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