Added javascript tasks

src/main/js/com_github_leetcode/treenode.js

@@ -0,0 +1,33 @@
+class TreeNode { // NOSONAR
+    constructor(val = 0, left = null, right = null) {
+        this.val = val
+        this.left = left
+        this.right = right
+    }
+};
+
+function createTreeNode(treeValues) {
+    if (treeValues.length === 0) return null
+
+    const root = new TreeNode(treeValues[0])
+    const queue = [root]
+    let i = 1
+
+    while (i < treeValues.length) {
+        const curr = queue.shift()
+        if (treeValues[i] !== null) {
+            curr.left = new TreeNode(treeValues[i])
+            queue.push(curr.left)
+        }
+        i++
+        if (i < treeValues.length && treeValues[i] !== null) {
+            curr.right = new TreeNode(treeValues[i])
+            queue.push(curr.right)
+        }
+        i++
+    }
+
+    return root
+};
+
+export { TreeNode, createTreeNode }

src/main/js/g0001_0100/s0021_merge_two_sorted_lists/solution.js

@@ -3,7 +3,7 @@
 // #Level_1_Day_3_Linked_List #Udemy_Linked_List #Big_O_Time_O(m+n)_Space_O(m+n)
 // #2024_12_04_Time_0_ms_(100.00%)_Space_52.3_MB_(20.64%)
 
-import { ListNode } from "../../com_github_leetcode/listnode";
+import { ListNode } from '../../com_github_leetcode/listnode';
 
 /**
  * Definition for singly-linked list.

src/main/js/g0001_0100/s0056_merge_intervals/readme.md

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+56\. Merge Intervals
+
+Medium
+
+Given an array of `intervals` where <code>intervals[i] = [start<sub>i</sub>, end<sub>i</sub>]</code>, merge all overlapping intervals, and return _an array of the non-overlapping intervals that cover all the intervals in the input_.
+
+**Example 1:**
+
+**Input:** intervals = [[1,3],[2,6],[8,10],[15,18]]
+
+**Output:** [[1,6],[8,10],[15,18]]
+
+**Explanation:** Since intervals [1,3] and [2,6] overlap, merge them into [1,6].
+
+**Example 2:**
+
+**Input:** intervals = [[1,4],[4,5]]
+
+**Output:** [[1,5]]
+
+**Explanation:** Intervals [1,4] and [4,5] are considered overlapping.
+
+**Constraints:**
+
+*   <code>1 <= intervals.length <= 10<sup>4</sup></code>
+*   `intervals[i].length == 2`
+*   <code>0 <= start<sub>i</sub> <= end<sub>i</sub> <= 10<sup>4</sup></code>

src/main/js/g0001_0100/s0056_merge_intervals/solution.js

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+// #Medium #Top_100_Liked_Questions #Top_Interview_Questions #Array #Sorting
+// #Data_Structure_II_Day_2_Array #Level_2_Day_17_Interval #Udemy_2D_Arrays/Matrix
+// #Big_O_Time_O(n_log_n)_Space_O(n) #2024_12_10_Time_7_ms_(81.68%)_Space_59.2_MB_(41.78%)
+
+/**
+ * @param {number[][]} intervals
+ * @return {number[][]}
+ */
+var merge = function(intervals) {
+    // Sort intervals based on the starting points
+    intervals.sort((a, b) => a[0] - b[0])
+
+    const result = []
+    let current = intervals[0]
+    result.push(current)
+
+    for (const next of intervals) {
+        if (current[1] >= next[0]) {
+            // Merge intervals
+            current[1] = Math.max(current[1], next[1])
+        } else {
+            // Move to the next interval
+            current = next;
+            result.push(current)
+        }
+    }
+
+    return result;
+};
+
+export { merge }

