Added js tasks

Author: javadev

src/main/js/g0301_0400/s0322_coin_change/readme.md

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+322\. Coin Change
+
+Medium
+
+You are given an integer array `coins` representing coins of different denominations and an integer `amount` representing a total amount of money.
+
+Return _the fewest number of coins that you need to make up that amount_. If that amount of money cannot be made up by any combination of the coins, return `-1`.
+
+You may assume that you have an infinite number of each kind of coin.
+
+**Example 1:**
+
+**Input:** coins = [1,2,5], amount = 11
+
+**Output:** 3
+
+**Explanation:** 11 = 5 + 5 + 1
+
+**Example 2:**
+
+**Input:** coins = [2], amount = 3
+
+**Output:** -1
+
+**Example 3:**
+
+**Input:** coins = [1], amount = 0
+
+**Output:** 0
+
+**Constraints:**
+
+*   `1 <= coins.length <= 12`
+*   <code>1 <= coins[i] <= 2<sup>31</sup> - 1</code>
+*   <code>0 <= amount <= 10<sup>4</sup></code>

src/main/js/g0301_0400/s0322_coin_change/solution.js

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+// #Medium #Top_100_Liked_Questions #Array #Dynamic_Programming #Breadth_First_Search
+// #Algorithm_II_Day_18_Dynamic_Programming #Dynamic_Programming_I_Day_20
+// #Level_2_Day_12_Dynamic_Programming #Big_O_Time_O(m*n)_Space_O(amount)
+// #2024_12_22_Time_26_ms_(95.33%)_Space_54.1_MB_(79.36%)
+
+/**
+ * @param {number[]} coins
+ * @param {number} amount
+ * @return {number}
+ */
+var coinChange = function(coins, amount) {
+    const dp = new Array(amount + 1).fill(0)
+    dp[0] = 1
+
+    for (const coin of coins) {
+        for (let i = coin; i <= amount; i++) {
+            const prev = dp[i - coin]
+            if (prev > 0) {
+                if (dp[i] === 0) {
+                    dp[i] = prev + 1
+                } else {
+                    dp[i] = Math.min(dp[i], prev + 1)
+                }
+            }
+        }
+    }
+
+    return dp[amount] - 1
+};
+
+export { coinChange }

src/main/js/g0301_0400/s0338_counting_bits/readme.md

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+338\. Counting Bits
+
+Easy
+
+Given an integer `n`, return _an array_ `ans` _of length_ `n + 1` _such that for each_ `i` (`0 <= i <= n`)_,_ `ans[i]` _is the **number of**_ `1`_**'s** in the binary representation of_ `i`.
+
+**Example 1:**
+
+**Input:** n = 2
+
+**Output:** [0,1,1]
+
+**Explanation:**
+
+    0 --> 0
+    1 --> 1
+    2 --> 10 
+
+**Example 2:**
+
+**Input:** n = 5
+
+**Output:** [0,1,1,2,1,2]
+
+**Explanation:**
+
+    0 --> 0
+    1 --> 1
+    2 --> 10
+    3 --> 11
+    4 --> 100
+    5 --> 101 
+
+**Constraints:**
+
+*   <code>0 <= n <= 10<sup>5</sup></code>
+
+**Follow up:**
+
+*   It is very easy to come up with a solution with a runtime of `O(n log n)`. Can you do it in linear time `O(n)` and possibly in a single pass?
+*   Can you do it without using any built-in function (i.e., like `__builtin_popcount` in C++)?

