TheAlgorithms
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diff --git a/‎Dynamic-Programming/Sliding-Window/PermutationinString.js‎
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@@ -10,4 +10,4 @@ jobs:
         with:
           # file types to ignore
           skip: "*.json,*.yml,DIRECTORY.md,PermutateString.test.js,SubsequenceRecursive.js"
-          
+          ignore_words_list: "ba"
@@ -0,0 +1,49 @@
+/**
+ *  @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 LongestSubstringWithoutRepeatingCharacters
+ * @description Get the length of the longest substring without repeating characters
+ * @param {String} s - The input string
+ */
+export function LongestSubstringWithoutRepeatingCharacters (s) {
+  let maxLength = 0
+  let start = 0
+  let end = 0
+  const map = {}
+  while (end < s.length) {
+    if (map[s[end]] === undefined) {
+      map[s[end]] = 1
+      maxLength = Math.max(maxLength, end - start + 1)
+      end++
+    } else {
+      while (s[start] !== s[end]) {
+        delete map[s[start]]
+        start++
+      }
+      delete map[s[start]]
+      start++
+    }
+  }
+  return maxLength
+}
+
+// Example 1:
+// Input: s = "abcabcbb"
+// Output: 3
+// Explanation: The answer is "abc", with the length of 3.
+
+// Example 2:
+// Input: s = "bbbbb"
+// Output: 1
+// Explanation: The answer is "b", with the length of 1.
+
+// Example 3:
+// Input: s = "pwwkew"
+// Output: 3
+// Explanation: The answer is "wke", with the length of 3.
+// Notice that the answer must be a substring, "pwke" is a subsequence and not a substring.
@@ -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
@@ -0,0 +1,11 @@
+import { LongestSubstringWithoutRepeatingCharacters } from '../LongestSubstringWithoutRepeatingCharacters.js'
+
+describe('LongestSubstringWithoutRepeatingCharacters', () => {
+  it('should return longest substring without repeating characters', () => {
+    expect(LongestSubstringWithoutRepeatingCharacters('abcabcbb')).toEqual(3)
+    expect(LongestSubstringWithoutRepeatingCharacters('bbbbb')).toEqual(1)
+    expect(LongestSubstringWithoutRepeatingCharacters('pwwkew')).toEqual(3)
+    expect(LongestSubstringWithoutRepeatingCharacters('a')).toEqual(1)
+    expect(LongestSubstringWithoutRepeatingCharacters('')).toEqual(0)
+  })
+})
@@ -0,0 +1,10 @@
+import { PermutationinString } from '../PermutationinString.js'
+
+describe('PermutationinString', () => {
+  it("should  return true if one of s1's permutations is the substring of s2", () => {
+    expect(PermutationinString('ab', 'eidbaooo')).toEqual(true)
+    expect(PermutationinString('abc', 'bcab')).toEqual(true)
+    expect(PermutationinString('ab', 'eidboaoo')).toEqual(false)
+    expect(PermutationinString('abc', '')).toEqual(false)
+  })
+})