobar
all your base
Case #1: test a is this
Case #2: foobar
Case #3: base your all

C++ Solution:

/*
    Author      :   Sreejith Sreekantan
    Description :   Problem B. Reverse Words https://code.google.com/codejam/contest/351101/dashboard#s=p1
                    
*/

#include 
#include 
#include 
#include 
#include 
#include 

using namespace std;


int main(int argc, char const *argv[])
{
    
    int numOfTestInstances;
    cin >> numOfTestInstances;
    
    for (int testInstanceNum = 0; testInstanceNum < numOfTestInstances; ++testInstanceNum)
    {
        istringstream in;
        string s;
        cin >> ws;
        getline(cin ,s);
        replace(s.begin(), s.end(), ' ', '\n');
        in.str(s);
        
        stack stack_s_rev;

        while (in >> s)
        {
            stack_s_rev.push(s);
        }
        cout << "case #" << testInstanceNum+1 << ": ";
        while(!stack_s_rev.empty())
        {
            cout << stack_s_rev.top() << " ";
            stack_s_rev.pop();
        }
        cout << endl;

    }
    return 0;
}


Google Code jam Solutions: Problem A. Alien Language

Problem

After years of study, scientists at Google Labs have discovered an alien language transmitted from a faraway planet. The alien language is very unique in that every word consists of exactly L lowercase letters. Also, there are exactly D words in this language.

Once the dictionary of all the words in the alien language was built, the next breakthrough was to discover that the aliens have been transmitting messages to Earth for the past decade. Unfortunately, these signals are weakened due to the distance between our two planets and some of the words may be misinterpreted. In order to help them decipher these messages, the scientists have asked you to devise an algorithm that will determine the number of possible interpretations for a given pattern.

A pattern consists of exactly L tokens. Each token is either a single lowercase letter (the scientists are very sure that this is the letter) or a group of unique lowercase letters surrounded by parenthesis ( and ). For example: (ab)d(dc) means the first letter is either a or b, the second letter is definitely d and the last letter is either d or c. Therefore, the pattern (ab)d(dc) can stand for either one of these 4 possibilities: add, adc, bdd, bdc.

Input

The first line of input contains 3 integers, L, D and N separated by a space. D lines follow, each containing one word of length L. These are the words that are known to exist in the alien language. N test cases then follow, each on its own line and each consisting of a pattern as described above. You may assume that all known words provided are unique.

Output

For each test case, output

Case #X: K

where X is the test case number, starting from 1, and K indicates how many words in the alien language match the pattern.

Limits

 

Small dataset

1 ≤ L ≤ 10
1 ≤ D ≤ 25
1 ≤ N ≤ 10

Large dataset

1 ≤ L ≤ 15
1 ≤ D ≤ 5000
1 ≤ N ≤ 500

Sample

Input Output
3 5 4
abc
bca
dac
dbc
cba
(ab)(bc)(ca)
abc
(abc)(abc)(abc)
(zyx)bc
Case #1: 2
Case #2: 1
Case #3: 3
Case #4: 0

C++ Solution:

/*
    Author      :   Sreejith Sreekantan
    Description :   Problem A. Alien Language (https://code.google.com/codejam/contest/90101/dashboard#s=p0)

*/

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 

using namespace std;


#define rep(v,N) for(int v=0; v<n; v++)="" #define="" <span="" class="hiddenSpellError" pre="define " data-mce-bogus="1">rep2(v,M,N) for(int v=M; v<n; v++)="" #define="" for(v,c)="" for(__typeof(c.begin())="" v="C.begin();" v!="C.end();" ++v)="" vi="" std::vector<int="">
#define vll std::vector
#define pb push_back
#define Sort(C) std::sort(C.begin(), C.end())
#define RSort(C) std::sort(C.rbegin(), C.rend())
#define Copy(ans,out) copy(ans.begin(),ans.end(), ostream_iterator<__typeof(ans[0])>(out, " "))

// #define SMALL
#define LARGE

int main(int argc, char const *argv[])
{
    freopen("a.in", "rt", stdin);
    // freopen("a.out", "wt", stdout);
#ifdef SMALL
    freopen("as.in", "rt", stdin);
    freopen("as.out", "wt", stdout);
#endif

