Knuth Morris Pratt

Prefix function: Given a string, p[i] : longest proper prefix which is also suffix of the string s[0…i]

Efficient Algorithm (Online):

int p[n];

void prefixfunc(string &s){
        
        int n = sz(s);
        
    	for(int i=1; i<n; i++){

		int j = lps[i-1];

		while(j>0 && str[i]!=str[j]) j = lps[j-1];

		if(str[i] == str[j]) lps[i] = j+1;

		else lps[i] = 0;
		
		// log(i, str[i], lps[i]);

	}
}

</br> Analysis of Prefix Function: </br> </br>

Bound: </br> </br> pi[x] <= pi[x-1]+1, else pi[x-1] can be made larger (contradiction). Each time j is decremented in the while:j loop, it corresponds to some previous(unique) increment of pi. There are max N elements, so maximum iteration of while:j can be N. So complexity is O(N+N) = O(N). About uniqueness of the increment, think like, everytime j uses as pi[x] to decrement j, it exhausts it and no longer can be used, as to be build again uniquely in the upcoming iterations.

Problems: </br> </br>

1. Substring Seach

Given a pattern t and a string s, print occurences of t in s.

Idea: Consider x = t#s, x = t1 t2 t3....tN # s1 s2 ....sM For all i, if pi[i] = N, pos = i-2N

Time : O(N+M) </br> </br>

1. String Periodicity

Given a string s, and s(i+T) = s(i), for all i. find the fundamental period T.</br> In other words, s = [a(0)…a(k-1)][a(k)…..a(2k-1)]…[a(N-k)….a(N-1)]. </br>

Idea: If u = N - pi[N-1], and u divides N, s[0...u-1] is the compressed periodic string, otherwise no period exists.

Periodicity Lemma: If a string has period T1 and T2, then gcd(T1,T2) is also a period of the string!

Z-function

int z[N];

void z_function(string &s){
    int l=0,r=0;
    for(int i=1;i<n;++i){
        if(i<=r) z[i] = min(r-i-1,z[i-l]);
        while(i+z[i]<n && s[z[i]]==s[i+z[i]]) ++z[i];
        if(i+z[i]-1>r) l=i, r=i+z[i]-1;
    }
}

1. String Periodicity

period = min i such that: i+z[i]=n

1. Min addition at start of a string to make it palindrome

Prefix function Trick: Form a string t by concating string s with reverse of s. find the lps of t. Ans=n-pi(t)

String Stream : To read a whole line and break it into strings. </br>

string s,t;
getline(cin,s);
stringstream x(s);

while(getline(x,t,' ')){
    cout << t << endl;
}

1. No of substrings of a string which is equal to its prefix

/*Consider at index::i, a[0...i]. Now what is the max length substring ending at i which is a prefix. Ans = pi[i] and the pref is a[0...pi[i]-1].
What is the next such string, a[0....pi[pi[i]-1]-1] and so on. 

But realize the structure here.

Any prefix of length x i.e a[0...x-1] will contribute to pi[x-1], pi[pi[x-1]-1], ......

Use dp to solve this*/

vector<int> cnt(n+1,0);

for(int i=0;i<n;i++) cnt[pi[i]]++;
for(int i=len;i>=0;i--) cnt[pi[i]-1] += cnt[i];
for(int i=1;i<=len;i++) cnt[i]++;
              

eg. https://codeforces.com/contest/432/problem/D

Nice KMP problem : https://www.codechef.com/problems/CENS20C