Roadway to Ubuntu

This post is neither a solution nor a suggestion. This is what I experienced during installation of ubuntu in my system.

    Since the first day I used a computer, I was using MS Windows as default and primary operating system. Though, its important for a student to cope with different operating systems. During this time, I have used different Unix/Linux based live OS. Deciding which OS I should use is pretty tough decision because there are bunch of choices and all of those are free. :P
    My first attempt to install Linux was on a AMD A55 machine. It was a failure because of hardware configuration. ( I have tried it with both 32bit and 64bit different OSs. )
Then I got my laptop, an Intel system with pre-installed Windows 7 in it. Well, I removed it with Windows 8 from DreamSpark account at 0.00$ cost. :D Living together with Windows 8 was quiet fun. xD However, when I took Operating System course at University, it was a necessity having a Linux OS in my system. But, I failed again. Ubuntu live OS ran fine, installation started every time, but it could not modify the karnel about installation! I tried with Fedora, OpenSuse, Ubuntu, BackTrack and LinuxMint. Each time it started fine, but ends with error. Error! Error!! Error!!! ( Now I have figured it why.. Windows hibernate/hybrid-shutdown option was not allowing any other OS to modify the storage. Later I turned it off using the following command. )

powercfg /h off

    Enough installing Linux for me! So, I kept it cool with Windows.
Meanwhile, my pc was attacked by SIREFEF virus. It turned off, turned off(!) Windows Defender permanently and converted system directories into junctions. Thats why I could not turn Windows Defender on. It ate all the website certificates on all browses. Whenever I turned on the pc, a VB script was creating executable file and some null processes on background. I got rid off this virus using HitmanPro.
    Another thing this virus did was making disk partitions undetectable in disk management tool or windows setup. So, I was thinking on a new set up. Why not Linux now? Yes, it was time for Ubuntu to take all the disk spaces. :D This time I was stuck with, UEFI abd Legacy Mode. Certainly I wanted Ubuntu in UEFI mode but did not find any option for Windows but Legacy mode. Oh, I surely wanted a duel boot.
    I created partitions, installed Ubuntu. Then another problem did arise! Windows could not format GPT partitions. I erased the whole disk using diskpart.

diskpart
list disk
select disk *
clean
convert gpt
exit

    Well, windows was installed. Finally, I installed Ubuntu by formatting unallocated spaces. Installation Finish! Restart!! Windows Boot!!!
    Ah, the MBR, windows boot manager. Again I inserted Ubuntu live USB to install GRUB.

sudo add-apt-repository ppa:yannubuntu/boot-repair && sudo apt-get update
sudo apt-get install -y boot-repair
sudo boot-repair

    At last... Got my dual boot system with Ubuntu 12.04.2 and Windows 8. Decided to use Ubuntu as primary and Windows as supporting OS.
Finally, Enjoying Ubuntu!!

Maximum Interval Product

Previously, I have posted about maximum interval sum. Now, unlike summation, product of an interval or segment is bit different.
Well, the problem is, from a given array of integers we have to determine the maximum product which can be found by any segment of the given array of integers.
Example:
Let, we have an array of these integers:
1 -2 4 -3 -1
So, the maximum interval product is 24 from 1 to -3 ( 1*-2*4*-3 ).
1 -2 4 0 -3 -1
Here, the maximum interval product is 4 from only 4.
Iteration:

	initial	1	-2	4	-3	-1
mp	inf	1	inf	4	24	12
mn	inf	inf	-2	-8	-12	-24
mxProduct	-1	1	1	4	24	24

Code:

// Maximum Interval Product
// Complexity: O(n)
long long mxProduct  = -1;
// maximum product is stored. if unchanged, no valid product.
long long t1, t2;
long long mp = INF, mn = INF;
// two counters for maxPositiveProduct and maxNegativeProduct
for( i=0; i<N; i++ ) {
    scanf( "%d", &arr[i] );

    if( arr[i] == 0 ) {
        if( mp != INF ) mxProduct = MAX( mxProduct, mp );
        mp = mn = INF;
        continue;
    }

    if( mp==INF && mn==INF ) {
        arr[i] > 0 ? mp = arr[i] : mn = arr[i];
        if( mp != INF ) mxProduct = MAX( mxProduct, mp );
        continue;
    }

