Updated readmes

Author: javadev

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

@@ -40,8 +40,6 @@ You can return the answer in any order.
 
 **Follow-up:** Can you come up with an algorithm that is less than <code>O(n<sup>2</sup>)</code> time complexity?
 
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 ## Solution
 
 ```python

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

@@ -37,8 +37,6 @@ You may assume the two numbers do not contain any leading zero, except the numbe
 *   `0 <= Node.val <= 9`
 *   It is guaranteed that the list represents a number that does not have leading zeros.
 
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 ## Solution
 
 ```python

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

@@ -42,8 +42,6 @@ Given a string `s`, find the length of the **longest substring** without repeati
 *   <code>0 <= s.length <= 5 * 10<sup>4</sup></code>
 *   `s` consists of English letters, digits, symbols and spaces.
 
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 ## Solution
 
 ```python

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

@@ -52,8 +52,6 @@ The overall run time complexity should be `O(log (m+n))`.
 *   `1 <= m + n <= 2000`
 *   <code>-10<sup>6</sup> <= nums1[i], nums2[i] <= 10<sup>6</sup></code>
 
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 ## Solution
 
 ```python

src/main/python/g0001_0100/s0005_longest_palindromic_substring/readme.md

@@ -36,8 +36,6 @@ Given a string `s`, return _the longest palindromic substring_ in `s`.
 *   `1 <= s.length <= 1000`
 *   `s` consist of only digits and English letters.
 
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 ## Solution
 
 ```python

src/main/python/g0001_0100/s0006_zigzag_conversion/readme.md

@@ -41,8 +41,6 @@ string convert(string s, int numRows);
 *   `s` consists of English letters (lower-case and upper-case), `','` and `'.'`.
 *   `1 <= numRows <= 1000`
 
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 ## Solution
 
 ```python

src/main/python/g0001_0100/s0007_reverse_integer/readme.md

@@ -37,8 +37,6 @@ Given a signed 32-bit integer `x`, return `x` _with its digits reversed_. If rev
 
 *   <code>-2<sup>31</sup> <= x <= 2<sup>31</sup> - 1</code>
 
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 ## Solution
 
 ```python

src/main/python/g0001_0100/s0008_string_to_integer_atoi/readme.md

@@ -115,8 +115,6 @@ Since -91283472332 is less than the lower bound of the range [-2<sup>31</sup>, 2
 *   `0 <= s.length <= 200`
 *   `s` consists of English letters (lower-case and upper-case), digits (`0-9`), `' '`, `'+'`, `'-'`, and `'.'`.
 
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 ## Solution
 
 ```python

src/main/python/g0001_0100/s0009_palindrome_number/readme.md

@@ -43,8 +43,6 @@ An integer is a **palindrome** when it reads the same backward as forward. For e
 
 **Follow up:** Could you solve it without converting the integer to a string?
 
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 ## Solution
 
 ```python

src/main/python/g0001_0100/s0010_regular_expression_matching/readme.md

@@ -58,8 +58,6 @@ The matching should cover the **entire** input string (not partial).
 *   `p` contains only lowercase English letters, `'.'`, and `'*'`.
 *   It is guaranteed for each appearance of the character `'*'`, there will be a previous valid character to match.
 
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 ## Solution
 
 ```python

src/main/python/g0001_0100/s0011_container_with_most_water/readme.md

@@ -43,8 +43,6 @@ Given `n` non-negative integers <code>a<sub>1</sub>, a<sub>2</sub>, ..., a<sub>n
 *   <code>2 <= n <= 10<sup>5</sup></code>
 *   <code>0 <= height[i] <= 10<sup>4</sup></code>
 
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 ## Solution
 
 ```python

src/main/python/g0001_0100/s0015_3sum/readme.md

@@ -32,8 +32,6 @@ Notice that the solution set must not contain duplicate triplets.
 *   `0 <= nums.length <= 3000`
 *   <code>-10<sup>5</sup> <= nums[i] <= 10<sup>5</sup></code>
 
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 ## Solution
 
 ```python

src/main/python/g0001_0100/s0017_letter_combinations_of_a_phone_number/readme.md

@@ -34,8 +34,6 @@ A mapping of digit to letters (just like on the telephone buttons) is given belo
 *   `0 <= digits.length <= 4`
 *   `digits[i]` is a digit in the range `['2', '9']`.
 
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 ## Solution
 
 ```python

src/main/python/g0001_0100/s0019_remove_nth_node_from_end_of_list/readme.md

@@ -36,8 +36,6 @@ Given the `head` of a linked list, remove the `nth` node from the end of the lis
 
 **Follow up:** Could you do this in one pass?
 
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 ## Solution
 
 ```python

src/main/python/g0001_0100/s0020_valid_parentheses/readme.md

@@ -47,8 +47,6 @@ An input string is valid if:
 *   <code>1 <= s.length <= 10<sup>4</sup></code>
 *   `s` consists of parentheses only `'()[]{}'`.
 
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 ## Solution
 
 ```python

src/main/python/g0001_0100/s0021_merge_two_sorted_lists/readme.md

@@ -33,8 +33,6 @@ Merge two sorted linked lists and return it as a **sorted** list. The list shoul
 *   `-100 <= Node.val <= 100`
 *   Both `l1` and `l2` are sorted in **non-decreasing** order.
 
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 ## Solution
 
 ```python

src/main/python/g0001_0100/s0022_generate_parentheses/readme.md

@@ -23,8 +23,6 @@ Given `n` pairs of parentheses, write a function to _generate all combinations o
 
 *   `1 <= n <= 8`
 
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 ## Solution
 
 ```python

src/main/python/g0001_0100/s0023_merge_k_sorted_lists/readme.md

@@ -38,8 +38,6 @@ _Merge all the linked-lists into one sorted linked-list and return it._
 *   `lists[i]` is sorted in **ascending order**.
 *   The sum of `lists[i].length` won't exceed `10^4`.
 
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 ## Solution
 
 ```python

src/main/python/g0001_0100/s0024_swap_nodes_in_pairs/readme.md

@@ -32,8 +32,6 @@ Given a linked list, swap every two adjacent nodes and return its head. You must
 *   The number of nodes in the list is in the range `[0, 100]`.
 *   `0 <= Node.val <= 100`
 
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 ## Solution
 
 ```python

src/main/python/g0001_0100/s0025_reverse_nodes_in_k_group/readme.md

@@ -48,8 +48,6 @@ You may not alter the values in the list's nodes, only nodes themselves may be c
 
 **Follow-up:** Can you solve the problem in O(1) extra memory space?
 
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 ## Solution
 
 ```python

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

@@ -40,8 +40,6 @@ The replacement must be **[in place](http://en.wikipedia.org/wiki/In-place_algor
 *   `1 <= nums.length <= 100`
 *   `0 <= nums[i] <= 100`
 
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 ## Solution
 
 ```python

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

@@ -34,8 +34,6 @@ Given a string containing just the characters `'('` and `')'`, find the length o
 *   <code>0 <= s.length <= 3 * 10<sup>4</sup></code>
 *   `s[i]` is `'('`, or `')'`.
 
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 ## Solution
 
 ```python

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

@@ -39,8 +39,6 @@ You must write an algorithm with `O(log n)` runtime complexity.
 *   `nums` is an ascending array that is possibly rotated.
 *   <code>-10<sup>4</sup> <= target <= 10<sup>4</sup></code>
 
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 ## Solution
 
 ```python

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

@@ -36,7 +36,73 @@ You must write an algorithm with `O(log n)` runtime complexity.
 *   `nums` is a non-decreasing array.
 *   <code>-10<sup>9</sup> <= target <= 10<sup>9</sup></code>
 
-To solve this problem efficiently with a runtime complexity of O(log n), we can use binary search to find the starting and ending positions of the target value in the sorted array. 
+To solve this problem efficiently with a runtime complexity of O(log n), we can use binary search to find the starting and ending positions of the target value in the sorted array. Here are the steps:
+
+### Approach:
+
+1. **Binary Search for Starting Position:**
+   - Use binary search to find the starting position of the target value.
+   - When the target value is found, continue searching towards the left to find the leftmost occurrence.
+   - Update the left boundary to narrow down the search space.
+
+2. **Binary Search for Ending Position:**
+   - Use binary search to find the ending position of the target value.
+   - When the target value is found, continue searching towards the right to find the rightmost occurrence.
+   - Update the right boundary to narrow down the search space.
+
+3. **Handle Missing Target:**
+   - If the target value is not found, return [-1, -1].
+
+### Python Code:
+
+```python
+class Solution:
+    def searchRange(self, nums, target):
+        def search_boundary(nums, target, is_left):
+            left, right = 0, len(nums) - 1
+            boundary = -1
+
+            while left <= right:
+                mid = (left + right) // 2
+
+                if nums[mid] == target:
+                    boundary = mid
+                    if is_left:
+                        right = mid - 1
+                    else:
+                        left = mid + 1
+                elif nums[mid] < target:
+                    left = mid + 1
+                else:
+                    right = mid - 1
+
+            return boundary
+
+        left_boundary = search_boundary(nums, target, is_left=True)
+        right_boundary = search_boundary(nums, target, is_left=False)
+
+        return [left_boundary, right_boundary]
+
+# Example Usage:
+solution = Solution()
+
+# Example 1:
+nums1 = [5, 7, 7, 8, 8, 10]
+target1 = 8
+print(solution.searchRange(nums1, target1))  # Output: [3, 4]
+
+# Example 2:
+nums2 = [5, 7, 7, 8, 8, 10]
+target2 = 6
+print(solution.searchRange(nums2, target2))  # Output: [-1, -1]
+
+# Example 3:
+nums3 = []
+target3 = 0
+print(solution.searchRange(nums3, target3))  # Output: [-1, -1]
+```
+
+This code defines a `Solution` class with a `searchRange` method to find the starting and ending positions of the target value in the given array. The example usage demonstrates how to create an instance of the `Solution` class and call the `searchRange` method with different inputs.
 
 ## Solution
 

src/main/python/g0001_0100/s0035_search_insert_position/readme.md

@@ -46,7 +46,65 @@ You must write an algorithm with `O(log n)` runtime complexity.
 *   `nums` contains **distinct** values sorted in **ascending** order.
 *   <code>-10<sup>4</sup> <= target <= 10<sup>4</sup></code>
 
-To solve this problem efficiently with a runtime complexity of O(log n), we can use binary search to find the insertion position of the target value in the sorted array. 
+To solve this problem efficiently with a runtime complexity of O(log n), we can use binary search to find the insertion position of the target value in the sorted array. Here are the steps:
+
+### Approach:
+
+1. **Binary Search:**
+   - Implement binary search to find the insertion position of the target value.
+   - If the target value is found, return its index.
+   - If the target value is not found, binary search will determine the index where the target would be inserted to maintain the sorted order.
+
+### Python Code:
+
+```python
+class Solution:
+    def searchInsert(self, nums, target):
+        left, right = 0, len(nums) - 1
+
+        while left <= right:
+            mid = (left + right) // 2
+
+            if nums[mid] == target:
+                return mid
+            elif nums[mid] < target:
+                left = mid + 1
+            else:
+                right = mid - 1
+
+        # If the target is not found, return the insertion position (left).
+        return left
+
+# Example Usage:
+solution = Solution()
+
+# Example 1:
+nums1 = [1, 3, 5, 6]
+target1 = 5
+print(solution.searchInsert(nums1, target1))  # Output: 2
+
+# Example 2:
+nums2 = [1, 3, 5, 6]
+target2 = 2
+print(solution.searchInsert(nums2, target2))  # Output: 1
+
+# Example 3:
+nums3 = [1, 3, 5, 6]
+target3 = 7
+print(solution.searchInsert(nums3, target3))  # Output: 4
+
+# Example 4:
+nums4 = [1, 3, 5, 6]
+target4 = 0
+print(solution.searchInsert(nums4, target4))  # Output: 0
+
+# Example 5:
+nums5 = [1]
+target5 = 0
+print(solution.searchInsert(nums5, target5))  # Output: 0
+```
+
+This code defines a `Solution` class with a `searchInsert` method to find the insertion position of the target value in the given sorted array. The example usage demonstrates how to create an instance of the `Solution` class and call the `searchInsert` method with different inputs.
 
 ## Solution
 

src/main/python/g0001_0100/s0039_combination_sum/readme.md

@@ -54,8 +54,6 @@ It is **guaranteed** that the number of unique combinations that sum up to `targ
 *   All elements of `candidates` are **distinct**.
 *   `1 <= target <= 500`
 
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 ## Solution
 
 ```python

src/main/python/g0001_0100/s0041_first_missing_positive/readme.md

@@ -32,8 +32,6 @@ You must implement an algorithm that runs in `O(n)` time and uses constant extra
 *   <code>1 <= nums.length <= 5 * 10<sup>5</sup></code>
 *   <code>-2<sup>31</sup> <= nums[i] <= 2<sup>31</sup> - 1</code>
 
-
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 ## Solution
 
 ```python

src/main/python/g0001_0100/s0042_trapping_rain_water/readme.md

@@ -29,8 +29,6 @@ Given `n` non-negative integers representing an elevation map where the width of
 *   <code>1 <= n <= 2 * 10<sup>4</sup></code>
 *   <code>0 <= height[i] <= 10<sup>5</sup></code>
 
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 ## Solution
 
 ```python

src/main/python/g0001_0100/s0045_jump_game_ii/readme.md

@@ -32,8 +32,6 @@ You can assume that you can always reach the last index.
 *   <code>1 <= nums.length <= 10<sup>4</sup></code>
 *   `0 <= nums[i] <= 1000`
 
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 ## Solution
 
 ```python

src/main/python/g0001_0100/s0046_permutations/readme.md

@@ -31,8 +31,6 @@ Given an array `nums` of distinct integers, return _all the possible permutation
 *   `-10 <= nums[i] <= 10`
 *   All the integers of `nums` are **unique**.
 
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 ## Solution
 
 ```python