From e4faf7a85fc759a369b1d7a49df51f7a96ce2cc0 Mon Sep 17 00:00:00 2001
From: Sujan Chhetri <44917558+sujanchhetri@users.noreply.github.com>
Date: Sat, 3 Oct 2020 23:34:13 +0530
Subject: [PATCH] Added StringSearch (#385)

* Added StringSearch

* fixed the standard issue
---
 Search/StringSearch.js | 83 ++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 83 insertions(+)
 create mode 100644 Search/StringSearch.js

diff --git a/Search/StringSearch.js b/Search/StringSearch.js
new file mode 100644
index 000000000..3cd9c1e3e
--- /dev/null
+++ b/Search/StringSearch.js
@@ -0,0 +1,83 @@
+/*
+ * 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 chracter
+      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`
+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
+}
+
+console.log(stringSearch('Hello search the position of me', 'pos'))