Updated readme

Author: javadev

README.md

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+[![](https://img.shields.io/github/stars/LeetCode-in-Ruby/LeetCode-in-Ruby?label=Stars&style=flat-square)](https://github.com/LeetCode-in-Ruby/LeetCode-in-Ruby)
+[![](https://img.shields.io/github/forks/LeetCode-in-Ruby/LeetCode-in-Ruby?label=Fork%20me%20on%20GitHub%20&style=flat-square)](https://github.com/LeetCode-in-Ruby/LeetCode-in-Ruby/fork)
+
+## 1\. Two Sum
+
+Easy
+
+Given an array of integers `nums` and an integer `target`, return _indices of the two numbers such that they add up to `target`_.
+
+You may assume that each input would have **_exactly_ one solution**, and you may not use the _same_ element twice.
+
+You can return the answer in any order.
+
+**Example 1:**
+
+**Input:** nums = [2,7,11,15], target = 9
+
+**Output:** [0,1]
+
+**Explanation:** Because nums[0] + nums[1] == 9, we return [0, 1]. 
+
+**Example 2:**
+
+**Input:** nums = [3,2,4], target = 6
+
+**Output:** [1,2] 
+
+**Example 3:**
+
+**Input:** nums = [3,3], target = 6
+
+**Output:** [0,1] 
+
+**Constraints:**
+
+*   <code>2 <= nums.length <= 10<sup>4</sup></code>
+*   <code>-10<sup>9</sup> <= nums[i] <= 10<sup>9</sup></code>
+*   <code>-10<sup>9</sup> <= target <= 10<sup>9</sup></code>
+*   **Only one valid answer exists.**
+
+**Follow-up:** Can you come up with an algorithm that is less than <code>O(n<sup>2</sup>)</code> time complexity?
+
+## Solution
+
+```ruby
+# @param {Integer[]} nums
+# @param {Integer} target
+# @return {Integer[]}
+def two_sum(nums, target)
+  dict = {}
+  nums.each_with_index do |n, i|
+    if dict[target - n]
+      return dict[target - n], i
+    end
+    dict[n] = i
+  end
+  [-1, -1]
+end
+```

src/main/ruby/g0001_0100/s0001_two_sum/readme.md

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+[![](https://img.shields.io/github/stars/LeetCode-in-Ruby/LeetCode-in-Ruby?label=Stars&style=flat-square)](https://github.com/LeetCode-in-Ruby/LeetCode-in-Ruby)
+[![](https://img.shields.io/github/forks/LeetCode-in-Ruby/LeetCode-in-Ruby?label=Fork%20me%20on%20GitHub%20&style=flat-square)](https://github.com/LeetCode-in-Ruby/LeetCode-in-Ruby/fork)
+
+## 2\. Add Two Numbers
+
+Medium
+
+You are given two **non-empty** linked lists representing two non-negative integers. The digits are stored in **reverse order**, and each of their nodes contains a single digit. Add the two numbers and return the sum as a linked list.
+
+You may assume the two numbers do not contain any leading zero, except the number 0 itself.
+
+**Example 1:**
+
+![](https://assets.leetcode.com/uploads/2020/10/02/addtwonumber1.jpg)
+
+**Input:** l1 = [2,4,3], l2 = [5,6,4]
+
+**Output:** [7,0,8]
+
+**Explanation:** 342 + 465 = 807. 
+
+**Example 2:**
+
+**Input:** l1 = [0], l2 = [0]
+
+**Output:** [0] 
+
+**Example 3:**
+
+**Input:** l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
+
+**Output:** [8,9,9,9,0,0,0,1] 
+
+**Constraints:**
+
+*   The number of nodes in each linked list is in the range `[1, 100]`.
+*   `0 <= Node.val <= 9`
+*   It is guaranteed that the list represents a number that does not have leading zeros.
+
+## Solution
+
+```ruby
+require_relative '../../com_github_leetcode/list_node'
+
+# Definition for singly-linked list.
+# class ListNode
+#     attr_accessor :val, :next
+#     def initialize(val = 0, _next = nil)
+#         @val = val
+#         @next = _next
+#     end
+# end
+# @param {ListNode} l1
+# @param {ListNode} l2
+# @return {ListNode}
+def add_two_numbers(l1, l2)
+  dummy_head = ListNode.new(0)
+  p = l1
+  q = l2
+  curr = dummy_head
+  carry = 0
+
+  while p || q
+    x = p ? p.val : 0
+    y = q ? q.val : 0
+    sum = carry + x + y
+    carry = sum / 10
+    curr.next = ListNode.new(sum % 10)
+    curr = curr.next
+
+    p = p.next if p
+    q = q.next if q
+  end
+
+  curr.next = ListNode.new(carry) if carry > 0
+
+  dummy_head.next
+end
+```

