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‎README.md

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@@ -30,6 +30,7 @@ JavaScript-based LeetCode algorithm problem solutions, regularly updated.
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| <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- -->
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|-|-|-|-|-|-
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| 0035 |[Search Insert Position](src/main/js/g0001_0100/s0035_search_insert_position/solution.js)| Easy | Top_100_Liked_Questions, Array, Binary_Search, Big_O_Time_O(log_n)_Space_O(1) | 0 | 100.00
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#### Day 2 Two Pointers
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| <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- -->
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|-|-|-|-|-|-
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| 0034 |[Find First and Last Position of Element in Sorted Array](src/main/js/g0001_0100/s0034_find_first_and_last_position_of_element_in_sorted_array/solution.js)| Medium | Top_100_Liked_Questions, Top_Interview_Questions, Array, Binary_Search, Big_O_Time_O(log_n)_Space_O(1) | 0 | 100.00
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| 0033 |[Search in Rotated Sorted Array](src/main/js/g0001_0100/s0033_search_in_rotated_sorted_array/solution.js)| Medium | Top_100_Liked_Questions, Top_Interview_Questions, Array, Binary_Search, Big_O_Time_O(log_n)_Space_O(1) | 0 | 100.00
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#### Day 2 Binary Search
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| <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- -->
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|-|-|-|-|-|-
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| 0035 |[Search Insert Position](src/main/js/g0001_0100/s0035_search_insert_position/solution.js)| Easy | Top_100_Liked_Questions, Array, Binary_Search, Big_O_Time_O(log_n)_Space_O(1) | 0 | 100.00
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#### Day 3
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| <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- -->
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|-|-|-|-|-|-
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| 0034 |[Find First and Last Position of Element in Sorted Array](src/main/js/g0001_0100/s0034_find_first_and_last_position_of_element_in_sorted_array/solution.js)| Medium | Top_100_Liked_Questions, Top_Interview_Questions, Array, Binary_Search, Big_O_Time_O(log_n)_Space_O(1) | 0 | 100.00
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#### Day 6
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| <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- -->
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|-|-|-|-|-|-
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| 0033 |[Search in Rotated Sorted Array](src/main/js/g0001_0100/s0033_search_in_rotated_sorted_array/solution.js)| Medium | Top_100_Liked_Questions, Top_Interview_Questions, Array, Binary_Search, Big_O_Time_O(log_n)_Space_O(1) | 0 | 100.00
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#### Day 12
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| <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- -->
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|-|-|-|-|-|-
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| 0033 |[Search in Rotated Sorted Array](src/main/js/g0001_0100/s0033_search_in_rotated_sorted_array/solution.js)| Medium | Top_100_Liked_Questions, Top_Interview_Questions, Array, Binary_Search, Big_O_Time_O(log_n)_Space_O(1) | 0 | 100.00
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#### Day 9 Binary Search Tree
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| <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- -->
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|-|-|-|-|-|-
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| 0007 |[Reverse Integer](src/main/js/g0001_0100/s0007_reverse_integer/solution.js)| Medium | Top_Interview_Questions, Math | 66 | 87.91
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| 0007 |[Reverse Integer](src/main/js/g0001_0100/s0007_reverse_integer/solution.js)| Medium | Top_Interview_Questions, Math | 67 | 85.38
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| 0009 |[Palindrome Number](src/main/js/g0001_0100/s0009_palindrome_number/solution.js)| Easy | Math | 4 | 97.14
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#### Udemy Strings
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| <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- -->
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| 0033 |[Search in Rotated Sorted Array](src/main/js/g0001_0100/s0033_search_in_rotated_sorted_array/solution.js)| Medium | Top_100_Liked_Questions, Top_Interview_Questions, Array, Binary_Search, Big_O_Time_O(log_n)_Space_O(1) | 0 | 100.00
