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# Z-algorithm
# Z Algorithm
The Z-algorithm finds occurrences of a "word" `W`
within a main "text string" `T` in linear time.
within a main "text string" `T` in linear time `O(|W| + |T|)`.
Given a string `S` of length `n`, the algorithm produces
an array, `Z` where `Z[i]` represents the ongest substring
an array, `Z` where `Z[i]` represents the longest substring
starting from `S[i]` which is also a prefix of `S`. Finding
`Z` for the string obtained by concatenating the word, `W`
with a nonce character, say `$` followed by the text, `T`,
@ -12,7 +12,7 @@ helps with pattern matching, for if there is some index `i`
such that `Z[i]` equals the pattern length, then the pattern
must be present at that point.
While the `Z` array can be computed with two nested loops, the
While the `Z` array can be computed with two nested loops in `O(|W| * |T|)` time, the
following strategy shows how to obtain it in linear time, based
on the idea that as we iterate over the letters in the string
(index `i` from `1` to `n-1`), we maintain an interval `[L,R]`
@ -21,7 +21,42 @@ and `S[L...R]` is a prefix that is also a substring (if no such
interval exists, just let `L=R=-1`). For `i=1`, we can
simply compute `L` and `R` by comparing `S[0...]` to `S[1...]`.
**Example of Z array**
```
Index 0 1 2 3 4 5 6 7 8 9 10 11
Text a a b c a a b x a a a z
Z values X 1 0 0 3 1 0 0 2 2 1 0
```
Other examples
```
str = a a a a a a
Z[] = x 5 4 3 2 1
```
```
str = a a b a a c d
Z[] = x 1 0 2 1 0 0
```
```
str = a b a b a b a b
Z[] = x 0 6 0 4 0 2 0
```
**Example of Z box**
![z-box](https://ivanyu.me/wp-content/uploads/2014/09/zalg1.png)
## Complexity
- **Time:** `O(|W| + |T|)`
- **Space:** `O(|W|)`
## References
- [GeeksForGeeks](https://www.geeksforgeeks.org/z-algorithm-linear-time-pattern-searching-algorithm/)
- [YouTube](https://www.youtube.com/watch?v=CpZh4eF8QBw&t=0s&list=PLLXdhg_r2hKA7DPDsunoDZ-Z769jWn4R8&index=70)
- [Z Algorithm by Ivan Yurchenko](https://ivanyu.me/blog/2013/10/15/z-algorithm/)