/*
 * String Search
 */

function makeTable(str) {
  // create a table of size equal to the length of `str`
  // table[i] will store the prefix of the longest prefix of the substring str[0..i]
  const table = new Array(str.length)
  let maxPrefix = 0
  // the longest prefix of the substring str[0] has length
  table[0] = 0

  // for the substrings the following substrings, we have two cases
  for (let i = 1; i < str.length; i++) {
    // case 1. the current character doesn't match the last character of the longest prefix
    while (maxPrefix > 0 && str.charAt(i) !== str.charAt(maxPrefix)) {
      // if that is the case, we have to backtrack, and try find a character  that will be equal to the current character
      // if we reach 0, then we couldn't find a character
      maxPrefix = table[maxPrefix - 1]
    }
    // case 2. The last character of the longest prefix matches the current character in `str`
    if (str.charAt(maxPrefix) === str.charAt(i)) {
      // if that is the case, we know that the longest prefix at position i has one more character.
      // for example consider `.` be any character not contained in the set [a.c]
      // str = abc....abc
      // consider `i` to be the last character `c` in `str`
      // maxPrefix = will be 2 (the first `c` in `str`)
      // maxPrefix now will be 3
      maxPrefix++
      // so the max prefix for table[9] is 3
    }
    table[i] = maxPrefix
  }
  return table
}

// Find all the words that matches in a given string `str`
export function stringSearch(str, word) {
  // find the prefix table in O(n)
  const prefixes = makeTable(word)
  const matches = []

  // `j` is the index in `P`
  let j = 0
  // `i` is the index in `S`
  let i = 0
  while (i < str.length) {
    // Case 1.  S[i] == P[j] so we move to the next index in `S` and `P`
    if (str.charAt(i) === word.charAt(j)) {
      i++
      j++
    }
    // Case 2.  `j` is equal to the length of `P`
    // that means that we reached the end of `P` and thus we found a match
    // Next we have to update `j` because we want to save some time
    // instead of updating to j = 0 , we can jump to the last character of the longest prefix well known so far.
    // j-1 means the last character of `P` because j is actually `P.length`
    // e.g.
    // S =  a b a b d e
    // P = `a b`a b
    // we will jump to `a b` and we will compare d and a in the next iteration
    // a b a b `d` e
    //     a b `a` b
    if (j === word.length) {
      matches.push(i - j)
      j = prefixes[j - 1]
      // Case 3.
      // S[i] != P[j] There's a mismatch!
    } else if (str.charAt(i) !== word.charAt(j)) {
      // if we  found at least a character in common, do the same thing as in case 2
      if (j !== 0) {
        j = prefixes[j - 1]
      } else {
        // else j = 0, and we can move to the next character S[i+1]
        i++
      }
    }
  }

  return matches
}

// stringSearch('Hello search the position of me', 'pos')