1. Hey guys, Hope you all are doing great. I am back with Some Python Programming tips for

     CODE GOLF Challenges.

     So, What is CODE GOLF??
    As per Wikipedia, it is a type of recreational computer programming competition in which participants strive to achieve the shortest possible source code (not to be confused with binary Sizecoding) that implements a certain algorithm.

     In simple terms, your goal is to write a code is shortest way, using minimum number of characters to implement a logic in correct way.

     You can use any language Java, C++, Python, Perl, Ruby, JavaScript, and anything to write it. But scripting language have a upper-hand over descriptive language as you can write programs in concise way in scripting language like Ruby, Python, etc, while it takes an extra character each time to write the code in Java, C#, etc.

     So Preferred code golf languages are GolfScript, Flogscript, Perl, Python, Erlang, Ruby, Haskell, Bash, PHP, Clojure, K, Javascript, Vim and the list go on....

     I prefer Python or Python3 myself for code golfing

     So we will discuss dome tips as generics for code golf programming in Python:

     1. Make the assignments shorter.

    Use a=b=c=0 instead of a,b,c=0,0,0.
Use a,b,c='123' instead of a,b,c='1','2','3'.

    2. Tweak the if-else condition.

     
    b=2*a if a<0 else 3*a          # proper python
b=a<0 and 2*a or 3*a           # codegolf1
b=a*(3,2)[a<0]                 # codegolf2

    3. Combine multiple conditionals. Instead of "1<x and x<2".

     
    Use 1<x<2

    4. Using a built-in function repeatedly, it might be more space-efficient to give it a 
    new name, if using different arguments.

     
    r=range
    for x in r(10):
for y in r(100):print x,y

    5. Read only two lines from standard in (without the carriage return) and store then in an array of strings.

     
    r=raw_input
a=r(),r()

    6. Simulating C's ternary operator (p?a:b)

     
    r=[b,a][p]


    Stay tuned for more Tips and Tricks in the next Post!!

     Till Then Cheers!!




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  2. Hey guys, Hope you all are doing great. I am back with with a byte of python performance.

     I have been playing on codingame.com a lot now a days during my weekends.




     Its cool site for the player aka coders who want to try there hands into gaming domain. You have plenty to select from






    So stay tuned for the coming new gen puzzles and coding tips to increase your efficiency and performance!!


    Hope you find this post useful and interesting. Please let me know your feedback, on how i can improve and make it better. Till then!! Cheers!!
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  3. Hey guys, Hope you all are doing great. I am back with another problem of Project Euler on
    Project Euler #11: Largest product in a grid

    In the 20×20 grid below, four numbers along a diagonal line have been marked in bold.
    89 90 95 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08  
49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00  
81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65  
52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91  
22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80  
24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50  
32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70  
67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21  
24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72  
21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95  
78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92  
16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57  
86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58  
19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40  
04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66  
88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69  
04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36  
20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16  
20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54  
01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48
    The product of these numbers is 26 × 63 × 78 × 14 = 1788696.
    What is the greatest product of four adjacent numbers in the same direction (up, down, left, right, or diagonally) in the 20×20 grid?
    Input Format
    Input consists of 20 lines each containing 20 integers.
    Output Format
    Print the required answer.
    Constraints
    0 ≤ Each integer in the grid ≤ 100
    Sample Input
    89 90 95 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08  
49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00  
81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65  
52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91  
22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80  
24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50  
32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70  
67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21  
24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72  
21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95  
78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92  
16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57  
86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58  
19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40  
04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66  
88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69  
04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36  
20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16  
20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54  
01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48
    Sample Output
    73812150

    Algorithm:


    This problem ask for a simple approach to check for the product of numbers in:
    1. Horizontal - Left to Right
    2. Vertical - Top to Bottom
    3. Diagonal - Top Left to Bottom Right
    In this way we can cover all the possible 4 number sets:
    1. Horizontal: 17 * 20 (17 because we are taking sets of 4).
    2. Vertical: 17 * 20 (17 because we are taking sets of 4).
    3. Diagonal: 17*17
    So total of 969 calculations.
    I will jump to the program which is pretty straight forward.

