Details
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Improvement
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Status: Closed
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Major
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Resolution: Fixed
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3.4
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Patch
Description
Before reading the description below, please note that there was an issue created in 10/Jul/2011 titled "CharSetUtils.squeeze() speedup" LANG-715. This solution does not prevent the repetitive calls to the expensive contains() method explained below. In addition, I updated this to (major) because of the high CPU cost of the current solution.
A readable form of the description is attached in Description_Squeeze.docx.
Below is a text format of the description (in case the .docx does not work):
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Performance problems of current CharSetUtils.squeeze() algorithm:
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FIRST PROBLEM:
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-Method org.apache.commons.lang3.CharSet.contains() is called in each character iteration for a consecutive set of duplicated characters in str (starting from the second character in the duplicated set). This large number of calls to contains() are determined to increase the performance overhead for squeeze() calls given the high computing cost of executing contains() method. Note that only one call to contains() is needed for the second character in this set of duplicated characters. There is no need to repeat the calls to contains() method for the same duplicated character that we have already checked in the second occurrence. For example, if str=”abbbbbc” and set=”b”, then the algorithm should call contains() only once for the second b character occurrence and prevent calling contains() again given that it has already been called before for this block. The current algorithm adds high performance overhead by calling contains() repetitively for each duplicated b character starting from the second occurrence. For this example where str=”abbbbbc” and set=”b”, the suggested solution will call contains() method only once instead of 4 times which will largely reduce the performance overhead of squeeze() method.
-Consider the following example, System.currentTimeMillis() testing shows a performance improvement by around 83% if we run the suggested enhanced solution with the following sample input data
str = “abbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbc”
set = “b”
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PROPOSED SOLUTION FOR FIRST PROBLEM:
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-Using two character flags inChars and notInChars:
inChars: flag to record the recent character that has been determined to be in chars CharSet
notInChars: flag to record the recent character that has been determined to NOT be in chars CharSet
-Adding the following conditions to prevent unnecessary/expensive calls to contains() method. Only one call to contains() method is needed starting from the second character in the consecutive set of duplicated characters:
if ((inChars != null) && (ch == inChars))
this if-statement checks if this ch char has recently been determined to be in chars CharSet, this prevents calling contains() more than once for a consecutive set of duplicated characters in str
if one of these conditions fails, it means that the consecutive set of duplicated characters for this ch character has not yet been checked for this block of duplicated characters (i.e. has not been checked = we have not yet done the one-time call to contains method for this particular set of duplicated characters)
if ((notInChars == null) || (ch != notInChars))
if both conditions are false, it means that ch has recently been determined to NOT be in chars CharSet
if (chars.contains(ch))
here we call the expensive chars.contains() only once for this consecutive set of duplicated characters.
-Note: this suggested solution does not break the functionality of org.apache.commons.lang3.CharSetUtils.squeeze(String str, String… set). This suggested solution is executed against the latest UnitTest CharSetUtilsTest.java written by Apache.
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SECOND PROBLEM:
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the current algorithm starts the for-loop with i=0 which is not a necessary step. This adds the overhead of executing the if condition (i != 0). In this logic, the algorithm will always append the first character from str to the buffer. So there is no need to execute (i != 0) for every character in the consecutive set of duplicated characters.
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PROPOSED SOLUTION FOR SECOND PROBLEM:
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lastChar is assigned the first character in str and appended to buffer before executing the for loop. Then the for loop begins with i=1 removing the unnecessary (i != 0) executions.
END OF DESCRIPTION
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