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#### Udemy Arrays
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| # | Title | Difficulty | Tag | Time, ms | Time, %
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|------|----------------|-------------|-------------|----------|---------
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| 0035 |[Search Insert Position](src/main/js/g0001_0100/s0035_search_insert_position/solution.js)| Easy | Top_100_Liked_Questions, Array, Binary_Search, Algorithm_I_Day_1_Binary_Search, Binary_Search_I_Day_2, Big_O_Time_O(log_n)_Space_O(1) | 0 | 100.00
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| 0034 |[Find First and Last Position of Element in Sorted Array](src/main/js/g0001_0100/s0034_find_first_and_last_position_of_element_in_sorted_array/solution.js)| Medium | Top_100_Liked_Questions, Top_Interview_Questions, Array, Binary_Search, Algorithm_II_Day_1_Binary_Search, Binary_Search_I_Day_5, Big_O_Time_O(log_n)_Space_O(1) | 0 | 100.00
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| 0033 |[Search in Rotated Sorted Array](src/main/js/g0001_0100/s0033_search_in_rotated_sorted_array/solution.js)| Medium | Top_100_Liked_Questions, Top_Interview_Questions, Array, Binary_Search, Algorithm_II_Day_1_Binary_Search, Binary_Search_I_Day_11, Level_2_Day_8_Binary_Search, Udemy_Binary_Search, Big_O_Time_O(log_n)_Space_O(1) | 0 | 100.00
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| 0032 |[Longest Valid Parentheses](src/main/js/g0001_0100/s0032_longest_valid_parentheses/solution.js)| Hard | Top_100_Liked_Questions, String, Dynamic_Programming, Stack, Big_O_Time_O(n)_Space_O(1) | 1 | 96.21
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| 0031 |[Next Permutation](src/main/js/g0001_0100/s0031_next_permutation/solution.js)| Medium | Top_100_Liked_Questions, Array, Two_Pointers, Big_O_Time_O(n)_Space_O(1) | 0 | 100.00
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| 0025 |[Reverse Nodes in k-Group](src/main/js/g0001_0100/s0025_reverse_nodes_in_k_group/solution.js)| Hard | Top_100_Liked_Questions, Linked_List, Recursion, Data_Structure_II_Day_13_Linked_List, Udemy_Linked_List, Big_O_Time_O(n)_Space_O(k) | 0 | 100.00
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| 0024 |[Swap Nodes in Pairs](src/main/js/g0001_0100/s0024_swap_nodes_in_pairs/solution.js)| Medium | Top_100_Liked_Questions, Linked_List, Recursion, Data_Structure_II_Day_12_Linked_List, Udemy_Linked_List, Big_O_Time_O(n)_Space_O(1) | 1 | 100.00
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| 0023 |[Merge k Sorted Lists](src/main/js/g0001_0100/s0023_merge_k_sorted_lists/solution.js)| Hard | Top_100_Liked_Questions, Top_Interview_Questions, Heap_Priority_Queue, Linked_List, Divide_and_Conquer, Merge_Sort, Big_O_Time_O(k\*n\*log(k))_Space_O(log(k)) | 3 | 99.39
@@ -1254,7 +1267,7 @@ JavaScript-based LeetCode algorithm problem solutions, regularly updated.
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| 0010 |[Regular Expression Matching](src/main/js/g0001_0100/s0010_regular_expression_matching/solution.js)| Hard | Top_Interview_Questions, String, Dynamic_Programming, Recursion, Udemy_Dynamic_Programming, Big_O_Time_O(m\*n)_Space_O(m\*n) | 4 | 95.52
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| 0009 |[Palindrome Number](src/main/js/g0001_0100/s0009_palindrome_number/solution.js)| Easy | Math, Udemy_Integers | 4 | 97.14
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| 0008 |[String to Integer (atoi)](src/main/js/g0001_0100/s0008_string_to_integer_atoi/solution.js)| Medium | Top_Interview_Questions, String | 1 | 96.63
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| 0007 |[Reverse Integer](src/main/js/g0001_0100/s0007_reverse_integer/solution.js)| Medium | Top_Interview_Questions, Math, Udemy_Integers | 66 | 87.91
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| 0007 |[Reverse Integer](src/main/js/g0001_0100/s0007_reverse_integer/solution.js)| Medium | Top_Interview_Questions, Math, Udemy_Integers | 67 | 85.38
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| 0006 |[Zigzag Conversion](src/main/js/g0001_0100/s0006_zigzag_conversion/solution.js)| Medium | String | 2 | 99.71
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| 0005 |[Longest Palindromic Substring](src/main/js/g0001_0100/s0005_longest_palindromic_substring/solution.js)| Medium | Top_100_Liked_Questions, Top_Interview_Questions, String, Dynamic_Programming, Data_Structure_II_Day_9_String, Algorithm_II_Day_14_Dynamic_Programming, Dynamic_Programming_I_Day_17, Udemy_Strings, Big_O_Time_O(n)_Space_O(n) | 10 | 99.54
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| 0004 |[Median of Two Sorted Arrays](src/main/js/g0001_0100/s0004_median_of_two_sorted_arrays/solution.js)| Hard | Top_100_Liked_Questions, Top_Interview_Questions, Array, Binary_Search, Divide_and_Conquer, Big_O_Time_O(log(min(N,M)))_Space_O(1), AI_can_be_used_to_solve_the_task | 3 | 91.90