#ifdef LARGE
    freopen("al.in", "rt", stdin);
    freopen("al.out", "wt", stdout);
#endif

    unsigned int L, D, N;
    cin >> L >> D >> N;

    std::vector dict(D);
    std::vector inp(D);
    rep(i, D)
    {
        cin >> dict[i];
    }
    rep(i, N)
    {
        cin >> inp[i];
        replace(inp[i].begin(), inp[i].end(), '(', '[');
        replace(inp[i].begin(), inp[i].end(), ')', ']');
        // cout << inp[i]<< endl;
        int c = 0;
        rep(j, D)
        {
            if (regex_match(dict[j], regex(inp[i])) ) c++;
        }
        cout << "Case #" << i+1 << ": " << c << endl;
        
    }
    return 0;
}

Brain Teaser

T : Number of test cases

M: value of M

N: Number of strings

For each string, ascii value of each character of the string is raised to M and multiplied together. For each test case the sum of above values  is ODD or EVEN

Input:

1

10    2

ac ab

Output:

ODD

 

TIP: Challenge lies in finding the solution without doing all the mathematical operations mentioned in the question. (yes, also called optimizing..!)

 

 

Solution:

#!/usr/bin/python

T = int(raw_input())
while T>0:
    T = T - 1
    tmp = raw_input()
    M = int(tmp.split()[0])
    K = int(tmp.split()[1])
    tmp = raw_input()
    final_odd = False
    for s in tmp.split():
        odd = None
        for i in s:
            if ord(i)%2==0:        # the power is even
                odd = False
            else:                # the power is odd
                if odd==None:            # if odd is True, odd*odd=odd
                    odd = True
        # odd + odd = even
        # odd + even = odd
        # even + even = even
        if final_odd != odd:
            final_odd = True
        else:
            final_odd = False
    if final_odd:
        print "ODD"
    else:
        print "EVEN"

Regular Expression in Python: Greedy match

When you use re.findall(“(.*):” to get the strings ending with full-colon(:), you may not always get the expected result if your string has multiple full-colons. Lets run the command before we talk.

Concentrate on the commands put inside single colon. (You can neglect the rest as it is part of my effort to avoid creating a python file for running this code)

Screen Shot 2014-10-15 at 6.12.15 PMOutput:Screen Shot 2014-10-15 at 6.12.21 PM

What happened here is python did a greedy search and found maximum string ending with a full-colon. If that is what you desired, stop reading further.. 🙂

If you expected a list of all the strings ending with full-colon, change your command to re.findall(“(.*?):”

As you can see, I added a question mark which will force python to avoid greedy approach.

Before we part, lets make sure python is ok with this change..

 

Screen Shot 2014-10-15 at 6.11.40 PM

Output:

Screen Shot 2014-10-15 at 6.12.00 PMLOVE YOU PYTHON…

Find the biggest black square from an N*N matrix of black or white squares

Question:

Find the biggest black square from an N*N matrix of squares. 1 means it’s black, 0 means it’s white. Output should be the list of squares which forms the largest black square.

Input     : ‘{0#1#1#1#0#1#0#1,1#0#1#0#0#0#0#1,0#0#0#1#0#1#0#0,1#1#1#1#1#0#0#1,1#1#1#1#0#1#1#1,1#1#1#1#0#1#1#1,1#1#1#1#1#1#1#1,1#1#0#1#0#0#1#1}’

Output    :    {(3#0,3#1,3#2,3#3),(4#0,4#1,4#2,4#3),(5#0,5#1,5#2,5#3),(6#0,6#1,6#2,6#3)}

 

Solution(Python):

    # Author            :        Sreejith Sreekantan
    # Description        :        

    #                         Find the biggest black square from an N*N matrix of squares. 1 means it's black, 0 means it's white
    #                         Input     : '{0#1#1#1#0#1#0#1,1#0#1#0#0#0#0#1,0#0#0#1#0#1#0#0,1#1#1#1#1#0#0#1,1#1#1#1#0#1#1#1,1#1#1#1#0#1#1#1,1#1#1#1#1#1#1#1,1#1#0#1#0#0#1#1}'
    #                         Output    :    {(3#0,3#1,3#2,3#3),(4#0,4#1,4#2,4#3),(5#0,5#1,5#2,5#3),(6#0,6#1,6#2,6#3)}
    