src/main/ruby/g0001_0100/s0002_add_two_numbers/readme.md

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+[![](https://img.shields.io/github/stars/LeetCode-in-Ruby/LeetCode-in-Ruby?label=Stars&style=flat-square)](https://github.com/LeetCode-in-Ruby/LeetCode-in-Ruby)
+[![](https://img.shields.io/github/forks/LeetCode-in-Ruby/LeetCode-in-Ruby?label=Fork%20me%20on%20GitHub%20&style=flat-square)](https://github.com/LeetCode-in-Ruby/LeetCode-in-Ruby/fork)
+
+## 3\. Longest Substring Without Repeating Characters
+
+Medium
+
+Given a string `s`, find the length of the **longest substring** without repeating characters.
+
+**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. 
+
+**Example 4:**
+
+**Input:** s = ""
+
+**Output:** 0 
+
+**Constraints:**
+
+*   <code>0 <= s.length <= 5 * 10<sup>4</sup></code>
+*   `s` consists of English letters, digits, symbols and spaces.
+
+## Solution
+
+```ruby
+# #Medium #Top_100_Liked_Questions #Top_Interview_Questions #String #Hash_Table #Sliding_Window
+
+# @param {String} s
+# @return {Integer}
+def length_of_longest_substring(s)
+  last_indices = Array.new(256, -1)
+  max_len = 0
+  cur_len = 0
+  start = 0
+
+  s.each_char.with_index do |cur, i|
+    cur = cur.ord
+    if last_indices[cur] < start
+      last_indices[cur] = i
+      cur_len += 1
+    else
+      last_index = last_indices[cur]
+      start = last_index + 1
+      cur_len = i - start + 1
+      last_indices[cur] = i
+    end
+
+    max_len = cur_len if cur_len > max_len
+  end
+
+  max_len
+end
+```

src/main/ruby/g0001_0100/s0003_longest_substring_without_repeating_characters/readme.md

