PYTHON

kmp algorithm

#include <string>
#include <iostream>
#include <vector>
using namespace std;
int main()
{
    string l;
    string s;
    cin >> l;
    cin >> s;
    vector<long long>v(s.size(),0);
    int i, j;
    i = 0;
    j = 0;

    for (int i = 0; i < s.size(); i++) {
        i++;
        if (s[j] == s[i]) {
            v[i] = v[i - 1] + 1;
            j++;
        }
        else {
            if (j == 0) {
                j = v[j ];
            }
            else {
                j = v[j - 1];
            }
        }
    }
    j = 0;
    for (int i = 0; i < l.size(); i++) {
        if (l[i] == s[j]) {
            j++;
            if (j = s.size()) {
                cout << "start= " << i - s.size() << " end= " << i ;
                i = l.size();
            }

        }
        else {
            if (j == 0) {
                j = 0;
            }
            else {
                j = v[j - 1];
            }
        }
    }
}

kmp algorithm

#define ll long long int
#define vll vector<long long int>

ll kmp(string s)
{
    ll n = s.size();
    vll lps(n, 0); // longest prefix suffix
    for (ll i = 1; i < n; i++)
    {
        ll j = lps[i - 1];
        while (j > 0 && s[i] != s[j])
        {
            j = lps[j - 1];
        }
        if (s[i] == s[j])
        {
            j++;
        }
        lps[i] = j;
    }
    return lps[n - 1];
}

kmp algorithm

"""
This implementation demonstrates how 
to efficiently determine if a pattern 
string is a substring of some bigger,
target string.

Example: 
For following input values:
substring = "aefcd"
string = "dcaefcdcdaed"
Output would be: True

Let m be the size of the pattern and 
n be the size of the target string.

Time complexity: O(n+m)
Space complexity: O(m)
"""


def kmp_algorithm(string, substring):
    i, j = 0, 0
    """
    preprocess the pattern string by computing
    a failure function mapping indexes to shifts
    with the aim of reusing previously performed
    comparisons.
    """
    failure = compute_failure_function(substring)
    str_len, substr_len = len(string), len(substring)
    while i < str_len:
        if string[i] == substring[j]:
            # Pattern is found when its last char reached
            if j == substr_len - 1:
                return True
            i += 1
            j += 1
        elif j > 0:
            # Determine next continuation index in pattern
            # by consulting the failure function.
            j = failure[j-1]
        else:
            i += 1
    return False


def compute_failure_function(substring):
    # The failure function maps each index j
    # in pattern P to length of longest prefix
    # of P that is a suffix of P[1:j]
    j, i = 0, 1
    substr_len = len(substring)
    failure = [0] * substr_len
    while i < substr_len:
        if substring[j] == substring[i]:
            # We have matched j + 1 characters
            failure[i] = j + 1
            i += 1
            j += 1
        elif j > 0:
            # Place j just after a prefix that matches
            j = failure[j-1]
        else:
            i += 1
    return failure


print(kmp_algorithm("dcaefcdcdaed", "aefcd"))  # True
print(kmp_algorithm("dcaefccdaed", "aefcd"))  # False

kmp algorithm

COMPUTE- PREFIX- FUNCTION (P)
 1. m ←length [P]		//'p' pattern to be matched
 2. Π [1] ← 0
 3. k ← 0
 4. for q ← 2 to m
 5. do while k > 0 and P [k + 1] ≠ P [q]
 6. do k ← Π [k]
 7. If P [k + 1] = P [q]
 8. then k← k + 1
 9. Π [q] ← k
 10. Return Π

kmp algorithm

KMP-MATCHER (T, P)
 1. n ← length [T]
 2. m ← length [P]
 3. Π← COMPUTE-PREFIX-FUNCTION (P)
 4. q ← 0		// numbers of characters matched
 5. for i ← 1 to n	// scan S from left to right 
 6. do while q > 0 and P [q + 1] ≠ T [i]
 7. do q ← Π [q]		// next character does not match
 8. If P [q + 1] = T [i]
 9. then q ← q + 1		// next character matches
 10. If q = m			           // is all of p matched?
 11. then print "Pattern occurs with shift" i - m
 12. q ← Π [q]				// look for the next match