src/main/js/g0001_0100/s0062_unique_paths/readme.md

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+62\. Unique Paths
+
+Medium
+
+There is a robot on an `m x n` grid. The robot is initially located at the **top-left corner** (i.e., `grid[0][0]`). The robot tries to move to the **bottom-right corner** (i.e., `grid[m - 1][n - 1]`). The robot can only move either down or right at any point in time.
+
+Given the two integers `m` and `n`, return _the number of possible unique paths that the robot can take to reach the bottom-right corner_.
+
+The test cases are generated so that the answer will be less than or equal to <code>2 * 10<sup>9</sup></code>.
+
+**Example 1:**
+
+![](https://assets.leetcode.com/uploads/2018/10/22/robot_maze.png)
+
+**Input:** m = 3, n = 7
+
+**Output:** 28
+
+**Example 2:**
+
+**Input:** m = 3, n = 2
+
+**Output:** 3
+
+**Explanation:** From the top-left corner, there are a total of 3 ways to reach the bottom-right corner: 
+
+1. Right -> Down -> Down 
+
+2. Down -> Down -> Right 
+
+3. Down -> Right -> Down
+
+**Constraints:**
+
+*   `1 <= m, n <= 100`

src/main/js/g0001_0100/s0062_unique_paths/solution.js

@@ -0,0 +1,34 @@
+// #Medium #Top_100_Liked_Questions #Top_Interview_Questions #Dynamic_Programming #Math
+// #Combinatorics #Algorithm_II_Day_13_Dynamic_Programming #Dynamic_Programming_I_Day_15
+// #Level_1_Day_11_Dynamic_Programming #Big_O_Time_O(m*n)_Space_O(m*n)
+// #2024_12_10_Time_0_ms_(100.00%)_Space_49.1_MB_(57.14%)
+
+/**
+ * @param {number} m
+ * @param {number} n
+ * @return {number}
+ */
+var uniquePaths = function(m, n) {
+    // Initialize a 2D array with all values set to 0
+    const dp = Array.from({ length: m }, () => Array(n).fill(0))
+
+    // Fill the first row and first column with 1
+    for (let i = 0; i < m; i++) {
+        dp[i][0] = 1
+    }
+    for (let j = 0; j < n; j++) {
+        dp[0][j] = 1
+    }
+
+    // Fill the rest of the dp table
+    for (let i = 1; i < m; i++) {
+        for (let j = 1; j < n; j++) {
+            dp[i][j] = dp[i - 1][j] + dp[i][j - 1]
+        }
+    }
+
+    // The answer is in the bottom-right corner
+    return dp[m - 1][n - 1]
+};
+
+export { uniquePaths }

src/main/js/g0001_0100/s0064_minimum_path_sum/readme.md

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+64\. Minimum Path Sum
+
+Medium
+
+Given a `m x n` `grid` filled with non-negative numbers, find a path from top left to bottom right, which minimizes the sum of all numbers along its path.
+
+**Note:** You can only move either down or right at any point in time.
+
+**Example 1:**
+
+![](https://assets.leetcode.com/uploads/2020/11/05/minpath.jpg)
+
+**Input:** grid = [[1,3,1],[1,5,1],[4,2,1]]
+
+**Output:** 7
+
+**Explanation:** Because the path 1 → 3 → 1 → 1 → 1 minimizes the sum.
+
+**Example 2:**
+
+**Input:** grid = [[1,2,3],[4,5,6]]
+
+**Output:** 12
+
+**Constraints:**
+
+*   `m == grid.length`
+*   `n == grid[i].length`
+*   `1 <= m, n <= 200`
+*   `0 <= grid[i][j] <= 100`