    if( mp == INF ) {
        if( arr[i] > 0 ) {
            mp = arr[i];
            mn = arr[i] * mn;
            mxProduct = MAX( mxProduct, mp );
            continue;
        }
        if( arr[i] < 0 ) {
            mp = arr[i] * mn;
            if( arr[i] >= mn ) mn = arr[i];
            else mn = INF;
            mxProduct = MAX( mxProduct, mp );
            continue;
        }
    }

    if( mn == INF ) {
        if( arr[i] < 0 ) {
            mn = arr[i] * mp;
            mp = INF;
            continue;
        }
        if( arr[i] > 0 ) {
            mp = arr[i] * mp;
            mn = INF;
            mxProduct = MAX( mxProduct, mp );
            continue;
        }
    }

    t1 = arr[i] * mp;
    t2 = arr[i] * mn;
    if( arr[i] > 0 ) mp = t1, mn = t2;
    if( arr[i] < 0 ) mp = t2, mn = t1;
    mxProduct = MAX( mxProduct, mp );
}

Source:
http://acm.uva.es/board/viewtopic.php?f=33&t=11466&hilit=11059&sid=6ed77239e43f67e7ca9999e1412b0180

Step Ahead in JAVA

Hello everyone, this is my solution for "Life, the Universe, and Everyting" in SPOJ. It was my first submission in JAVA. Take here a look.

/*
Task: Life, the Universe, and Everything (SPOJ)
Code: TEST
*/

import java.io.*;   // java input\output library 
import java.util.*; // using for Scanner class

class LifeSPOJ {
    public static void main ( String [] args ) {
        Scanner ss = new Scanner(System.in);
        int num;
        num = ss.nextInt(); // nest intejar
        while( num != 42 ) {
            System.out.println( num ); // printing that number
            num = ss.nextInt();
        }
    }
}

Longest Palindromic Subsequence or Maximum Length of Palindrome

Definition:
              A sequence is called "Palindrome" if it is read same left to right and right to left. The sequence remains same as before after reversing it. The sequence "MADAM" is a Palindrome.
More about Palindrome.

A sequence which is a Palindrome is called a Palindromic Sequence. In this problem, given any sequence (not necessarily palindromic sequence), and we have to determine the Longest Palindromic Sub-sequence of Maximum Length of Palindrome of it.
A dynamic approach is not mandatory for this problem, because there is no overlapping sub-problems and so no optimal sub-structure available for Longest Palindromic Sub-sequence.
But if we go recursively, it can be easily determined. How? Obviously there is no overlapping sub-problems but we can divide it into sub-problems (that how we do it :p).
Here is a C++ solution for LPS.

#include <iostream>
#include <string>

using namespace std;

template<class T>inline T MAX( T a, T b ){ return a>b ? a:b; } // Note: returns maximum of a and b

string A;

int LPS( int i, int j ) {
    if( i == j ) return 1; // A[i] & A[j] are same, cost is 1
    if( i > j ) return 0;
    if( A[i] == A[j] )
        return 2 + LPS( i+1, j-1 ); // A[i] & A[j] are different, cost is 2, 1 for each.
    else
        return MAX( LPS(i+1, j), LPS(i, j-1) );
}

int main( ){
    while( cin >> A ){
        int len = A.length( );
        int res = LPS(0, len);
        cout << "Longest Palindromic Subsequence is: " << res << endl;
    }
    return 0;
}

Obviously, a memoization make the procedure lot more faster. Here is a memoized faster version.

#include <iostream>
#include <string>
#include <cstring>

using namespace std;

template<class T>inline T MAX( T a, T b ){ return a>b ? a:b; } // Note: returns maximum of a and b
const int SIZ = 105;

#define mset2d(x,n) memset(x,n,sizeof(int [SIZ][SIZ]))

string A;
int memo[SIZ][SIZ];

int LPS( int i, int j ) {
    if( i == j ) return 1; // A[i] & A[j] are same, cost is 1
    if( i > j ) return 0;
    if( memo[i][j] != -1 ) return memo[i][j];
    if( A[i] == A[j] )
        memo[i][j] = 2 + LPS( i+1, j-1 ); // A[i] & A[j] are different, cost is 2, 1 for each.
    else
        memo[i][j] = MAX( LPS(i+1, j), LPS(i, j-1) );
    return memo[i][j];
}

int main( ){
    while( cin >> A ){
        mset2d( memo, -1 );
        int len = A.length( );
        int res = LPS(0, len);
        cout << "Longest Palindromic Subsequence is: " << res << endl;
    }
    return 0;
}

Addition, Subtraction & Multiplication in .ASM

Addition, Subtraction & Multiplication of two numbers submitted, until the program terminates by user. The termination process is actually maintained by 0 and 1, 0 terminates the program and 1 continues.
Limitations of the code is the initial two numbers and the resultant numbers must be between 0 to 9. "emu8086" compiler was used. This is my 5th code in assembly language I have done in my lab.

