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Aakash Panchal
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articles/Akash Articles/md/Z-algorithm.md
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articles/Akash Articles/md/Z-algorithm.md
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Z-function and z-algorithm
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# Z-Algorithm
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Z-function for a given string $s$ of length $n$ is an array of length $n$, where $z[i]$ represents length of longest common prefix of string $s$ and suffix of s starting at $i$ i.e. $s[i,n-1]$.
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**Note:** $s[l,r]$ represents substring of $S$ starting at index $l$ and ending at index $r$. Here, we are taking zero based indices.
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Note that value of $z[0]$ is not properly defined so we take it as zero($0$).
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For example,
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1. $z("cccc") = [0,3,2,1]$
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Why $z[1]=3$?
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Because $s[0,2] = s[1,3] = "ccc"$.
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2. $z("ababab")=[0,0,4,0,2,0]$
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Why z[2] = 4?
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Because $s[0,3] = s[2,5] = "abab"$.
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3. $z("abacaba") = [0,0,1,0,3,0,1]$
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Why z[4] = 3?
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Because $s[0,2] = s[4,6] = "aba"$.
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Can you figure out how do we find value of z-function?
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## Trivial Algorithm
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Basic way to find value of z-function is to do brute force. For index - $i$, we find it following way.
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```
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z[i] = 0;
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while(i + z[i] < n && s[z[i]] == s[i + z[i]])
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z[i]++;
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```
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Simply, do this for every indices.
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```cpp
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vector<int> z_function(string s) {
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int n = (int) s.size();
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vector<int> z(n);
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for (int i = 1; i < n; ++i)
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while (i + z[i] < n && s[z[i]] == s[i + z[i]])
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z[i]++;
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return z;
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}
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```
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## Efficient Algorithm
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Now, we will take advantage of previously computed values as much as possible.
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**Note:** $s[l,r]$ represents substring of $s$ starting at index $l$ and ending at index $r$.
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Suppose we are given two indices $l$ and $r$ and also we are informed that $s[0,r-l]$ and $s[l,r]$ are equal. And we are finding value of z[i] such that $l<=i <= r$.
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How can we take advantage of that information to find $z[i]$?
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We can see that $s[i,r]$ and $s[i-l,r-l]$ are equal. Now, look at $z[i-l]$ and think how can we take advantage of it to find $z[i]$?
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$z[i-l]$ tells us that $s[0,z[i-l]-1]$ and $s[i-l,i-l+z[i-l]-1]$ are equal and therefore $s[0,z[i-l]-1]$ and $s[i,i+z[i-l]-1]$ are equal, which means that $z[i]=z[i-l]$.
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**But if $i+z[i-l]-1>r$, then it is ambiguous as we don't know anything about characters beyond $r$.**
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And therefore we simply take $z[i]=min(z[i-l],r-i+1)$, which does not go beyond $r$.
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Now, we will run brute force algorithm:
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```
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// As per the discussion
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z[i] = min(z[i-l],r-i+1);
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while(i + z[i] < n && s[z[i]] == s[i + z[i]])
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z[i]++;
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```
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After that if $i+z[i]$ is going beyond $r$, then we simply update indices $[l,r]$ to maintain **rightmost segment match** to take advantage of previous values as much as possible for next indices as well.
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**Note that initially $[l,r]$ segment is taken as $[0,0]$**. So, we basically start by doing brute force, or generally for an index $i$,
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1. If $i<=r$, then we wiil take advantage of previous value and then do brute force.
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2. Else if $i>r$, we directly do brute force as we can't take advantage of any previous value.
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```cpp
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vector<int> z_function(string s) {
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int n = (int) s.size();
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vector<int> z(n);
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int l = 0, r = 0;
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for (int i = 1; i < n; ++i) {
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// Take advantage of previous value
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if (i <= r)
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z[i] = min (r - i + 1, z[i - l]);
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// Now do it usual brute-force way
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while (i + z[i] < n && s[z[i]] == s[i + z[i]])
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++z[i];
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// Set new range [l,r]
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if (i + z[i] - 1 > r) {
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l = i;
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r = i + z[i] - 1;
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}
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}
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return z;
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}
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```
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### Time complexity
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$O(N)$, as at each step of the algorithm $r$ at least increases one step and maximum possible value of r is $n-1$.
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## Search for a string
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Z-algorithm is used to search all occurrences of pattern-string $p$ in a string $s$ in $O(N)$.
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For example, `p = "ab"` and `s = "abbbabab"`, then Z-algorithm will find us `[0,4,6]` because $s$ has 3 occurrences of `"ab"`.
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Basic idea here is to create a new string having $p$ as a prefix and $s$ as a suffix i.e. `new_str = p + '#' + s`.
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**To make sure that the value of Z-function does not exceed length of $p$, we will add an additional character which is never going to appear in string $s$**.
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Now, we will find Z-function of `new_str`.
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Let say $m$ is the length of $p$.
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$Z[i] = m$, means that `new_str[0..m-1]` is equal to `new_str[i...i+m-1]`, which is bacially means $p$(=`new_str[0...m-1]`) is equal to `new_str[i...i+m-1]`.
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And therefore **all indices-$i$ where the values of Z-function $Z[i]$ equals to the length of $p$ means it is an occurrence of $p$ in $s$.**
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```cpp
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int main()
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{
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string s,p;
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s = "abbbabab";
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p = "ab";
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int n = s.size(), m = p.size();
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// To save memory concatenate
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// s in p
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p += "#";
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p += s;
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// p = "ab#abbbabab";
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vector<int> z = z_function(p);
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// p = "ab#abbbabab";
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// ^
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// m+1
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cout << "occurences in s at the following indices: ";
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for(int i = m + 1; i < z.size(); i++) {
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if(z[i] == m) {
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cout << i - m - 1 << " ";
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}
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}
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return 0;
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}
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```
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## To find period of string
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Period of string is the shortest length such that a larger string $s$ can be represented as a concatenation of one or more copies of a substring($t$).