src/main/js/g0301_0400/s0338_counting_bits/solution.js

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+// #Easy #Dynamic_Programming #Bit_Manipulation #Udemy_Bit_Manipulation
+// #Big_O_Time_O(num)_Space_O(num) #2024_12_22_Time_0_ms_(100.00%)_Space_57_MB_(33.09%)
+
+/**
+ * @param {number} n
+ * @return {number[]}
+ */
+var countBits = function(num) {
+    const result = new Array(num + 1).fill(0)
+    let borderPos = 1
+    let incrPos = 1
+
+    for (let i = 1; i <= num; i++) {
+        // When we reach a power of 2, reset `borderPos` and `incrPos`
+        if (incrPos === borderPos) {
+            result[i] = 1
+            incrPos = 1
+            borderPos = i
+        } else {
+            result[i] = 1 + result[incrPos++]
+        }
+    }
+
+    return result
+}
+
+export { countBits }

src/main/js/g0301_0400/s0347_top_k_frequent_elements/readme.md

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+347\. Top K Frequent Elements
+
+Medium
+
+Given an integer array `nums` and an integer `k`, return _the_ `k` _most frequent elements_. You may return the answer in **any order**.
+
+**Example 1:**
+
+**Input:** nums = [1,1,1,2,2,3], k = 2
+
+**Output:** [1,2]
+
+**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>
+*   `k` is in the range `[1, the number of unique elements in the array]`.
+*   It is **guaranteed** that the answer is **unique**.
+
+**Follow up:** Your algorithm's time complexity must be better than `O(n log n)`, where n is the array's size.

src/main/js/g0301_0400/s0347_top_k_frequent_elements/solution.js

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+// #Medium #Top_100_Liked_Questions #Array #Hash_Table #Sorting #Heap_Priority_Queue #Counting
+// #Divide_and_Conquer #Quickselect #Bucket_Sort #Data_Structure_II_Day_20_Heap_Priority_Queue
+// #Big_O_Time_O(n*log(n))_Space_O(k) #2024_12_22_Time_6_ms_(95.00%)_Space_54.3_MB_(53.50%)
+
+/**
+ * @param {number[]} nums
+ * @param {number} k
+ * @return {number[]}
+ */
+var topKFrequent = function (nums, k) {
+    let obj = {}, result = []
+    for (let item of nums) {
+        obj[item] = (obj[item] ? obj[item] : 0) + 1
+    }
+    let temp = Object.entries(obj).sort((a, b) => b[1] - a[1])
+    for (let i = 0; i < k; i++) {
+        result.push(Number(temp[i][0]))
+    }
+    return result
+};
+
+export { topKFrequent }

src/main/js/g0301_0400/s0394_decode_string/readme.md

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+394\. Decode String
+
+Medium
+
+Given an encoded string, return its decoded string.
+
+The encoding rule is: `k[encoded_string]`, where the `encoded_string` inside the square brackets is being repeated exactly `k` times. Note that `k` is guaranteed to be a positive integer.
+
+You may assume that the input string is always valid; there are no extra white spaces, square brackets are well-formed, etc. Furthermore, you may assume that the original data does not contain any digits and that digits are only for those repeat numbers, `k`. For example, there will not be input like `3a` or `2[4]`.
+
+The test cases are generated so that the length of the output will never exceed <code>10<sup>5</sup></code>.
+
+**Example 1:**
+
+**Input:** s = "3[a]2[bc]"
+
+**Output:** "aaabcbc"
+
+**Example 2:**
+
+**Input:** s = "3[a2[c]]"
+
+**Output:** "accaccacc"
+
+**Example 3:**
+
+**Input:** s = "2[abc]3[cd]ef"
+
+**Output:** "abcabccdcdcdef"
+
+**Constraints:**
+
+*   `1 <= s.length <= 30`
+*   `s` consists of lowercase English letters, digits, and square brackets `'[]'`.
+*   `s` is guaranteed to be **a valid** input.
+*   All the integers in `s` are in the range `[1, 300]`.