#!/usr/bin/python

def largestSquareAt(sq, i, j, k=1):
    if not ( i+k<len(sq) and="" j+k<len(sq[i])="" ):=""   =""    =""  return="" 0=""  for="" x="" in="" range(0,k):=""  if="" sq[i+x][j+k-1]="='0':" y="" sq[i+k-1][j+y]="='0':" 1="" +="" largestsquareat(sq,="" i,="" j,="" k+1)  =""  ="" def="" biggestsquare(input1):="" len(input1)="=0" or="" not="" (="" input1.count('{')="=input1.count('}')" )="" :="" ''=""  i="str(input1)"  j="i" for="" j]=""  resx="-1"  rexy="-1"  ressize="-1"  res="[]" range(0,len(j)):=""  res.append([])="" range(0,len(j[x])):=""  res[x].append(largestsquareat(j,="" x,="" y))="" ressize<res[x][y]:=""  resy="y"  resstring="" ressize="">0:                                
        resstring = "{"
        for x in range(resx,resx+ressize):
            if x>resx:
                resstring += ","
            resstring += "("
            for y in range(resy,resy+ressize):
                if y>resy:
                    resstring += ","
                resstring += (str(x)+"#"+str(y))
            resstring += ")"    
        resstring += "}"
    
    return resstring  

input = '{0#1#1#0,0#1#1#1,0#1#1#1,1#1#1#1}'
input1 = '{0#1#1#1#0#1#0#1,1#0#1#0#0#0#0#1,0#0#0#1#0#1#0#0,1#1#1#1#1#0#0#1,1#1#1#1#0#1#1#1,1#1#1#1#0#1#1#1,1#1#1#1#1#1#1#1,1#1#0#1#0#0#1#1}'
input2 = '{0#1#1#1#0#1#0#1,1#0#1#0#0#0#0#1,0#0#0#1#0#1#0#0,1#1#1#1#1#0#0#1,1#1#0#1#0#1#1#1,1#1#1#1#0#1#1#1,1#1#1#1#1#1#1#1,1#1#0#1#0#0#1#1}'
input3= '{0#1#1#1#0#1#0#1,1#0#1#0#0#0#0#1,0#0#0#1#0#1#0#0,1#1#1#1#1#0#0#1,0#1#0#1#0#1#1#1,1#1#1#1#0#1#1#1,1#1#1#1#1#1#1#1,1#1#0#1#0#0#1#1}'
input4 = '{0#0#0#0,0#0#0#0,0#0#0#0,0#0#0#0}'
print biggestSquare('')    
print biggestSquare(input1)    
print biggestSquare(input2)        
print biggestSquare(input3)        
print biggestSquare(input4)

Given a string of length N, output “Correct” if brackets close in correct order else output “Incorrect”

Question:

Given a string of length N, output “Correct” if brackets close in correct order else output “Incorrect”

Input     :   ({}[((({{}})[{()}]))])

Solution(Python):

    # Author            :        Sreejith Sreekantan
    # Description        :        

    #                         Given a string of length N, output "Correct" if brackets close in correct order else output "Incorrect"
    #                         Input     : ({}[((({{}})[{()}]))])
    #                         

#!/usr/bin/python
def validString(input1):
    stk = []
    flag = True
    for x in input1:
        if x in ['{', '(', '[']:
            stk.append(x)
        else:
            if  len(stk)==0:
                flag = False
            elif x == '}' and not stk[len(stk)-1]=='{' :
                flag = False    
            elif x == ')' and not stk[len(stk)-1]=='(' :
                flag = False 
            elif x == '}' and not stk[len(stk)-1]=='{' :
                flag = False    
            if not flag:
                return "Incorrect"
            else:
                stk.pop()
    return "Correct"
 

input1 = '({}[((({{}})[{()}]))])'
input2 = '({}((({{}})[{()}]))])'
print validString(input1)    
print validString(input2)