8/18/2019 Coding Problem

1/2

In its effort to push the limits of file compression, Dropbox recently developed a lossless compression algorithm for H.264 and JPEG files. Since you are thinking about applying for a job at Dropbox, you decided to experiment with simple lossless compression as part of your interview prep.

One of the most widely known approaches in the field of compression algorithms is sliding window compression. It works as follows:

Consider characters one by one. Let the current character index be i.Take the last width characters before the current one (i.e. s[i - width, i - 1], where s[i, j] means the substring of s from index i to index j, both inclusive), and call it the window.Find such startIndex and length that s[i, i + length - 1] = s[startIndex, startIndex + length - 1] and s[startIndex, startIndex + length - 1] is contained within the window. If there are several such pairs, choose the one with the largest length. If there still remains more than one option, choose the one with the smallest startIndex.If the search was successful, append "(startIndex,length)" to the result and move to the index i + length.

Otherwise, append the current character to the result and move on to the next one.Given a string, apply sliding window compression to it.

Example

For inputString = "abacabadabacaba" and width = 7the answer is

losslessDataCompression(inputString, width) = "ab(0,1)c(0,3)d(4,3)c(8,3)".

Step 1: i = 0, inputString[i] = 'a', window = "". 'a' is not contained within the window, so it is appended to the result.

Step 2: i = 1, inputString[i] = 'b', window = "a". 'b' is not contained within the window, so it is appended to the result.Step 3: i = 2, inputString[i] = 'a', window = "ab". 'a' can be found in the window. 'a' in the window corresponds to the inputString[0], so (0,1) representing the substring "a" is appended to the result.Step 4: i = 3, inputString[i] = 'c', window = "aba". The same situation as in the first two steps.Step 5: i = 4, inputString[i] = 'a', window = "abac". Consider startIndex = 0, length = 3. inputString[startIndex, startIndex + length - 1] = "aba" and it is contained within the window, inputString[i, i + length - 1] = "aba". Therefore, "(0,3)" should be added to the result. i += length.

Step 6: i = 7, inputString[i] = 'd', window = inputString[0, 6] = "abacaba". The same situation as in the first two steps.Step 7: i = 8, inputString[i] = 'a', window = inputString[1, 7] = "bacabad". Consider length = 3 again. inputString[i, i + b - 1] = "aba", window[3, 5] = "aba", and it corresponds to inputString[4, 6] since inputString[0, 2] is no longer within the window. So, "(4,3)" should be appended. i += length.Step 8: i = 11, inputString[i] = 'c', window = "abadaba". The same situation asat the first two steps.Step 9: i = 12, inputString[i] = 'a', window = "badabac". length = 3, inputString[i, i + length - 1] = "aba", window[3, 5] = "aba", and it corresponds to inputString[8, 10]. So, "(8,3)" should be appended. i += length.

For inputString = "abacabadabacaba" and width = 8

the answer is

losslessDataCompression(inputString, width) = "ab(0,1)c(0,3)d(0,7)".

8/18/2019 Coding Problem

2/2

In both of the above examples the resulting "compressed" string becomes even longer than the initial one. In fact, sliding window compression proves to be efficient for longer inputs. E.g.:

For inputString = "aaaaaaaaaaaaaaaaaaaaaaaaaaaa" and width = 12the answer is

losslessDataCompression(inputString, width) = "a(0,1)(0,2)(0,4)(0,8)(4,12)".In the last example the resulting string is one character shorter than the input one. It is the shortest possible example of the efficient work of sliding window compression. If the input contained even more letters 'a', then the effect ofthis approach would be even more considerable.

[input] string inputString

A non-empty string of lowercase letters.[input] integer width

A positive integer.[output] string

Compressed inputString.