Longest matching pattern

Given two strings find the longest matching pattern such that the pattern in the first string and the second string are permutation of each other.

Eg. - 
S1 = “ABCDEFG”
S2 = “DBCAPFG”
Then the output should be DBCA (length 4)
Assume there are no repeating characters.

Solution -
The idea is to transform S1 into clusters of indices of S2. So, for the example above we will get
Cluster 1 - 3, 1, 2, 0
Cluster 2 - 5, 6

Now, for each cluster find the longest portion (window) that satisfies the following property - 

• Max - Min + 1 = No of Elements

The longest portion (window) of the cluster will represent the indices of S2 that are contiguous and since the cluster itself is a a contiguous portion of transformed S1, the final solution will be the characters in S2 at the indices represented by such longest portion.

To find the longest portion satisfying the above property, we can recursively implement it as shown below in method getMaxCommonWindow. The idea is to find the maximum window starting with the first element satisfying the above property and then recusively find the maximum window not starting with the first element. The greater of these two windows will be the desired longest portion.

String rotation

Given a string of length N, rotate the string by K units. Eg. rotation of abcdef by 2 units will result in cdefab.

Solution approaches:-
1) Using a temp Array
2) 1 shift at a time 
3) Juggling algorithm - uses GCD(N,K) outer loop and inner loop till cycle is encountered
4) Gries algorithm - divide and conquer
5) Reversal algorithm - revert individual blocks and then revert the full string
 
Links:-
Source - http://www.cs.bell-labs.com/cm/cs/pearls/s02b.pdf
Solution - http://www.geeksforgeeks.org/archives/2398
Solution - http://www.geeksforgeeks.org/archives/2878
Solution - http://www.geeksforgeeks.org/archives/2838
Solution and Performance - http://www.drdobbs.com/article/print?articleId=232900395&siteSectionName=parallel
 
Juggling algorithm explanation:-
Taken from http://eli.thegreenplace.net/2008/08/29/space-efficient-list-rotation/

Now, are you asking yourself "Just when will the process come back to x[0], and how many elements will have been moved by that stage ?". So did I, and the answer turns out to be an exciting application of the greatest common divisor (GCD) function.

In the first iteration, the "juggling pointer" jumps i elements at a time and stops when it reaches 0. How many steps will this take ? Ignoring the modulo for a moment, To reach 0, the pointer must be a multiple of n, so 0 will be reached at an index that is a multiple of both i and n. The first such multiple, in fact.

The first multiple (also known as LCM – least common multiple) is easy to compute from the GCD.

The amount of steps is lcm(n, i) / i. This is n / gcd(n, i). Therefore, in each iteration, n / gcd(n, i) elements will be rotated. The next iteration will pick up at x[1], an keep hopping in steps of i from there, moving another n / gcd(n, i) elements. In special cases, like when n and i are coprime, the first iteration will run over all the elements in the vector, without the need for a second one.

In any case, the whole process will always make n steps in total, moving all the elements to their correct positions.

Repeating substring

Q: Given a string S find the smallest sub-string X such that S = (X)+, i.e. S is a repetition of one or more X's.

Soln approach: 

• Iterate string from start to end keeping 2 pointers pointing to certain indexes in the string
• The first one indicating endIndex of the sub-sequence X found till now
• The second one indicating the index with which we must compare the character read in the next iteration
• Update both the indexes in each iteration based on whether the character read matches the sub-sequence 
• At the end X will be the sub-string from 0 to the first index mentioned in first point above

Java code: