Added js tasks

Author: javadev

src/main/js/g0201_0300/s0230_kth_smallest_element_in_a_bst/readme.md

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+230\. Kth Smallest Element in a BST
+
+Medium
+
+Given the `root` of a binary search tree, and an integer `k`, return _the_ <code>k<sup>th</sup></code> _smallest value (**1-indexed**) of all the values of the nodes in the tree_.
+
+**Example 1:**
+
+![](https://assets.leetcode.com/uploads/2021/01/28/kthtree1.jpg)
+
+**Input:** root = [3,1,4,null,2], k = 1
+
+**Output:** 1
+
+**Example 2:**
+
+![](https://assets.leetcode.com/uploads/2021/01/28/kthtree2.jpg)
+
+**Input:** root = [5,3,6,2,4,null,null,1], k = 3
+
+**Output:** 3
+
+**Constraints:**
+
+*   The number of nodes in the tree is `n`.
+*   <code>1 <= k <= n <= 10<sup>4</sup></code>
+*   <code>0 <= Node.val <= 10<sup>4</sup></code>
+
+**Follow up:** If the BST is modified often (i.e., we can do insert and delete operations) and you need to find the kth smallest frequently, how would you optimize?

src/main/js/g0201_0300/s0230_kth_smallest_element_in_a_bst/solution.js

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+// #Medium #Top_100_Liked_Questions #Depth_First_Search #Tree #Binary_Tree #Binary_Search_Tree
+// #Data_Structure_II_Day_17_Tree #Level_2_Day_9_Binary_Search_Tree #Big_O_Time_O(n)_Space_O(n)
+// #2024_12_21_Time_0_ms_(100.00%)_Space_55.4_MB_(48.16%)
+
+/**
+ * Definition for a binary tree node.
+ * function TreeNode(val, left, right) {
+ *     this.val = (val===undefined ? 0 : val)
+ *     this.left = (left===undefined ? null : left)
+ *     this.right = (right===undefined ? null : right)
+ * }
+ */
+/**
+ * @param {TreeNode} root
+ * @param {number} k
+ * @return {number}
+ */
+var kthSmallest = function(root, k) {
+    let count = 0
+    let result = null
+
+    const calculate = (node) => {
+        if (!node) return
+        if (node.left) calculate(node.left)
+
+        count++
+        if (count === k) {
+            result = node.val
+            return
+        }
+
+        if (node.right) calculate(node.right)
+    }
+
+    calculate(root)
+    return result
+};
+
+export { kthSmallest }

src/main/js/g0201_0300/s0234_palindrome_linked_list/readme.md

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+234\. Palindrome Linked List
+
+Easy
+
+Given the `head` of a singly linked list, return `true` _if it is a palindrome or_ `false` _otherwise_.
+
+**Example 1:**
+
+![](https://assets.leetcode.com/uploads/2021/03/03/pal1linked-list.jpg)
+
+**Input:** head = [1,2,2,1]
+
+**Output:** true
+
+**Example 2:**
+
+![](https://assets.leetcode.com/uploads/2021/03/03/pal2linked-list.jpg)
+
+**Input:** head = [1,2]
+
+**Output:** false
+
+**Constraints:**
+
+*   The number of nodes in the list is in the range <code>[1, 10<sup>5</sup>]</code>.
+*   `0 <= Node.val <= 9`
+
+**Follow up:** Could you do it in `O(n)` time and `O(1)` space?

src/main/js/g0201_0300/s0234_palindrome_linked_list/solution.js

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+// #Easy #Top_100_Liked_Questions #Two_Pointers #Stack #Linked_List #Recursion
+// #Level_2_Day_3_Linked_List #Udemy_Linked_List #Big_O_Time_O(n)_Space_O(1)
+// #2024_12_21_Time_3_ms_(93.71%)_Space_69.8_MB_(76.50%)
+
+/**
+ * Definition for singly-linked list.
+ * function ListNode(val, next) {
+ *     this.val = (val===undefined ? 0 : val)
+ *     this.next = (next===undefined ? null : next)
+ * }
+ */
+/**
+ * @param {ListNode} head
+ * @return {boolean}
+ */
+var isPalindrome = function(head) {
+    let len = 0
+    let right = head
+
+    // Calculate the length of the linked list
+    while (right !== null) {
+        right = right.next
+        len++
+    }
+
+    // Move to the start of the second half of the list
+    len = Math.floor(len / 2)
+    right = head
+    for (let i = 0; i < len; i++) {
+        right = right.next
+    }
+
+    // Reverse the right half of the list
+    let prev = null
+    while (right !== null) {
+        let next = right.next
+        right.next = prev
+        prev = right
+        right = next
+    }
+
+    // Compare the left half and the reversed right half
+    for (let i = 0; i < len; i++) {
+        if (prev !== null && head.val === prev.val) {
+            head = head.next
+            prev = prev.next
+        } else {
+            return false
+        }
+    }
+    return true
+};
+
+export { isPalindrome }

src/main/js/g0201_0300/s0236_lowest_common_ancestor_of_a_binary_tree/readme.md

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+236\. Lowest Common Ancestor of a Binary Tree
+
+Medium
+
+Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.
+
+According to the [definition of LCA on Wikipedia](https://en.wikipedia.org/wiki/Lowest_common_ancestor): “The lowest common ancestor is defined between two nodes `p` and `q` as the lowest node in `T` that has both `p` and `q` as descendants (where we allow **a node to be a descendant of itself**).”
+
+**Example 1:**
+
+![](https://assets.leetcode.com/uploads/2018/12/14/binarytree.png)
+
+**Input:** root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
+
+**Output:** 3
+
+**Explanation:** The LCA of nodes 5 and 1 is 3.
+
+**Example 2:**
+
+![](https://assets.leetcode.com/uploads/2018/12/14/binarytree.png)
+
+**Input:** root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
+
+**Output:** 5
+
+**Explanation:** The LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.
+
+**Example 3:**
+
+**Input:** root = [1,2], p = 1, q = 2
+
+**Output:** 1
+
+**Constraints:**
+
+*   The number of nodes in the tree is in the range <code>[2, 10<sup>5</sup>]</code>.
+*   <code>-10<sup>9</sup> <= Node.val <= 10<sup>9</sup></code>
+*   All `Node.val` are **unique**.
+*   `p != q`
+*   `p` and `q` will exist in the tree.

src/main/js/g0201_0300/s0236_lowest_common_ancestor_of_a_binary_tree/solution.js

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+// #Medium #Top_100_Liked_Questions #Depth_First_Search #Tree #Binary_Tree
+// #Data_Structure_II_Day_18_Tree #Udemy_Tree_Stack_Queue #Big_O_Time_O(n)_Space_O(n)
+// #2024_12_21_Time_53_ms_(98.59%)_Space_58.7_MB_(88.33%)
+
+/**
+ * Definition for a binary tree node.
+ * function TreeNode(val) {
+ *     this.val = val;
+ *     this.left = this.right = null;
+ * }
+ */
+/**
+ * @param {TreeNode} root
+ * @param {TreeNode} p
+ * @param {TreeNode} q
+ * @return {TreeNode}
+ */
+var lowestCommonAncestor = function(root, p, q) {
+    if(!root) {
+        return null
+    }
+    if(p.val === root.val || q.val === root.val) {
+        return root
+    }
+    const leftNode = lowestCommonAncestor(root.left, p, q)
+    const rightNode = lowestCommonAncestor(root.right, p, q)
+    if(leftNode && rightNode) {
+        return root
+    }
+    return leftNode || rightNode
+};
+
+export { lowestCommonAncestor }

src/main/js/g0201_0300/s0238_product_of_array_except_self/readme.md

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+238\. Product of Array Except Self
+
+Medium
+
+Given an integer array `nums`, return _an array_ `answer` _such that_ `answer[i]` _is equal to the product of all the elements of_ `nums` _except_ `nums[i]`.
+
+The product of any prefix or suffix of `nums` is **guaranteed** to fit in a **32-bit** integer.
+
+You must write an algorithm that runs in `O(n)` time and without using the division operation.
+
+**Example 1:**
+
+**Input:** nums = [1,2,3,4]
+
+**Output:** [24,12,8,6]
+
+**Example 2:**
+
+**Input:** nums = [-1,1,0,-3,3]
+
+**Output:** [0,0,9,0,0]
+
+**Constraints:**
+
+*   <code>2 <= nums.length <= 10<sup>5</sup></code>
+*   `-30 <= nums[i] <= 30`
+*   The product of any prefix or suffix of `nums` is **guaranteed** to fit in a **32-bit** integer.
+
+**Follow up:** Can you solve the problem in `O(1) `extra space complexity? (The output array **does not** count as extra space for space complexity analysis.)

