Python 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

#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];
}

void computeLPSArray(char* pat, int M, int* lps)
{
    int len = 0;
    lps[0] = 0;
    int i = 1;
    while (i < M) {
        if (pat[i] == pat[len]) {
            len++;
            lps[i] = len;
            i++;
        }
        else
        {
            if (len != 0) {
                len = lps[len - 1];
            }
            else
            {
                lps[i] = 0;
                i++;
            }
        }
    }
}
int matchFound=0;
void KMPSearch(char* pat, char* txt)
{
    matchFound=0;
    int M = strlen(pat);
    int N = strlen(txt);
    int lps[M];
    computeLPSArray(pat, M, lps);
    int i = 0;
    int j = 0;
    while (i < N) {
        if (pat[j] == txt[i]) {
            j++;
            i++;
        }
        if (j == M) {
            matchFound++;
//            printf("Found pattern at index %d ", i - j);
            j = lps[j - 1];
        }
        else if (i < N && pat[j] != txt[i]) {
            if (j != 0)
                j = lps[j - 1];
            else
                i = i + 1;
        }
    }
}

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-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