## Topcoder Problem archive – Creating a palindrome

0 / 1801 Problem Statement:

Source Topcoder – Palindrome and BitFlipper

Here are a couple of examples of the types of problems you can expect from a TopCoder competition. Once you become a TopCoder member, you can view all of the problems that TopCoder has ever used in the statistics section of our web site. If you’re looking to practice, you’ll find that after each TopCoder competition the problem sets that were used will be added as “practice rooms” in the TopCoder competition Arena. The TopCoder practice rooms are a great place to test your skills without the pressure of live competition and at the same time, get a feel for the competition environment. During a TopCoder competition, you’ll find that the problem set for division two rooms is easier then the one for division one rooms. All first-time competitors will be placed in a division two competition room. Once your TopCoder rating reaches 1200, you’ll be placed into the more difficult division one rooms. Below is an example of a problem statement from each of the two divisions.

Example Division Two Problem Statement
(From Single Round Match 96)

1) A palindrome is a number that is the same whether it is read from left-to-right or right-to-left. For example, 121 and 34543 are both palindromes. It turns out that nearly every integer can be transformed into a palindrome by reversing its digits and adding it to the original number. If that does not create a palindrome, add the reverse of the new number to itself. A palindrome is created by repeating the process of reversing the number and adding it to itself until the number is a palindrome.

Create a class Transform that contains the method palindrome, which takes a number N that is to be transformed and returns a number that is the resultant palindrome from this process. Of course if N is already a palindrome, return it without changing it. Though it is theorized that all numbers can be transformed to palindromes in this way, some numbers do not converge in a reasonable amount of time. For instance, 196 has been carried out to 26,000 digits without finding a palindrome. So if the method finds that the resultant palindrome must be greater than 1,000,000,000, return the special value -1 instead.

DEFINITION
Class: Transform
Method: palindrome
Parameters: int
Returns: int
Method signature (be sure your method is public): int palindrome(int N);

NOTES
– Leading zeroes are never considered part of a number when it is reversed. For instance, 12’s reverse will always be 21 regardless of whether it is represented as 12, 012, or 0012. Examples with leading zeroes use the leading zeroes for clarity only.

TopCoder will ensure the validity of the inputs. Inputs are valid if all of the following criteria are met:
– N will be between 1 and 10000 inclusive.

EXAMPLES
Worked examples:
Example 1: N = 28
28 + 82 = 110
110 + 011 = 121, a palindrome. Return 121

Example 2: N = 51
51 + 15 = 66, a palindrome. Return 66

Further examples:
Example 3: N = 11, return 11
Example 4: N = 607, return 4444
Example 5: N = 196, return -1

Example Division One Problem Statement

(From Single Round Match 102)

2) Binary strings, as most of us know, are composed solely of 0’s and 1’s. Sometimes it is necessary to turn all the bits either on (1) or off (0). However, sometimes it is not possible to just pick and flip individual bits. In this hypothetical scenario, it is only possible to flip bits in a contiguous segment which is a subset of the contiguous segment which was flipped immediately prior to it. For example, if bits 2-4 are flipped, it is not legal to flip bits 3-5 next, or bits 1-3. However, bits 3-4 or bits 2-3 would be legal. The first segment to be flipped can be located anywhere in the sequence.

Create a class BitFlipper which contains a method minFlip which determines the minimum number of bits that must be flipped under the above restriction in order to get all the bits set to 0. For purposes of this problem, to flip a bit means to change it from 0 to 1 or from 1 to 0.

DEFINITION

Class: BitFlipper
Method: minFlip
Parameters: String
Returns: int

Method signature (be sure it is declared public): int minFlip (String bits);

TopCoder will ensure the validity of the inputs. Inputs are valid if all of the following criteria are met:
* bits will be between 0 and 50 characters in length, inclusive
* bits will contain only 1’s and 0’s.

EXAMPLES

Example 1:
bits = “00110”.
By flipping bits 3-4, we get “00000”. Method returns 2.

Example 2:
bits = “10110”
If we flip bits 1-4, we get “01000”. Now we flip bit 2 and get “00000”.
Method returns 4 + 1 = 5.

Example 3:
bits = “1001110001”
Flipping bits 1-10 yields “0110001110”
Now, flipping bits 2-9 yields “0001110000”
Again, flipping bits 4-6 yields “0000000000”
Method returns 10 + 8 + 3 = 21.

Example 4:
bits = “10001”
Method returns 8.

Example 5:
bits = “101010101”
Method returns 25.

Example 6:
bits = “”
Method returns 0.

_____________________________________________________________________________________________________________________________________________________
Let’s try to solve the above 2 problems using Javascript so that they can be executed and tested directly on the browser

Below is the code for problem 1 (Palindrome):

Ouput of the above code:-

topcoder.html:56 11
topcoder.html:57 4444
topcoder.html:58 121
topcoder.html:59 66
topcoder.html:60 -1

Let’s dig deeper into this code.

Transform() loops through until number checks positive for palindrome or loop count exceeds the said limit (say 26,000 here).sumDigits() returns the sum of digits of given number. This is achieved by extracting digits one by one from number  by modulo 10 and diving number by 10 until it reaches 0.

reverseNumber() reverses the number passed to it as argument by extracting each digit using the modulo 10 technique and multiplying that number with equivalent powers of 10 to get the reverse number at the end.

Below is the code for problem  2 (BitFlipper):

Ouput of the above code:-

Number of flips required to flip 1001110001 to all 0’s : 21
Number of flips required to flip 101010101 to all 0’s : 25
Number of flips required to flip 0000000000111100000 to all 0’s : 4
Number of flips required to flip 0000 to all 0’s : 0
Number of flips required to flip ‘ ‘ to all 0’s : 0
Number of flips required to flip 10110 to all 0’s : 5
Number of flips required to flip 00110 to all 0’s : 2
Number of flips required to flip 0000111000 to all 0’s : 3
Number of flips required to flip 10001 to all 0’s : 8

Please comment if you have any doubts regarding the above algorithm and stay tuned for more interesting stuff.

-Team Golibrary  🙂

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