src/main/js/g0201_0300/s0238_product_of_array_except_self/solution.js

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+// #Medium #Top_100_Liked_Questions #Array #Prefix_Sum #Data_Structure_II_Day_5_Array #Udemy_Arrays
+// #Big_O_Time_O(n^2)_Space_O(n) #2024_12_21_Time_3_ms_(93.60%)_Space_64.9_MB_(45.67%)
+
+/**
+ * @param {number[]} nums
+ * @return {number[]}
+ */
+var productExceptSelf = function(nums) {
+    const res = new Array(nums.length).fill(1)
+
+    // Compute prefix product
+    let prefixProduct = 1
+    for (let i = 0; i < nums.length; i++) {
+        res[i] = prefixProduct
+        prefixProduct *= nums[i]
+    }
+
+    // Compute suffix product and multiply with prefix product
+    let suffixProduct = 1
+    for (let i = nums.length - 1; i >= 0; i--) {
+        res[i] *= suffixProduct
+        suffixProduct *= nums[i]
+    }
+
+    return res
+};
+
+export { productExceptSelf }

src/main/js/g0201_0300/s0239_sliding_window_maximum/readme.md

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+239\. Sliding Window Maximum
+
+Hard
+
+You are given an array of integers `nums`, there is a sliding window of size `k` which is moving from the very left of the array to the very right. You can only see the `k` numbers in the window. Each time the sliding window moves right by one position.
+
+Return _the max sliding window_.
+
+**Example 1:**
+
+**Input:** nums = [1,3,-1,-3,5,3,6,7], k = 3
+
+**Output:** [3,3,5,5,6,7]
+
+**Explanation:** 
+
+Window position Max 
+
+--------------- ----- 
+
+[1 3 -1] -3 5 3 6 7 **3** 
+
+1 [3 -1 -3] 5 3 6 7 **3** 
+
+1 3 [-1 -3 5] 3 6 7 **5** 
+
+1 3 -1 [-3 5 3] 6 7 **5** 
+
+1 3 -1 -3 [5 3 6] 7 **6** 
+
+1 3 -1 -3 5 [3 6 7] **7**
+
+**Example 2:**
+
+**Input:** nums = [1], k = 1
+
+**Output:** [1]
+
+**Constraints:**
+
+*   <code>1 <= nums.length <= 10<sup>5</sup></code>
+*   <code>-10<sup>4</sup> <= nums[i] <= 10<sup>4</sup></code>
+*   `1 <= k <= nums.length`

src/main/js/g0201_0300/s0239_sliding_window_maximum/solution.js

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+// #Hard #Top_100_Liked_Questions #Array #Heap_Priority_Queue #Sliding_Window #Queue
+// #Monotonic_Queue #Udemy_Arrays #Big_O_Time_O(n*k)_Space_O(n+k)
+// #2024_12_21_Time_28_ms_(98.27%)_Space_83.4_MB_(17.86%)
+
+/**
+ * @param {number[]} nums
+ * @param {number} k
+ * @return {number[]}
+ */
+var maxSlidingWindow = function (nums, k) {
+    // initialize an empty deqeue
+    let dq = []
+    // initailize two pointers with 0
+    let i = 0, j = 0
+
+    // intiailize an empty result variable
+    let res = []
+
+    // loop till j reaches to the end of loop
+    while (j < nums.length) {
+
+        // if dq is empty, push j
+        if (dq.length === 0) {
+            dq.push(j)
+        } else {
+            // if dq is non empty & back element in deqeue is less or equals to nums[j],
+            // pop back element from deqeue
+            while (dq.length !== 0 && nums[dq[dq.length - 1]] <= nums[j]) {
+                dq.pop()
+            }
+
+            // push j in deqeue
+            dq.push(j)
+        }
+
+        // if current window size is equals to k,
+        // push front deqeue element in res
+        if ((j - i + 1) === k) {
+            res.push(nums[dq[0]])
+
+            // if front deqeue element is equals to i,
+            // remove front deqeue element
+            if (dq[0] === i) {
+                dq.shift()
+            }
+
+            // increment i by 1
+            i++
+        }
+
+        // increment j by 1
+        j++
+    }
+
+    // return result
+    return res 
+};
+
+export { maxSlidingWindow }