Substring Search - II

Given two strings s1 and s2, find the index of the first occurrence of s2 in s1 as a substring.

If no such occurence exists, return -1.

This problem is also known as finding needle in a haystack.

Use the Rabin-Karp algorithm to solve this problem.

• rabin karp rolling hash solution

int findStartIndexOfSubstring(string s1, string s2) {
    int n = s1.length(), m = s2.length();
    if(n == m) return s1 == s2 ? 0 : -1;
    if(n < m) return -1;

    long long base = 1010101, mod = 1e9 + 7, d = 1;
    long long target = 0, hash = 0;
    
     for(int i = 0; i < m; i++) {
         target = (target * base + s2[i]) % mod;
         d = d * base % mod;
     }
     
     for(int i = 0; i < n; i++) {
         hash = (hash * base + s1[i]) % mod;
         if(i >= m) hash = (hash + mod - d * s1[i - m] % mod) % mod;
         if(i >= m - 1 and hash == target and s1.substr(i - m + 1, m) == s2) return i - m + 1;
     }
    
    
    return -1;
}

• kmp solution

vector<int> PI(string& s) {
    int n = s.length();
    vector<int> pi(n);
    for(int i = 1, j = 0; i < n; i++) {
        while(j > 0 and s[i] != s[j]) j = pi[j - 1];
        if(s[i] == s[j]) pi[i] = ++j;
    }
    return pi;
}

int findStartIndexOfSubstring(string s1, string s2) {
    int n = s1.length(), m = s2.length();
    if(n == m) return s1 == s2 ? 0 : -1;
    if(n < m) return -1;

    auto pi = PI(s2);
    
    for(int i = 0, j = 0; i < n; i++) {
        while(j > 0 and s1[i] != s2[j]) j = pi[j - 1];
        if(s1[i] == s2[j]) {
            if(++j == m)
                return i - m + 1;
        }
    }

    return -1;
}