merge: Sliding window (#833)

* new algorithm 'Sliding-Window' added with example 'Longest Substring Without Repeating Characters'

* new example of sliding window 'Permutation in String' added

* chore: add `ignore_words_list`

* sliding-window moved into Dynamic-programming directory

Co-authored-by: Rak Laptudirm <raklaptudirm@gmail.com>

Dynamic-Programming/Sliding-Window/PermutationinString.js

@@ -0,0 +1,59 @@
+/**
+ *  @name The-Sliding-Window Algorithm is primarily used for the problems dealing with linear data structures like Arrays, Lists, Strings etc.
+ *  These problems can easily be solved using Brute Force techniques which result in quadratic or exponential time complexity.
+ *  Sliding window technique reduces the required time to linear O(n).
+ *  @see [The-Sliding-Window](https://www.geeksforgeeks.org/window-sliding-technique/)
+ */
+/**
+ * @function PermutationinString
+ * @description Given two strings s1 and s2, return true if s2 contains a permutation of s1, or false otherwise.
+ * @param {String} s1 - The input string
+ * @param {String} s2 - The input string
+ * @return {boolean} - Returns true if s2 contains a permutation of s1, or false otherwise.
+ */
+
+export function PermutationinString (s1, s2) {
+  if (s1.length > s2.length) return false
+  let start = 0
+  let end = s1.length - 1
+  const s1Set = SetHash()
+  const s2Set = SetHash()
+  for (let i = 0; i < s1.length; i++) {
+    s1Set[s1[i]]++
+    s2Set[s2[i]]++
+  }
+  if (equals(s1Set, s2Set)) return true
+  while (end < s2.length - 1) {
+    if (equals(s1Set, s2Set)) return true
+    end++
+    console.log(s2[start], s2[end], equals(s1Set, s2Set))
+    const c1 = s2[start]
+    const c2 = s2[end]
+    if (s2Set[c1] > 0) s2Set[c1]--
+    s2Set[c2]++
+    start++
+    if (equals(s1Set, s2Set)) return true
+  }
+  return false
+}
+function equals (a, b) {
+  return JSON.stringify(a) === JSON.stringify(b)
+}
+
+function SetHash () {
+  const set = new Set()
+  const alphabets = 'abcdefghijklmnopqrstuvwxyz'
+  for (let i = 0; i < alphabets.length; i++) {
+    set[alphabets[i]] = 0
+  }
+  return set
+}
+
+// Example 1:
+// Input: s1 = "ab", s2 = "eidbaooo"
+// Output: true
+// Explanation: s2 contains one permutation of s1 ("ba").
+
+// Example 2:
+// Input: s1 = "ab", s2 = "eidboaoo"
+// Output: false