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Knuth–Morris–Pratt (KMP) Pattern Matching Substring Search - First Occurrence Of Substring
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Question: Given a string s and a pattern p, determine if the pattern exists in the string. Return the index of where the first occurrence starts.
The Brute Force
The naive approach to solving this is looking in s for the first character in p.
If a match is found we begin a search from that index, call it i (for intersect).
We compare the 2nd character of p with index i + 1 of s
We compare the 3rd character of p with index i + 2 of s
...and so on until we have matched to all of p without either having overrun s or having found a mismatch between characters being compared.
We can then return i as our answer.
It doesn’t work well in cases where we see many matching characters followed by a mismatching character.
Complexities:
Time: O( len(s) * len(p) )
In a simple worst case we can have
s = "aaaaaab"
p = "aaab"
The problem is that for each first character match we have the potential to naively go into a search on a string that would never yield a correct answer repeatedly.
Other Algorithms
There are three linear time string matching algorithms: KMP (nuth–Morris–Pratt), Boyer-Moore, and Rabin-Karp.
Of these, Rabin-Karp is by far the simplest to understand and implement
Analysis
The time complexity of the KMP algorithm is O(len(s) + len(p)) "linear" in the worst case.
The key behind KMP is that it takes advantage of the succesful character checks during an unsuccessful pattern comparison subroutine.
We may have a series of many comparisons that succeed and then even if one fails at the end, we should not repeat the comparison work done since we already saw that a series matched.
What we will do is very similar to the naive algorithm, it is just that we save comparisons by tracking the longest propert prefixes of pattern that are also suffixes.
The key is that everytime we have a mismatch we try our best to prevent going backwards in s and repeating comparisons.
Algorithm Steps
We will preprocess the pattern string and create an array that indicates the longest proper prefix which is also suffix at each point in the pattern string.
A proper prefix does not include the original string.
For example, prefixes of “ABC” are “”, “A”, “AB” and “ABC”. Proper prefixes are “”, “A” and “AB”.
For example, suffixes of "ABC" are, "", "C", "BC", and "ABC". Proper prefixes are "", "C", and "BC".
Why do we care about these??
We know all characters behind our mismatch character match.
If we can find the length of the longest prefix that matches a suffix to that point, we can skip len(prefix) comparisons at the beginning.
The key reason we care about the prefix to suffix is because we want to "teleport" back to as early in the string to the point that we still know that there is a match.
Our goal is to minimize going backwards in our search string.
Complexities:
Time: O( len(p) + len(s) )
We spend len(p) time to build the prefix-suffix table and we spend len(s) time for the traversal on average.
Space: O( len(p) )
Our prefix-suffix table is going to be the length of the pattern string.
++++++++++++++++++++++++++++++++++++++++++++++++++
++++++++++++++++++++++++++++++++++++++++++++++++++
This question is number 7.13 in "Elements of Programming Interviews" by Adnan Aziz (they do Rabin-Karp but same problem, different algorithm)
🏃 Run Code As You Learn
💾 Save Progress
❓New Unseen Questions
🔎 Get All Solutions
Question: Given a string s and a pattern p, determine if the pattern exists in the string. Return the index of where the first occurrence starts.
The Brute Force
The naive approach to solving this is looking in s for the first character in p.
If a match is found we begin a search from that index, call it i (for intersect).
We compare the 2nd character of p with index i + 1 of s
We compare the 3rd character of p with index i + 2 of s
...and so on until we have matched to all of p without either having overrun s or having found a mismatch between characters being compared.
We can then return i as our answer.
It doesn’t work well in cases where we see many matching characters followed by a mismatching character.
Complexities:
Time: O( len(s) * len(p) )
In a simple worst case we can have
s = "aaaaaab"
p = "aaab"
The problem is that for each first character match we have the potential to naively go into a search on a string that would never yield a correct answer repeatedly.
Other Algorithms
There are three linear time string matching algorithms: KMP (nuth–Morris–Pratt), Boyer-Moore, and Rabin-Karp.
Of these, Rabin-Karp is by far the simplest to understand and implement
Analysis
The time complexity of the KMP algorithm is O(len(s) + len(p)) "linear" in the worst case.
The key behind KMP is that it takes advantage of the succesful character checks during an unsuccessful pattern comparison subroutine.
We may have a series of many comparisons that succeed and then even if one fails at the end, we should not repeat the comparison work done since we already saw that a series matched.
What we will do is very similar to the naive algorithm, it is just that we save comparisons by tracking the longest propert prefixes of pattern that are also suffixes.
The key is that everytime we have a mismatch we try our best to prevent going backwards in s and repeating comparisons.
Algorithm Steps
We will preprocess the pattern string and create an array that indicates the longest proper prefix which is also suffix at each point in the pattern string.
A proper prefix does not include the original string.
For example, prefixes of “ABC” are “”, “A”, “AB” and “ABC”. Proper prefixes are “”, “A” and “AB”.
For example, suffixes of "ABC" are, "", "C", "BC", and "ABC". Proper prefixes are "", "C", and "BC".
Why do we care about these??
We know all characters behind our mismatch character match.
If we can find the length of the longest prefix that matches a suffix to that point, we can skip len(prefix) comparisons at the beginning.
The key reason we care about the prefix to suffix is because we want to "teleport" back to as early in the string to the point that we still know that there is a match.
Our goal is to minimize going backwards in our search string.
Complexities:
Time: O( len(p) + len(s) )
We spend len(p) time to build the prefix-suffix table and we spend len(s) time for the traversal on average.
Space: O( len(p) )
Our prefix-suffix table is going to be the length of the pattern string.
++++++++++++++++++++++++++++++++++++++++++++++++++
++++++++++++++++++++++++++++++++++++++++++++++++++
This question is number 7.13 in "Elements of Programming Interviews" by Adnan Aziz (they do Rabin-Karp but same problem, different algorithm)
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