    Program:

    grid=list()
for i in range(20):
    grid.append(map(int, raw_input().split()))
ans=0
for i in range(20):
    for j in range(20 - 4 + 1):
        phv = max(grid[i][j] * grid[i][j+1] * grid[i][j+2] * grid[i][j+3], grid[j][i] * grid[j+1][i] * grid[j+2][i] * grid[j+3][i])
        if (i < 20 - 4):
            pdd = max(grid[i][j] * grid[i+1][j+1] * grid[i+2][j+2] * grid[i+3][j+3], grid[i][j+3] * grid[i+1][j+2] * grid[i+2][j+1] * grid[i+3][j])
        ans = max(ans, phv, pdd)
print ans
    Output:
    73812150

    And Presto!! Its Done!!

     Please let me know how i can improve my post and any other feedback are welcome.

     Till then Cheers!!
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  4. Hey guys, Hope you all are doing great.
    Today i came across an intruding method, which is not used in generally but is very effective and we encounter its application a lot of .

     That's 

    Tabibitosan method


    Tabibitosan: is a Japanese word which literally means "Pilgrim". This came up when people thought it as a method there you transverse through a consecutive pattern (same as going from temple to temple).

    I came across this method while exploring Oracle Community Forum written by Aketi Jyuuzou who explained it beautifully.

    Thanks Aketi Jyuuzou!!

    Tabibitosan method is useful in below application:
    1. For counting consecutive value sets.
    2. For getting distinct attributes based on continued start and end dates
    3. Grouping of continuous dates
    4. Counting continues years, months, weeks.
    5. To optimize cases where there are duplicate records. (In Next Post we will discuss on optimum duplicate detection and removal techniques.)
    6. There can be many more application depending on your need.
    Now, understanding how this method works, I will pic an example as a problem in Hackerrank Advanced Join section 

    Projects


    You are given a table, Projects, containing three columns: Task_ID, Start_Date and End_Date. It is guaranteed that the difference between the End_Date and the Start_Date is equal to 1 day for each row in the table.
    If the End_Date of the tasks are consecutive, then they are part of the same project. Samantha is interested in finding the total number of different projects completed.
    Write a query to output the start and end dates of projects listed by the number of days it took to complete the project in ascending order. If there is more than one project that have the same number of completion days, then order by the start date of the project.
    Sample Input
    Sample Output
    2015-10-28 2015-10-29
2015-10-30 2015-10-31
2015-10-13 2015-10-15
2015-10-01 2015-10-04

    What we need to do?
    The example describes following four projects:
    • Project 1: Tasks 1, 2 and 3 are completed on consecutive days, so these are part of the project. Thus start date of project is 2015-10-01 and end date is 2015-10-04, so it took 3 days to complete the project.
    • Project 2: Tasks 4 and 5 are completed on consecutive days, so these are part of the project. Thus, the start date of project is 2015-10-13 and end date is 2015-10-15, so it took 2 days to complete the project.
    • Project 3: Only task 6 is part of the project. Thus, the start date of project is 2015-10-28 and end date is 2015-10-29, so it took 1 day to complete the project.
    • Project 4: Only task 7 is part of the project. Thus, the start date of project is 2015-10-30 and end date is 2015-10-31, so it took 1 day to complete the project.
    So in all we need to find the projects which have continued start date and end date.
    We will use Tabibitosan method:

    So we will do like i move from last end date to next start date, if the time is 1 then its continued, else its a break in count, here is how we will realize that in SQL:


    SELECT  START_DATE, END_DATE, END_DATE - ROW_NUMBER() OVER (ORDER BY END_DATE) DIST
FROM PROJECTS


    Here we will have list of start date and end date with a third date that is common to the continued dates.

    Output to above problem's data for the SQL will be:

     
    START_DATE      END_DATE      DIST
2015-10-28     2015-10-29    2015-10-14
2015-10-30     2015-10-31    2015-10-15
2015-10-13     2015-10-15    2015-10-07
2015-10-14     2015-10-16    2015-10-07
2015-10-01     2015-10-04    2015-09-30
2015-10-01     2015-10-04    2015-09-30
2015-10-01     2015-10-04    2015-09-30
2015-10-01     2015-10-04    2015-09-30

    Now its easy to group the Start and End Date by the common Distance date, with lower limit as MIN(START_DATE) and upper limit as MAX(END_DATE).

     And Presto!! We are done.

