Updated tasks 73-102

Author: ThanhNIT

src/main/java/g0001_0100/s0073_set_matrix_zeroes/readme.md

@@ -2,7 +2,7 @@
 
 Medium
 
-Given an `m x n` integer matrix `matrix`, if an element is `0`, set its entire row and column to `0`'s, and return _the matrix_.
+Given an `m x n` integer matrix `matrix`, if an element is `0`, set its entire row and column to `0`'s.
 
 You must do it [in place](https://en.wikipedia.org/wiki/In-place_algorithm).
 

src/main/java/g0001_0100/s0074_search_a_2d_matrix/readme.md

@@ -2,11 +2,15 @@
 
 Medium
 
-Write an efficient algorithm that searches for a value in an `m x n` matrix. This matrix has the following properties:
+You are given an `m x n` integer matrix `matrix` with the following two properties:
 
-*   Integers in each row are sorted from left to right.
+*   Each row is sorted in non-decreasing order.
 *   The first integer of each row is greater than the last integer of the previous row.
 
+Given an integer `target`, return `true` _if_ `target` _is in_ `matrix` _or_ `false` _otherwise_.
+
+You must write a solution in `O(log(m * n))` time complexity.
+
 **Example 1:**
 
 ![](https://assets.leetcode.com/uploads/2020/10/05/mat.jpg)

src/main/java/g0001_0100/s0075_sort_colors/readme.md

@@ -20,23 +20,11 @@ You must solve this problem without using the library's sort function.
 
 **Output:** [0,1,2] 
 
-**Example 3:**
-
-**Input:** nums = [0]
-
-**Output:** [0] 
-
-**Example 4:**
-
-**Input:** nums = [1]
-
-**Output:** [1] 
-
 **Constraints:**
 
 *   `n == nums.length`
 *   `1 <= n <= 300`
-*   `nums[i]` is `0`, `1`, or `2`.
+*   `nums[i]` is either `0`, `1`, or `2`.
 
 **Follow up:** Could you come up with a one-pass algorithm using only constant extra space?
 

src/main/java/g0001_0100/s0076_minimum_window_substring/readme.md

@@ -2,12 +2,10 @@
 
 Hard
 
-Given two strings `s` and `t` of lengths `m` and `n` respectively, return _the **minimum window substring** of_ `s` _such that every character in_ `t` _(**including duplicates**) is included in the window. If there is no such substring__, return the empty string_ `""`_._
+Given two strings `s` and `t` of lengths `m` and `n` respectively, return _the **minimum window**_ **substring** _of_ `s` _such that every character in_ `t` _(**including duplicates**) is included in the window_. If there is no such substring, return _the empty string_ `""`.
 
 The testcases will be generated such that the answer is **unique**.
 
-A **substring** is a contiguous sequence of characters within the string.
-
 **Example 1:**
 
 **Input:** s = "ADOBECODEBANC", t = "ABC"

src/main/java/g0001_0100/s0077_combinations/readme.md

@@ -2,21 +2,25 @@
 
 Medium
 
-Given two integers `n` and `k`, return _all possible combinations of_ `k` _numbers out of the range_ `[1, n]`.
+Given two integers `n` and `k`, return _all possible combinations of_ `k` _numbers chosen from the range_ `[1, n]`.
 
 You may return the answer in **any order**.
 
 **Example 1:**
 
 **Input:** n = 4, k = 2
 
-**Output:** [ [2,4], [3,4], [2,3], [1,2], [1,3], [1,4], ] 
+**Output:** [[1,2],[1,3],[1,4],[2,3],[2,4],[3,4]]
+
+**Explanation:** There are 4 choose 2 = 6 total combinations. Note that combinations are unordered, i.e., [1,2] and [2,1] are considered to be the same combination. 
 
 **Example 2:**
 
 **Input:** n = 1, k = 1
 
-**Output:** [[1]] 
+**Output:** [[1]]
+
+**Explanation:** There is 1 choose 1 = 1 total combination. 
 
 **Constraints:**
 

src/main/java/g0001_0100/s0078_subsets/readme.md

@@ -2,7 +2,7 @@
 
 Medium
 
-Given an integer array `nums` of **unique** elements, return _all possible subsets (the power set)_.
+Given an integer array `nums` of **unique** elements, return _all possible_ **subset** _(the power set)_.
 
