String Searching by KMP algorithm (Knuth Morris Pratt algorithm)
Motivation Problem: Given $$2$$ strings $$P$$(pattern) and $$T$$(text), find the number of occurrences of $$P$$ in $$T$$.
Basic / Brute Force solution:
One obvious and easy to code solution that comes to mind is this: For each index of $$T$$, take it as a starting point and find if $$T_{i,i+1,...,i+|P|-1}$$ is equal to $$P$$.
for i = 0 to length(T)-length(P):
Found = true
for j = i to i + length(P) - 1:
if P[j-i] not equal to T[j]:
Found = False
if Found = True
answer = answer + 1
This brute force takes $$O(|P| \cdot |T|)$$ time in the worst case, which is obviously too slow for large strings.
Knuth Morris Pratt Algorithm:
Suppose for each index $$i$$ of some string $$Z$$, the longest suffix in $$Z_{0,1,...,i}$$ that is also a prefix of $$Z_{0,1,...,i}$$, be known. Formally, a length $$F_i$$ is known such that $$Z_{0,1,...,F_i-1}$$ = $$Z_{i-F_i+1,...,i}$$. Let these lengths be stored in array $$F$$. The suffix needs to be proper(whole string is not a proper suffix).
Then the solution to the motivation problem can be found as follows:
Define a string $$V = P + '#' + T$$, where $$'#'$$ is a delimiter that is not present in either of $$P$$ or $$T$$. Now, if the above information is known, all occurrences of $$P$$ in $$T$$ can be found as follows: If at some index $$i$$, $$F_i = |P|$$, then there is an occurrence of Pattern $$P$$ at position $$i-|P|+1$$. All such indices from $$|P|+1$$ [0 based indexing, the index just after '#'], need to be checked.
The main part of KMP algorithm calculates the array $$F$$, which is also called the prefix function. If calculation of $$F$$ or the prefix function can be done efficiently, then we have an efficient solution to the motivation problem. KMP algorithm finds the prefix function in $$O(length of String)$$ time.
To find the prefix function, best possible use of previous values of array $$F$$ is made, so that calculations aren't done again and again. Suppose all $$F_i$$ have been calculated, and now $$F_{i+1}$$ is to be calculated. It is to be noted that, value of $$F_{i+1}$$ can be at most 1 greater than $$F_i$$. Here is a proof by contradiction:
Suppose $$F_{i+1} > F_i + 1$$. Now, if the $$(i+1)^{th}$$ character is removed, we obtain a suffix ending at index $$i$$ that is of length $$F_{i+1} - 1$$, which is greater than $$F_i$$. This is a contradiction, hence proved.
Observe that if $$Z_{i+1} = Z_{F_i}$$, then the value of $$F_{i+1} = F_i + 1$$. If not, a smaller suffix ending at index $$i$$ is to be found, that is also a prefix of $$Z_{0,1...,i}$$. Let the length of such a suffix be $$j$$, then if $$Z_{i+1} = Z_{j}$$ then $$F_{i+1} = j + 1$$. If again the equality doesn't hold true, smaller and smaller suffixes that end at index $$i$$, which are also prefixes of $$Z_{0,1...,i}$$ need to be found.
The only thing remaining is, how to find the length of next smaller suffix ending at index $$i$$, that is also a prefix? This is also pretty simple. Observe that due to the property of $$F$$, the segment $$Z_{0,1,...,F_i-1}$$ is equal to the segment $$Z_{i-F_i+1,...,i}$$. So to find the next smaller suffix ending at index $$i$$, the longest suffix ending at $$F_i - 1$$ can be found which is $$F_{F_i-1}$$, and this suffix will be the next smaller suffix ending at index $$i$$. If this suffix also doesn't satisfy our criteria, then smaller suffixes can be found with the same process, here it will be $$F_{F_{F_i-1} - 1}$$. Note that, if at some point the length becomes $$0$$, the process is stopped.
This completes $$KMP$$ algorithm. Below is the code:
vector<int> prefix_function (string Z) {
int n = (int) Z.length();
vector<int> F (n);
F[0]=0;
for (int i=1; i<n; ++i) {
int j = F[i-1];
while (j > 0 && Z[i] != Z[j])
j = F[j-1];
if (Z[i] == Z[j]) ++j;
F[i] = j;
}
return F;
}
Finding the $$F$$ array for "ABCABC"
Initially, $$F_0 = 0$$.
Index $$1 \rightarrow F_{0} = 0$$, $$j$$ does not go into while loop and $$Z_j \neq Z_i$$, therefore value of $$F_i = 0$$.
Index $$2 \rightarrow F_{1} = 0$$, $$j$$ does not go into while loop and $$Z_j \neq Z_i$$, therefore value of $$F_i = 0$$.
Index $$3 \rightarrow F_{2} = 0$$, $$j$$ does not go into while loop and $$Z_j = Z_i$$, therefore value of $$F_i = 1$$.
Index $$4 \rightarrow F_{3} = 1$$, $$j$$ satisfies while loop condition but $$Z_j = Z_i$$, hence does not go into while loop, therefore value of $$F_i = 2$$.
Index $$5 \rightarrow F_{4} = 2$$, $$j$$ satisfies while loop condition but $$Z_j = Z_i$$, therefore value of $$F_i = 3$$.