src/main/js/g0001_0100/s0064_minimum_path_sum/solution.js

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+// #Medium #Top_100_Liked_Questions #Array #Dynamic_Programming #Matrix
+// #Dynamic_Programming_I_Day_16 #Udemy_Dynamic_Programming #Big_O_Time_O(m*n)_Space_O(m*n)
+// #2024_12_10_Time_3_ms_(83.07%)_Space_51.2_MB_(74.79%)
+
+/**
+ * @param {number[][]} grid
+ * @return {number}
+ */
+var minPathSum = function(grid) {
+    const rows = grid.length
+    const cols = grid[0].length
+
+    // Handle the special case where grid has only one cell
+    if (rows === 1 && cols === 1) {
+        return grid[0][0]
+    }
+
+    // Create a 2D array for dynamic programming
+    const dm = Array.from({ length: rows }, () => Array(cols).fill(0))
+
+    // Initialize the last column
+    let s = 0
+    for (let r = rows - 1; r >= 0; r--) {
+        dm[r][cols - 1] = grid[r][cols - 1] + s
+        s += grid[r][cols - 1]
+    }
+
+    // Initialize the last row
+    s = 0
+    for (let c = cols - 1; c >= 0; c--) {
+        dm[rows - 1][c] = grid[rows - 1][c] + s
+        s += grid[rows - 1][c]
+    }
+
+    // Recursive helper function
+    const recur = (r, c) => {
+        if (
+            dm[r][c] === 0 &&
+            r !== rows - 1 &&
+            c !== cols - 1
+        ) {
+            dm[r][c] =
+                grid[r][c] +
+                Math.min(
+                    recur(r + 1, c),
+                    recur(r, c + 1)
+                )
+        }
+        return dm[r][c]
+    }
+
+    // Start recursion from the top-left corner
+    return recur(0, 0)
+};
+
+export { minPathSum }

src/main/js/g0001_0100/s0070_climbing_stairs/readme.md

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+70\. Climbing Stairs
+
+Easy
+
+You are climbing a staircase. It takes `n` steps to reach the top.
+
+Each time you can either climb `1` or `2` steps. In how many distinct ways can you climb to the top?
+
+**Example 1:**
+
+**Input:** n = 2
+
+**Output:** 2
+
+**Explanation:** There are two ways to climb to the top. 
+
+1. 1 step + 1 step 
+
+2. 2 steps
+
+**Example 2:**
+
+**Input:** n = 3
+
+**Output:** 3
+
+**Explanation:** There are three ways to climb to the top. 
+
+1. 1 step + 1 step + 1 step 
+
+2. 1 step + 2 steps 
+
+3. 2 steps + 1 step
+
+**Constraints:**
+
+*   `1 <= n <= 45`

src/main/js/g0001_0100/s0070_climbing_stairs/solution.js

@@ -0,0 +1,31 @@
+// #Easy #Top_100_Liked_Questions #Top_Interview_Questions #Dynamic_Programming #Math #Memoization
+// #Algorithm_I_Day_12_Dynamic_Programming #Dynamic_Programming_I_Day_2
+// #Level_1_Day_10_Dynamic_Programming #Udemy_Dynamic_Programming #Big_O_Time_O(n)_Space_O(n)
+// #2024_12_10_Time_0_ms_(100.00%)_Space_49_MB_(22.45%)
+
+/**
+ * @param {number} n
+ * @return {number}
+ */
+var climbStairs = function(n) {
+    if (n < 2) {
+        return n
+    }
+
+    // Create a cache (DP array) to store results
+    const cache = new Array(n)
+
+    // Initialize base cases
+    cache[0] = 1
+    cache[1] = 2
+
+    // Fill the cache using the recurrence relation
+    for (let i = 2; i < n; i++) {
+        cache[i] = cache[i - 1] + cache[i - 2]
+    }
+
+    // Return the result for the nth step
+    return cache[n - 1]
+};
+
+export { climbStairs }

src/main/js/g0001_0100/s0072_edit_distance/readme.md

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+72\. Edit Distance
+
+Hard
+
+Given two strings `word1` and `word2`, return _the minimum number of operations required to convert `word1` to `word2`_.
+
+You have the following three operations permitted on a word:
+
+*   Insert a character
+*   Delete a character
+*   Replace a character
+
+**Example 1:**
+
+**Input:** word1 = "horse", word2 = "ros"
+
+**Output:** 3
+
+**Explanation:** 
+
+horse -> rorse (replace 'h' with 'r') 
+
+rorse -> rose (remove 'r') 
+
+rose -> ros (remove 'e')
+
+**Example 2:**
+
+**Input:** word1 = "intention", word2 = "execution"
+
+**Output:** 5
+
+**Explanation:** 
+
+intention -> inention (remove 't') 
+
+inention -> enention (replace 'i' with 'e') 
+
+enention -> exention (replace 'n' with 'x') 
+
+exention -> exection (replace 'n' with 'c') 
+
+exection -> execution (insert 'u')
+
+**Constraints:**
+
+*   `0 <= word1.length, word2.length <= 500`
+*   `word1` and `word2` consist of lowercase English letters.