 The solution set **must not** contain duplicate subsets. Return the solution in **any order**.
 

src/main/java/g0001_0100/s0080_remove_duplicates_from_sorted_array_ii/readme.md

@@ -18,7 +18,7 @@ The judge will test your solution with the following code:
     int[] expectedNums = [...]; // The expected answer with correct length
 
     int k = removeDuplicates(nums); // Calls your implementation
-    
+
     assert k == expectedNums.length;
     for (int i = 0; i < k; i++) {
         assert nums[i] == expectedNums[i];
@@ -46,4 +46,4 @@ If all assertions pass, then your solution will be **accepted**.
 
 *   <code>1 <= nums.length <= 3 * 10<sup>4</sup></code>
 *   <code>-10<sup>4</sup> <= nums[i] <= 10<sup>4</sup></code>
-*   `nums` is sorted in **non-decreasing** order.
+*   `nums` is sorted in **non-decreasing** order.

src/main/java/g0001_0100/s0088_merge_sorted_array/readme.md

@@ -14,7 +14,7 @@ The final sorted array should not be returned by the function, but instead be _s
 
 **Output:** [1,2,2,3,5,6]
 
-**Explanation:** The arrays we are merging are [1,2,3] and [2,5,6]. The result of the merge is [1,2,2,3,5,6] with the underlined elements coming from nums1. 
+**Explanation:** The arrays we are merging are [1,2,3] and [2,5,6]. The result of the merge is [<ins>1</ins>,<ins>2</ins>,2,<ins>3</ins>,5,6] with the underlined elements coming from nums1. 
 
 **Example 2:**
 

src/main/java/g0001_0100/s0094_binary_tree_inorder_traversal/readme.md

@@ -6,39 +6,35 @@ Given the `root` of a binary tree, return _the inorder traversal of its nodes' v
 
 **Example 1:**
 
-![](https://assets.leetcode.com/uploads/2020/09/15/inorder_1.jpg)
-
 **Input:** root = [1,null,2,3]
 
-**Output:** [1,3,2] 
-
-**Example 2:**
+**Output:** [1,3,2]
 
-**Input:** root = []
+**Explanation:**
 
-**Output:** [] 
+![](https://assets.leetcode.com/uploads/2024/08/29/screenshot-2024-08-29-202743.png)
 
-**Example 3:**
+**Example 2:**
 
-**Input:** root = [1]
+**Input:** root = [1,2,3,4,5,null,8,null,null,6,7,9]
 
-**Output:** [1] 
+**Output:** [4,2,6,5,7,1,3,9,8]
 
-**Example 4:**
+**Explanation:**
 
-![](https://assets.leetcode.com/uploads/2020/09/15/inorder_5.jpg)
+![](https://assets.leetcode.com/uploads/2024/08/29/tree_2.png)
 
-**Input:** root = [1,2]
+**Example 3:**
 
-**Output:** [2,1] 
+**Input:** root = []
 
-**Example 5:**
+**Output:** []
 
-![](https://assets.leetcode.com/uploads/2020/09/15/inorder_4.jpg)
+**Example 4:**
 
-**Input:** root = [1,null,2]
+**Input:** root = [1]
 
-**Output:** [1,2] 
+**Output:** [1]
 
 **Constraints:**
 

src/main/java/g0001_0100/s0097_interleaving_string/readme.md

@@ -4,7 +4,7 @@ Medium
 
 Given strings `s1`, `s2`, and `s3`, find whether `s3` is formed by an **interleaving** of `s1` and `s2`.
 
-An **interleaving** of two strings `s` and `t` is a configuration where they are divided into **non-empty** substrings such that:
+An **interleaving** of two strings `s` and `t` is a configuration where `s` and `t` are divided into `n` and `m` **substring** respectively, such that:
 
 *   <code>s = s<sub>1</sub> + s<sub>2</sub> + ... + s<sub>n</sub></code>
 *   <code>t = t<sub>1</sub> + t<sub>2</sub> + ... + t<sub>m</sub></code>
@@ -19,13 +19,17 @@ An **interleaving** of two strings `s` and `t` is a configuration where they are
 
 **Input:** s1 = "aabcc", s2 = "dbbca", s3 = "aadbbcbcac"
 
-**Output:** true 
+**Output:** true
+
+**Explanation:** One way to obtain s3 is: Split s1 into s1 = "aa" + "bc" + "c", and s2 into s2 = "dbbc" + "a". Interleaving the two splits, we get "aa" + "dbbc" + "bc" + "a" + "c" = "aadbbcbcac". Since s3 can be obtained by interleaving s1 and s2, we return true. 
 
 **Example 2:**
 
 **Input:** s1 = "aabcc", s2 = "dbbca", s3 = "aadbbbaccc"
 
-**Output:** false 
+**Output:** false
+
+**Explanation:** Notice how it is impossible to interleave s2 with any other string to obtain s3. 
 
 **Example 3:**