    Below is the final code:

     
    SELECT SD,ED
FROM
(SELECT MIN(START_DATE) SD, MAX(END_DATE) ED, ( MAX(END_DATE)-MIN(START_DATE)) dd
FROM
(
SELECT  START_DATE, END_DATE, END_DATE - ROW_NUMBER() OVER (ORDER BY END_DATE) DIST
FROM PROJECTS
 ORDER BY 1)
GROUP BY DIST
 ORDER BY 3,1);

    Output:


    SD         ED
2015-10-28 2015-10-29
2015-10-30 2015-10-31
2015-10-13 2015-10-15
2015-10-01 2015-10-04


    Hope you guys find it useful. Please let me know about your feed backs, areas where i can improve.

     Till then keep coding!!


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  5. Hey guys, 
    Hope you all are doing great. I am back with another problem of Project Euler on

    The sum of the primes below  is 
    Find the sum of all the primes not greater than given .
    Input Format
    The first line contains an integer  i.e. number of the test cases.
    The next  lines will contains an integer .
    Output Format
    Print the value corresponding to each test case in separate line.
    Constraints

    Sample Input
    2
5
10
    Sample Output
    10
17

    Algorithm:

    The Sieve of Eratosthenes is a simple algorithm that finds the prime numbers up to a given integer. Implement this algorithm, with the only allowed optimization that the outer loop can stop at the square root of the limit, and the inner loop may start at the square of the prime just found. That means especially that you shouldn't optimize by using pre computed wheels, i.e. don't assume you need only to cross out odd numbers (wheel based on 2), numbers equal to 1 or 5 modulo 6 (wheel based on 2 and 3), or similar wheels based on low primes.

    Odds-only version of the array sieve above
    The below code uses only odd composite operations (for a factor of over two) and because it has been optimized to use slice operations for composite number culling to avoid extra work by the interpreter

    Program:



    def primes2(limit):
    if limit < 2: return []
    if limit < 3: return [2]
    lmtbf = (limit - 3) // 2
    buf = [True] * (lmtbf + 1)
    for i in range((int(limit ** 0.5) - 3) // 2 + 1):
        if buf[i]:
            p = i + i + 3
            s = p * (i + 1) + i
            buf[s::p] = [False] * ((lmtbf - s) // p + 1)
    return sum([2] + [i + i + 3 for i, v in enumerate(buf) if v])

t=input()
while(t>0):
    print primes2(input())
    t-=1

    Hope you liked this post.
    Do let me know about any improvements i can incorporate to anything in the blog.

    Till Then...Cheers!!

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  6. Hey guys, Hope you all are doing great. I am back with a new domain SQL. So the problem goes like, you need to print prime numbers less than 1000.

    Seems easy right!! Just use a for loop check for the list of natural numbers for prime condition whether they are only divisible by 1 and itself.

    But if you are asked to do it only using plain SQL, i will be using Oracle in my case, how would you do that.

    In PL/SQL its possible with straight approach, but in SQL its bit tricky.Lets start....
    Write a query to print all prime numbers less than or equal to 1000. Print your result on a single line, and use the (,) character as your separator (instead of a space).
    For example, the output for all prime numbers  would be:
    2,3,5,7

    Algorithm:


    1. First we will generate the numbers from 1 to 1000. This can be achieved by using level connect in SQL i.e. Hierarchical SQL. A condition that identifies the relationship between parent rows and child rows of the hierarchy. So we will get number 1, 2, 3.....1000.
    This is how it will look on Oracle SQL format....

    select level l from dual connect by level  &lt;= 1000



    2. Now we can use above query to generate another set of numbers and proceed with our logic.
    So for checking the prime condition there should be a number which is when divided return 0 as remainder only 2 times, that is divided by 1 and divided by num (number itself).

     So we have a condition a SQL where we have Count = 2

     This is how we can implement it:


    select l prime_number
from (select level l from dual connect by level <= 1000)
, (select level m from dual connect by level < = 1000)
where m<=l
group by l
having count(case l/m when trunc(l/m) then 'Y' end) = 2;

    Output of above looks like:

    PRIME_NUMBER
43
83
151
167
173
179
227
269

     

     3. Now for the final step we need to connect the output is separate lines to single line output separated by comma (,). For this purpose we will use LISTAGG function in SQL

     Program/Script:
    So our final script looks like...