‎src/main/js/g0001_0100/s0031_next_permutation/readme.md

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31\. Next Permutation
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Medium
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A **permutation** of an array of integers is an arrangement of its members into a sequence or linear order.
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* For example, for `arr = [1,2,3]`, the following are all the permutations of `arr`: `[1,2,3], [1,3,2], [2, 1, 3], [2, 3, 1], [3,1,2], [3,2,1]`.
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The **next permutation** of an array of integers is the next lexicographically greater permutation of its integer. More formally, if all the permutations of the array are sorted in one container according to their lexicographical order, then the **next permutation** of that array is the permutation that follows it in the sorted container. If such arrangement is not possible, the array must be rearranged as the lowest possible order (i.e., sorted in ascending order).
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* For example, the next permutation of `arr = [1,2,3]` is `[1,3,2]`.
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* Similarly, the next permutation of `arr = [2,3,1]` is `[3,1,2]`.
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* While the next permutation of `arr = [3,2,1]` is `[1,2,3]` because `[3,2,1]` does not have a lexicographical larger rearrangement.
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Given an array of integers `nums`, _find the next permutation of_ `nums`.
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The replacement must be **[in place](http://en.wikipedia.org/wiki/In-place_algorithm)** and use only constant extra memory.
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**Example 1:**
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**Input:** nums = [1,2,3]
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**Output:** [1,3,2]
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**Example 2:**
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**Input:** nums = [3,2,1]
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**Output:** [1,2,3]
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**Example 3:**
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**Input:** nums = [1,1,5]
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**Output:** [1,5,1]
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**Constraints:**
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* `1 <= nums.length <= 100`
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* `0 <= nums[i] <= 100`

‎src/main/js/g0001_0100/s0031_next_permutation/solution.js

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// #Medium #Top_100_Liked_Questions #Array #Two_Pointers #Big_O_Time_O(n)_Space_O(1)
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// #2024_12_05_Time_0_ms_(100.00%)_Space_52.3_MB_(8.66%)
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/**
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* @param {number[]} nums
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* @return {void} Do not return anything, modify nums in-place instead.
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*/
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var nextPermutation = function(nums) {
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if (!nums || nums.length <= 1) {
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return
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}
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let i = nums.length - 2
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// Find the first index `i` where nums[i] < nums[i + 1]
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while (i >= 0 && nums[i] >= nums[i + 1]) {
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i--
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}
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if (i >= 0) {
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// Find the smallest number larger than nums[i] to swap with
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let j = nums.length - 1
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while (nums[j] <= nums[i]) {
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j--
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}
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swap(nums, i, j)
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}
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// Reverse the portion of the array from index `i + 1` to the end
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reverse(nums, i + 1, nums.length - 1)
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};
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function swap(nums, i, j) {
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const temp = nums[i]
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nums[i] = nums[j]
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nums[j] = temp
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}
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function reverse(nums, start, end) {
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while (start < end) {
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swap(nums, start++, end--)
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}
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}
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export { nextPermutation }

‎src/main/js/g0001_0100/s0032_longest_valid_parentheses/readme.md

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32\. Longest Valid Parentheses
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Hard
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Given a string containing just the characters `'('` and `')'`, find the length of the longest valid (well-formed) parentheses substring.
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**Example 1:**
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**Input:** s = "(()"
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**Output:** 2
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**Explanation:** The longest valid parentheses substring is "()".
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**Example 2:**
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**Input:** s = ")()())"
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**Output:** 4
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**Explanation:** The longest valid parentheses substring is "()()".
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**Example 3:**
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**Input:** s = ""
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**Output:** 0
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**Constraints:**
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* <code>0 <= s.length <= 3 * 10<sup>4</sup></code>
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* `s[i]` is `'('`, or `')'`.

‎src/main/js/g0001_0100/s0032_longest_valid_parentheses/solution.js

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// #Hard #Top_100_Liked_Questions #String #Dynamic_Programming #Stack #Big_O_Time_O(n)_Space_O(1)
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// #2024_12_05_Time_1_ms_(96.21%)_Space_52.5_MB_(15.11%)
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/**
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* @param {string} s
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* @return {number}
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*/
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var longestValidParentheses = function (s) {
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const n = s.length
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const f = new Array(n + 1).fill(0)
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for (let i = 2; i <= n; ++i) {
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if (s[i - 1] === ')') {
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if (s[i - 2] === '(') {
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f[i] = f[i - 2] + 2
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} else {
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const j = i - f[i - 1] - 1
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if (j && s[j - 1] === '(') {
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f[i] = f[i - 1] + 2 + f[j - 1]
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}
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}
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}
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}
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return Math.max(...f)
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};
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export { longestValidParentheses }