src/main/js/g0301_0400/s0394_decode_string/solution.js

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+// #Medium #Top_100_Liked_Questions #String #Stack #Recursion #Level_1_Day_14_Stack #Udemy_Strings
+// #Big_O_Time_O(n)_Space_O(n) #2024_12_22_Time_0_ms_(100.00%)_Space_49.4_MB_(10.78%)
+
+/**
+ * @param {string} s
+ * @return {string}
+ */
+var decodeString = function(s) {
+    let i = 0
+
+    const helper = () => {
+        let count = 0
+        let sb = ''
+
+        while (i < s.length) {
+            const c = s[i]
+            i++
+
+            if (/[a-zA-Z]/.test(c)) {
+                sb += c
+            } else if (/\d/.test(c)) {
+                count = count * 10 + Number(c)
+            } else if (c === ']') {
+                break
+            } else if (c === '[') {
+                // Recursive call for the substring
+                const repeat = helper()
+                while (count > 0) {
+                    sb += repeat
+                    count--
+                }
+            }
+        }
+
+        return sb
+    }
+
+    return helper()
+};
+
+export { decodeString }

src/main/js/g0401_0500/s0416_partition_equal_subset_sum/readme.md

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+416\. Partition Equal Subset Sum
+
+Medium
+
+Given a **non-empty** array `nums` containing **only positive integers**, find if the array can be partitioned into two subsets such that the sum of elements in both subsets is equal.
+
+**Example 1:**
+
+**Input:** nums = [1,5,11,5]
+
+**Output:** true
+
+**Explanation:** The array can be partitioned as [1, 5, 5] and [11].
+
+**Example 2:**
+
+**Input:** nums = [1,2,3,5]
+
+**Output:** false
+
+**Explanation:** The array cannot be partitioned into equal sum subsets.
+
+**Constraints:**
+
+*   `1 <= nums.length <= 200`
+*   `1 <= nums[i] <= 100`

src/main/js/g0401_0500/s0416_partition_equal_subset_sum/solution.js

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+// #Medium #Top_100_Liked_Questions #Array #Dynamic_Programming #Level_2_Day_13_Dynamic_Programming
+// #Big_O_Time_O(n*sums)_Space_O(n*sums) #2024_12_22_Time_18_ms_(97.98%)_Space_51.4_MB_(95.09%)
+
+/**
+ * @param {number[]} nums
+ * @return {boolean}
+ */
+var canPartition = function(nums) {
+    let sum = nums.reduce((acc, val) => acc + val, 0)
+    if (sum % 2 !== 0) {
+        return false
+    }
+    sum /= 2
+    
+    const set = new Array(sum + 1).fill(false)
+    const arr = new Array(sum + 2).fill(0)
+    let top = 0
+
+    for (let val of nums) {
+        for (let i = top; i >= 0; i--) {
+            const tempSum = val + arr[i]
+            if (tempSum <= sum && !set[tempSum]) {
+                if (tempSum === sum) {
+                    return true
+                }
+                set[tempSum] = true
+                arr[++top] = tempSum
+            }
+        }
+    }
+
+    return false
+};
+
+export { canPartition }

src/main/js/g0401_0500/s0437_path_sum_iii/readme.md

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+437\. Path Sum III
+
+Medium
+
+Given the `root` of a binary tree and an integer `targetSum`, return _the number of paths where the sum of the values along the path equals_ `targetSum`.
+
+The path does not need to start or end at the root or a leaf, but it must go downwards (i.e., traveling only from parent nodes to child nodes).
+
+**Example 1:**
+
+![](https://assets.leetcode.com/uploads/2021/04/09/pathsum3-1-tree.jpg)
+
+**Input:** root = [10,5,-3,3,2,null,11,3,-2,null,1], targetSum = 8
+
+**Output:** 3
+
+**Explanation:** The paths that sum to 8 are shown.
+
+**Example 2:**
+
+**Input:** root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22
+
+**Output:** 3
+
+**Constraints:**
+
+*   The number of nodes in the tree is in the range `[0, 1000]`.
+*   <code>-10<sup>9</sup> <= Node.val <= 10<sup>9</sup></code>
+*   `-1000 <= targetSum <= 1000`