; Date: October 10, 2012
; Author: Md. Nabid Imteaj
; Tag: Add, Substract, Multiplication as user want

TITLE ANOTHER_DAY_IN_CLASS  
.model small
.stack 100h
.code

.data
msg1 db 'ENTER Hex Value: $'
msg2 db 'Equivalent Decimal: $' 
msg3 db 'INVALID :($'
msg10 db '0 to exit, 1 to continue?: $'
msg11 db 'THANK YOU :D$' 

main proc
    
    mov ax, @data
    mov ds, ax 
    
    mov cx, 3

AGAIN_PAIN: 
; new line
    mov ah, 2
    mov dl, 0dh
    int 21h
    mov dl, 0ah
    int 21h 

cmp cl, 0
je TERMINATE
dec cx

; message print

    lea dx, msg1
    mov ah, 9
    int 21h
    
; enter character
    mov ah, 1
    int 21h 
    mov bl, al
    
    ; new line
    mov ah, 2
    mov dl, 0dh
    int 21h
    mov dl, 0ah
    int 21h
    
; cmpareing
    cmp bl, 'A'
    je print_A
    cmp bl, 'B'
    je print_B 
    cmp bl, 'C'
    je print_C
    cmp bl, 'D'
    je print_D
    cmp bl, 'E'
    je print_E
    cmp bl, 'F'
    je print_F 
    
    cmp bl, 30
    jge over_0
    
over_0:
    cmp bl, 39
    jle less_9 
    jmp print_invalid

less_9:
    mov ah, 2
    mov dl, bl
    int 21h 
    jmp EXIT_ 
    
print_A:  
lea dx, msg2
mov ah, 9
int 21h
   
    mov ah, 2
    mov dl, '1'
    int 21h
    mov ah, 2
    mov dl, '0'
    int 21h 
    jmp EXIT_ 
    
print_B:
lea dx, msg2
mov ah, 9
int 21h 
    mov ah, 2
    mov dl, '1'
    int 21h
    mov ah, 2
    mov dl, '1'
    int 21h 
    jmp EXIT_
    
print_C:
lea dx, msg2
mov ah, 9
int 21h 
    mov ah, 2
    mov dl, '1'
    int 21h 
    mov ah, 2
    mov dl, '2'
    int 21h 
    jmp EXIT_
    
print_D:
lea dx, msg2
mov ah, 9
int 21h 
    mov ah, 2
    mov dl, '1'
    int 21h 
    mov ah, 2
    mov dl, '3'
    int 21h 
    jmp EXIT_
    
print_E:
lea dx, msg2
mov ah, 9
int 21h 
    mov ah, 2
    mov dl, '1'
    int 21h
    mov ah, 2
    mov dl, '4'
    int 21h  
    jmp EXIT_     
    
print_F:
lea dx, msg2
mov ah, 9
int 21h 
    mov ah, 2
    mov dl, '1'
    int 21h
    mov ah, 2
    mov dl, '5'
    int 21h  
    jmp EXIT_ 
           
print_invalid:
    lea dx, msg3
    mov ah, 9
    int 21h
    jmp EXIT_
               
  
EXIT_: 

; new line
    mov ah, 2
    mov dl, 0dh
    int 21h
    mov dl, 0ah
    int 21h    
    
; operation message
    lea dx, msg10
    mov ah, 9
    int 21h 
    
; gets character
    mov bl, '1'
    mov ah, 1
    int 21h
    
    cmp al, bl
    je AGAIN_PAIN
    
    ; new line
    mov ah, 2
    mov dl, 0dh
    int 21h
    mov dl, 0ah
    int 21h 
    
TERMINATE:   
     ; operation message
    lea dx, msg11
    mov ah, 9
    int 21h 

mov ah, 4ch
int 21h
main endp
end main