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+[![](https://img.shields.io/github/stars/LeetCode-in-Ruby/LeetCode-in-Ruby?label=Stars&style=flat-square)](https://github.com/LeetCode-in-Ruby/LeetCode-in-Ruby)
+[![](https://img.shields.io/github/forks/LeetCode-in-Ruby/LeetCode-in-Ruby?label=Fork%20me%20on%20GitHub%20&style=flat-square)](https://github.com/LeetCode-in-Ruby/LeetCode-in-Ruby/fork)
+
+## 4\. Median of Two Sorted Arrays
+
+Hard
+
+Given two sorted arrays `nums1` and `nums2` of size `m` and `n` respectively, return **the median** of the two sorted arrays.
+
+The overall run time complexity should be `O(log (m+n))`.
+
+**Example 1:**
+
+**Input:** nums1 = [1,3], nums2 = [2]
+
+**Output:** 2.00000
+
+**Explanation:** merged array = [1,2,3] and median is 2. 
+
+**Example 2:**
+
+**Input:** nums1 = [1,2], nums2 = [3,4]
+
+**Output:** 2.50000
+
+**Explanation:** merged array = [1,2,3,4] and median is (2 + 3) / 2 = 2.5. 
+
+**Example 3:**
+
+**Input:** nums1 = [0,0], nums2 = [0,0]
+
+**Output:** 0.00000 
+
+**Example 4:**
+
+**Input:** nums1 = [], nums2 = [1]
+
+**Output:** 1.00000 
+
+**Example 5:**
+
+**Input:** nums1 = [2], nums2 = []
+
+**Output:** 2.00000 
+
+**Constraints:**
+
+*   `nums1.length == m`
+*   `nums2.length == n`
+*   `0 <= m <= 1000`
+*   `0 <= n <= 1000`
+*   `1 <= m + n <= 2000`
+*   <code>-10<sup>6</sup> <= nums1[i], nums2[i] <= 10<sup>6</sup></code>
+
+## Solution
+
+```ruby
+# @param {Integer[]} nums1
+# @param {Integer[]} nums2
+# @return {Float}
+def find_median_sorted_arrays(nums1, nums2)
+  combined = nums1.concat(nums2)
+  combined = combined.sort
+  var4 = 0
+  if combined.length.odd?
+    var1 = ((combined.length).div(2))
+    result = combined[var1]
+  end
+  if combined.length.even?
+    var2 = ((combined.length).div(2))
+    var3 = (var2) - 1
+    var4 = (combined[var2]) + (combined[var3])
+    result = var4 / (2.to_f)
+  end
+  result
+end
+```

src/main/ruby/g0001_0100/s0004_median_of_two_sorted_arrays/readme.md

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+[![](https://img.shields.io/github/stars/LeetCode-in-Ruby/LeetCode-in-Ruby?label=Stars&style=flat-square)](https://github.com/LeetCode-in-Ruby/LeetCode-in-Ruby)
+[![](https://img.shields.io/github/forks/LeetCode-in-Ruby/LeetCode-in-Ruby?label=Fork%20me%20on%20GitHub%20&style=flat-square)](https://github.com/LeetCode-in-Ruby/LeetCode-in-Ruby/fork)
+
+## 5\. Longest Palindromic Substring
+
+Medium
+
+Given a string `s`, return _the longest palindromic substring_ in `s`.
+
+**Example 1:**
+
+**Input:** s = "babad"
+
+**Output:** "bab" **Note:** "aba" is also a valid answer. 
+
+**Example 2:**
+
+**Input:** s = "cbbd"
+
+**Output:** "bb" 
+
+**Example 3:**
+
+**Input:** s = "a"
+
+**Output:** "a" 
+
+**Example 4:**
+
+**Input:** s = "ac"
+
+**Output:** "a" 
+
+**Constraints:**
+
+*   `1 <= s.length <= 1000`
+*   `s` consist of only digits and English letters.
+
+## Solution
+
+```ruby
+# @param {String} s
+# @return {String}
+def longest_palindrome(s)
+  new_str = Array.new(s.length * 2 + 1, 0)
+  new_str[0] = '#'
+
+  (0...s.length).each do |i|
+    new_str[2 * i + 1] = s[i]
+    new_str[2 * i + 2] = '#'
+  end
+
+  dp = Array.new(new_str.length, 0)
+  friend_center = 0
+  friend_radius = 0
+  lps_center = 0
+  lps_radius = 0
+
+  (0...new_str.length).each do |i|
+    dp[i] = friend_center + friend_radius > i ? [dp[2 * friend_center - i], friend_center + friend_radius - i].min : 1
+
+    while i + dp[i] < new_str.length && i - dp[i] >= 0 && new_str[i + dp[i]] == new_str[i - dp[i]]
+      dp[i] += 1
+    end
+
+    if friend_center + friend_radius < i + dp[i]
+      friend_center = i
+      friend_radius = dp[i]
+    end
+
+    lps_center, lps_radius = i, dp[i] if lps_radius < dp[i]
+  end
+
+  s[(lps_center - lps_radius + 1) / 2, lps_radius - 1]
+end
+```