    SELECT LISTAGG(prime_number,',') WITHIN GROUP (ORDER BY prime_number) AS NUMBERS
FROM(    
    select l prime_number
from (select level l from dual connect by level <= 1000)
, (select level m from dual connect by level <= 1000)
where m<=l
group by l
having count(case l/m when trunc(l/m) then 'Y' end) = 2);

    Output will be

     
    2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197,199,211,223,227,229,233,239,241,251,257,263,269,271,277,281,283,293,307,311,313,317,331,337,347,349,353,359,367,373,379,383,389,397,401,409,419,421,431,433,439,443,449,457,461,463,467,479,487,491,499,503,509,521,523,541,547,557,563,569,571,577,587,593,599,601,607,613,617,619,631,641,643,647,653,659,661,673,677,683,691,701,709,719,727,733,739,743,751,757,761,769,773,787,797,809,811,821,823,827,829,839,853,857,859,863,877,881,883,887,907,911,919,929,937,941,947,953,967,971,977,983,991,997

    Hope you guys find my post useful, let me know how i can improve....
    Till then. Take care!!
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  7. Hey guys,
    Hope you all are doing great. I am back with another problem of Project Euler on

     Project Euler #8: Largest product in a series
    Find the greatest product of K consecutive digits in the N digit number.
    Input Format
    First line contains T that denotes the number of test cases.
    First line of each test case will contain two integers N & K.
    Second line of each test case will contain a N digit integer. 
    Output Format
    Print the required answer for each test case.
    Constraints
    1≤T≤100
    1≤K≤7
    K≤N≤1000
    Sample Input
    2
10 5
3675356291
10 5
2709360626
    Sample Output
    3150
0
    Algorithm:
    As easy can it get, this is a simple problem to directly multiply the sequence of number.
    Just we need to take care of the thing as the numbers can be very large, so handling them as character sequence is preferable.

    Now its the trick how you convert it into a code. As we already know in Python we have a reduce function
    Ex:
    reduce(sum, range(5))
    Gives an output as 10, as sum is a operator(predefined function in python) which sums up the numbers in range 0 to 4

    Similarly we can use mul from operators in python to reduce a subsequnce of k digits.

    In my below code i have used a simple concept of product, but you can try the above approach to solve it as it worked for me.

    And Presto!! Pythoned!!

    Program:

    from operator import mul
t=input()
while(t>0):
    n,m = map(int,raw_input().split())
    num=list(raw_input())
    ans=0
    for i in range(0,n-m):
        x=1
        for k in range(0,m):
            x=x*int(num[i+k])
        ans=max(ans,x)
    print ans
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  8. Hey guys,
    Hope you all are doing great. I am back with another problem of Project Euler

    Project Euler #2: Even Fibonacci numbers


    Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:
    1,2,3,5,8,13,21,34,55,89,⋯

    By considering the terms in the Fibonacci sequence whose values do not exceed N, find the sum of the even-valued terms.
    Input Format
    First line contains T that denotes the number of test cases. This is followed by T lines, each containing an integer, N.
    Output Format
    Print the required answer for each test case.
    Constraints
    1≤T≤105
    10≤N≤4×1016
    Sample Input
    2
10
100
    Sample Output
    10
44
    Algorithm:
    Fibonacci Series is one of the most fundamental series in programming world.

    To proceed, you just need to add 2 numbers just prior to your current number. Starting with 1 and 1 it becomes:
    1. 1
    2. 1
    3. 1+1 = 2
    4. 1+2 = 3
    5. 2+3 = 5
    6. and so on....
    Here python comes to our rescue to code it efficiently. As we need only even numbers, will have a condition as num%2 == 2

    Now for the main part, taking a and b as the cursors to our program,
    Initialize them with a=1, b=1
    Then as in Python you can assign INLINE a number of variables so our mail condition becomes
    a , b = a , a+b
    In this way we are handling both previous numbers and calculating the nest number concurrently.


    And Presto!! Its Done!!

    Just we need to convert that into a code, which is easy ion Python, Check the below code:
     Program:


    t=input();
while (t>0):
    t-=1
    n=input()
    a, b = 1, 1;
    total = 0;
    while a <= n:
        if a % 2 == 0:
            total += a
        a, b = b, a+b  # the real formula for Fibonacci sequence
    print total;
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