‎src/main/js/g0001_0100/s0033_search_in_rotated_sorted_array/readme.md

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33\. Search in Rotated Sorted Array
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Medium
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There is an integer array `nums` sorted in ascending order (with **distinct** values).
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Prior to being passed to your function, `nums` is **possibly rotated** at an unknown pivot index `k` (`1 <= k < nums.length`) such that the resulting array is `[nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]]` (**0-indexed**). For example, `[0,1,2,4,5,6,7]` might be rotated at pivot index `3` and become `[4,5,6,7,0,1,2]`.
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Given the array `nums` **after** the possible rotation and an integer `target`, return _the index of_ `target` _if it is in_ `nums`_, or_ `-1` _if it is not in_ `nums`.
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You must write an algorithm with `O(log n)` runtime complexity.
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**Example 1:**
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**Input:** nums = [4,5,6,7,0,1,2], target = 0
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**Output:** 4
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**Example 2:**
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**Input:** nums = [4,5,6,7,0,1,2], target = 3
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**Output:** -1
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**Example 3:**
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**Input:** nums = [1], target = 0
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**Output:** -1
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**Constraints:**
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* `1 <= nums.length <= 5000`
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* <code>-10<sup>4</sup> <= nums[i] <= 10<sup>4</sup></code>
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* All values of `nums` are **unique**.
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* `nums` is an ascending array that is possibly rotated.
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* <code>-10<sup>4</sup> <= target <= 10<sup>4</sup></code>

‎src/main/js/g0001_0100/s0033_search_in_rotated_sorted_array/solution.js

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// #Medium #Top_100_Liked_Questions #Top_Interview_Questions #Array #Binary_Search
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// #Algorithm_II_Day_1_Binary_Search #Binary_Search_I_Day_11 #Level_2_Day_8_Binary_Search
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// #Udemy_Binary_Search #Big_O_Time_O(log_n)_Space_O(1)
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// #2024_12_05_Time_0_ms_(100.00%)_Space_49.3_MB_(21.97%)
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/**
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* @param {number[]} nums
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* @param {number} target
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* @return {number}
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*/
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var search = function(nums, target) {
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let lo = 0
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let hi = nums.length - 1
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while (lo <= hi) {
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const mid = Math.floor((hi - lo) / 2) + lo
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if (nums[mid] === target) {
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return mid
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}
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if (nums[lo] <= nums[mid]) {
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// Target is in the sorted left half
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if (nums[lo] <= target && target <= nums[mid]) {
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hi = mid - 1
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} else {
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lo = mid + 1
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}
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} else if (nums[mid] <= target && target <= nums[hi]) {
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lo = mid + 1
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} else {
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hi = mid - 1
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}
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}
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return -1
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};
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export { search }

‎src/main/js/g0001_0100/s0034_find_first_and_last_position_of_element_in_sorted_array/readme.md

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34\. Find First and Last Position of Element in Sorted Array
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Medium
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Given an array of integers `nums` sorted in non-decreasing order, find the starting and ending position of a given `target` value.
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If `target` is not found in the array, return `[-1, -1]`.
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You must write an algorithm with `O(log n)` runtime complexity.
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**Example 1:**
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**Input:** nums = [5,7,7,8,8,10], target = 8
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**Output:** [3,4]
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**Example 2:**
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**Input:** nums = [5,7,7,8,8,10], target = 6
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**Output:** [-1,-1]
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**Example 3:**
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**Input:** nums = [], target = 0
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**Output:** [-1,-1]
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**Constraints:**
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* <code>0 <= nums.length <= 10<sup>5</sup></code>
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* <code>-10<sup>9</sup> <= nums[i] <= 10<sup>9</sup></code>
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* `nums` is a non-decreasing array.
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* <code>-10<sup>9</sup> <= target <= 10<sup>9</sup></code>

‎src/main/js/g0001_0100/s0034_find_first_and_last_position_of_element_in_sorted_array/solution.js

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// #Medium #Top_100_Liked_Questions #Top_Interview_Questions #Array #Binary_Search
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// #Algorithm_II_Day_1_Binary_Search #Binary_Search_I_Day_5 #Big_O_Time_O(log_n)_Space_O(1)
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// #2024_12_05_Time_0_ms_(100.00%)_Space_49.9_MB_(40.62%)
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/**
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* @param {number[]} nums
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* @param {number} target
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* @return {number[]}
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*/
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var searchRange = function(nums, target) {
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return [
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helper(nums, target, false),
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helper(nums, target, true)
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]
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}
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function helper(nums, target, equals) {
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let l = 0, r = nums.length - 1
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let result = -1
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while (l <= r) {
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const mid = Math.floor((l + r) / 2)
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if (nums[mid] === target) {
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result = mid
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}
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if (nums[mid] < target || (nums[mid] === target && equals)) {
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l = mid + 1
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} else {
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r = mid - 1
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}
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}
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return result
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};
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export { searchRange }

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