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Sliding window #833
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2426700
new algorithm 'Sliding-Window' added with example 'Longest Substring ...
Aditya1942 b6d7962
new example of sliding window 'Permutation in String' added
Aditya1942 7bf3536
chore: add `ignore_words_list`
raklaptudirm ba42b93
sliding-window moved into Dynamic-programming directory
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49 changes: 49 additions & 0 deletions
Dynamic-Programming/Sliding-Window/LongestSubstringWithoutRepeatingCharacters.js
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|---|---|---|
| @@ -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/) | ||
| */ | ||
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| /** | ||
| * @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 | ||
| } | ||
|
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| // Example 1: | ||
| // Input: s = "abcabcbb" | ||
| // Output: 3 | ||
| // Explanation: The answer is "abc", with the length of 3. | ||
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| // Example 2: | ||
| // Input: s = "bbbbb" | ||
| // Output: 1 | ||
| // Explanation: The answer is "b", with the length of 1. | ||
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| // 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. |
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -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. | ||
| */ | ||
|
|
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| 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) | ||
| } | ||
|
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| function SetHash () { | ||
| const set = new Set() | ||
| const alphabets = 'abcdefghijklmnopqrstuvwxyz' | ||
| for (let i = 0; i < alphabets.length; i++) { | ||
| set[alphabets[i]] = 0 | ||
| } | ||
| return set | ||
| } | ||
|
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| // Example 1: | ||
| // Input: s1 = "ab", s2 = "eidbaooo" | ||
| // Output: true | ||
| // Explanation: s2 contains one permutation of s1 ("ba"). | ||
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| // Example 2: | ||
| // Input: s1 = "ab", s2 = "eidboaoo" | ||
| // Output: false |
11 changes: 11 additions & 0 deletions
Dynamic-Programming/Sliding-Window/test/LongestSubstringWithoutRepeatingCharacters.test.js
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,11 @@ | ||
| import { LongestSubstringWithoutRepeatingCharacters } from '../LongestSubstringWithoutRepeatingCharacters.js' | ||
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| 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) | ||
| }) | ||
| }) |
10 changes: 10 additions & 0 deletions
Dynamic-Programming/Sliding-Window/test/PermutationinString.test.js
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,10 @@ | ||
| import { PermutationinString } from '../PermutationinString.js' | ||
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| 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) | ||
| }) | ||
| }) |
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