TheAlgorithms · raklaptudirm · Oct 2, 2023 · Oct 1, 2023 · Oct 1, 2023 · Oct 1, 2023
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+/*
+ * @description : Given a string, the function finds the length of the longest substring without any repeating characters
+ * @param {String} str - The input string
+ * @returns {Number} The Length of the longest substring in a given string without repeating characters
+ * @example lengthOfLongestSubstring("abcabcbb") => 3
+ * @example lengthOfLongestSubstring("bbbbb") => 1
+ * @see https://leetcode.com/problems/longest-substring-without-repeating-characters/
+ */
+
+const lengthOfLongestSubstring = (s) => {
+  if (typeof s !== 'string') {
+    throw new TypeError('Invalid Input Type')
+  }
+  let maxLength = 0
+  let start = 0
+  const charMap = new Map()
+  for (let end = 0; end < s.length; end++) {
+    if (charMap.has(s[end])) {
+      start = Math.max(start, charMap.get(s[end]) + 1)
+    }
+    charMap.set(s[end], end)
+    maxLength = Math.max(maxLength, end - start + 1)
+  }
+  return maxLength
+}
+
+export { lengthOfLongestSubstring }
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+import { lengthOfLongestSubstring } from '../LengthofLongestSubstringWithoutRepetition'
+
+describe('LengthOfLongestSubstring', () => {
+  it('should throw error if parameter is not string', () => {
+    expect(() => lengthOfLongestSubstring(345)).toThrowError(TypeError)
+    expect(() => lengthOfLongestSubstring(true)).toThrowError(TypeError)
+    expect(() => lengthOfLongestSubstring(null)).toThrowError(TypeError)
+  })
+
+  it('should check substrings containing unique characters', () => {
+    expect(lengthOfLongestSubstring('abcabcbb')).toBe(3)
+    expect(lengthOfLongestSubstring('bbbbb')).toBe(1)
+    expect(lengthOfLongestSubstring('pwwkew')).toBe(3)
+    expect(lengthOfLongestSubstring(' ')).toBe(1)
+    expect(lengthOfLongestSubstring('abcdefghijklmnaaaaa')).toBe(13)
+  })
+
+  it('should give zero for empty strings', () => {
+    expect(lengthOfLongestSubstring('')).toBe(0)
+  })
+
+  it('should be case-sensitive', () => {
+    expect(lengthOfLongestSubstring('AaBbCc')).toBe(3)
+    expect(lengthOfLongestSubstring('AbCdEf')).toBe(6)
+  })
+})