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For example, `s = "ababab"` has a period of $2$, where `t = "ab"`.
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Let's see how to find period of $s$ using value of z-function of $s$.
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**First of all note that length of string $s$($n$) is divisible by period of string.** Therefore, we can divide string $s$ into multiple blocks of same length as period of $s$.
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First of all, we will find all divisors of $n$ and value of z-function of $s$. Now, we will need to find smallest divisor of $n$ for which $i+z[i] = n$, which is period of string $s$. Why?
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$z[i]$ represents length of the longest common prefix of $s[0,n-1]$ and $s[i,n-1]$. As $i$ is divisor of $n$, we can divide the whole string into blocks of length $i$.
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From the value of $z[i] = n-i$($\because i+z[i]=n$), we can see that the first block($s[0,i-1]$) is equal to the second block starting at $i$-$s[i,i+i-1]$, which is also equal to third block $s[2*i,3*i-1]$ and similarly all blocks turns out to be equal.
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Therefore, smallest $i$ such that $n\% i=0$ and $i+z[i]=n$, is period of string $s$. If there is no such $i$, then string is not periodic as we cannot divide string into equivalent blocks.
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```cpp
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vector<int> getDivisors(int n)
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{
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vector<int> v;
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for (int i=1; i<=sqrt(n); i++)
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if (n%i==0)
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{
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v.push_back(i);
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if (n != i*i)
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v.push_back(n/i);
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}
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return v;
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}
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int main()
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{
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string s,p;
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s = "abcabcabc";
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int n = (int) s.size();
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vector<int> divs = getDivisors(n);
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sort(divs.begin(),divs.end());
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vector<int> z = z_function(s);
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int period = 0;
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for(auto i:divs) {
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if(i < n && z[i] + i == n) {
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period = i;
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break;
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}
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}
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if(period)
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cout << period << endl;
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else
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cout << "String is not periodic" << endl;
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return 0;
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}
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```
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### String compression
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Now, we know how to find a period of a string and therefore we can compress string as only one block of size $i$ which repeats all over again and again in $s$.
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To retrive the string back from compressed version, we can attatch its real length i.e. length of $s$.
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```cpp
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int main()
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{
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string s,p;
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s = "abcabcabc";
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int n = (int) s.size();
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vector<int> divs = getDivisors(n);
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sort(divs.begin(),divs.end());
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vector<int> z = z_function(s);
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int period = 0;
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for(auto i:divs) {
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if(i < n && z[i] + i == n) {
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period = i;
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break;
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}
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}
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if(period != 0) {
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// A way to represent compressed string
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// Attatch real length of string to retrieve easily
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pair<string, int> compressed_str{s.substr(0,period), n};
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}
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else {
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cout << "can't be compressed by this method" << endl;
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}
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return 0;
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}
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```
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## Number of distinct substring in a string
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**Problem statement:** Find number of unique substrings in a given string $s$.
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**Brief idea:** Basic idea here is to take an empty string $t$ and add characters one by one from string $s$ and along with that check how many new substrings are created, due to addition of a character in $t$, using z-function.
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Let say we have already added some characters to $t$ from $s$ and $k$ is the number of distinct substrings currently. Now, we are a adding character $c$ to $t$, $t = t+c$.
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Note that total number of new substrings created by appending a character to any string($t$) is equal to the length of new string($t=t+c$) created. **For example, Appending `'d'` in `"abc"` creates 4 new substrings: `"d"`, `"cd"`, `"bcd"`, `"abcd"`.**
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But how to find number of new unique substrings created by addition of $c$ **using z-function**?
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**Hint:** Reverse $t$.
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By reversing $t$, our task burn down into computing how many prefixes there are that don't appear anywhere else in $t$, which can be done by finding z-function of $t$.
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After finding value of z-function, we will find maximum value $z_{max}$($z_{max} = max\{z[i]\}, \forall i$) in the z-function of reversed $t$, which shows the length of longest prefix which is already in $t$ as a substring and it also implies that all smaller prefixes are already present as substrings in $t$.
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Therefore, we will deduct this number of already present substrings i.e. $z_{max}$, from the total number of new substrings i.e. $|t|$.
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Where $|t|$ is the length of $t$.
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Finally, number of new unique substrings created by addition of a character turns out to be $|t|-z_{max}$.
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**Note that $|t|$ is the length of $t$ after adding a character.**
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```cpp
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// Returns maximum of z[i]
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int z_function(string& s) {
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int n = (int) s.size();
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vector<int> z(n);
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int l = 0, r = 0;
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int mx = 0;
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for (int i = 1; i < n; ++i) {
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// Take advantage of previous value
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if (i <= r)
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z[i] = min (r - i + 1, z[i - l]);
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// Now do it usual brute-force way
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while (i + z[i] < n && s[z[i]] == s[i + z[i]])
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++z[i];
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mx = max(z[i], mx);
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// Set new range [l,r]
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if (i + z[i] - 1 > r) {
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l = i;
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r = i + z[i] - 1;
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}
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}
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return mx;
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}
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int main()
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{
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string s,p;
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s = "abc";
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int n = s.size();
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string t, temp;
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int unique_substr = 0;
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for(int i=0; i < n; i++) {
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t += s[i];
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temp = t;
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reverse(temp.begin(), temp.end());
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// |t| - mx
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unique_substr += (int)t.size() - z_function(temp);
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}
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// Total number of unique substrings
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cout << unique_substr << endl;
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return 0;
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}
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```
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