Added js tasks

Author: javadev

README.md

@@ -66,7 +66,7 @@ class Solution {
     }
 
     // Recursive helper method to construct binary tree
-    private TreeNode build(int[] preorder, int[] inorder, int preStart, preEnd, int inStart, int inEnd) {
+    private TreeNode build(int[] preorder, int[] inorder, int preStart, int preEnd, int inStart, int inEnd) {
         if (preStart > preEnd || inStart > inEnd) return null; // Base case
 
         int rootValue = preorder[preStart]; // Root node value

src/main/java/g0101_0200/s0105_construct_binary_tree_from_preorder_and_inorder_traversal/readme.md

@@ -42,41 +42,34 @@ Given a string `s` and a dictionary of strings `wordDict`, return `true` if `s`
 ## Solution
 
 ```java
-import java.util.HashSet;
 import java.util.List;
-import java.util.Set;
 
 public class Solution {
+    private Boolean[] memo;
+
     public boolean wordBreak(String s, List<String> wordDict) {
-        Set<String> set = new HashSet<>();
-        int max = 0;
-        boolean[] flag = new boolean[s.length() + 1];
-        for (String st : wordDict) {
-            set.add(st);
-            if (max < st.length()) {
-                max = st.length();
-            }
-        }
-        for (int i = 1; i <= max; i++) {
-            if (dfs(s, 0, i, max, set, flag)) {
-                return true;
-            }
-        }
-        return false;
+        memo = new Boolean[s.length() + 1];
+        return dp(s, 0, wordDict);
     }
 
-    private boolean dfs(String s, int start, int end, int max, Set<String> set, boolean[] flag) {
-        if (!flag[end] && set.contains(s.substring(start, end))) {
-            flag[end] = true;
-            if (end == s.length()) {
-                return true;
+    public boolean dp(String s, int i, List<String> wordDict) {
+        if (i == s.length()) {
+            return true;
+        }
+        if (memo[i] != null) {
+            return memo[i];
+        }
+        for (String word : wordDict) {
+            int len = word.length();
+            if (i + len > s.length() || !s.substring(i, i + len).equals(word)) {
+                continue;
             }
-            for (int i = 1; i <= max; i++) {
-                if (end + i <= s.length() && dfs(s, end, end + i, max, set, flag)) {
-                    return true;
-                }
+            if (dp(s, i + len, wordDict)) {
+                memo[i] = true;
+                return true;
             }
         }
+        memo[i] = false;
         return false;
     }
 }

src/main/java/g0101_0200/s0139_word_break/readme.md

@@ -38,18 +38,20 @@ A **subarray** is a contiguous subsequence of the array.
 ```java
 public class Solution {
     public int maxProduct(int[] nums) {
-        int currentMaxProd = nums[0];
-        int currentMinProd = nums[0];
-        int overAllMaxProd = nums[0];
-        for (int i = 1; i < nums.length; i++) {
-            if (nums[i] < 0) {
-                int temp = currentMaxProd;
-                currentMaxProd = currentMinProd;
-                currentMinProd = temp;
+        int overAllMaxProd = Integer.MIN_VALUE;
+        int n = nums.length;
+        int start = 1;
+        int end = 1;
+        for (int i = 0; i < n; i++) {
+            if (start == 0) {
+                start = 1;
             }
-            currentMaxProd = Math.max(nums[i], nums[i] * currentMaxProd);
-            currentMinProd = Math.min(nums[i], nums[i] * currentMinProd);
-            overAllMaxProd = Math.max(overAllMaxProd, currentMaxProd);
+            if (end == 0) {
+                end = 1;
+            }
+            start = start * nums[i];
+            end = end * nums[n - i - 1];
+            overAllMaxProd = Math.max(overAllMaxProd, Math.max(start, end));
         }
         return overAllMaxProd;
     }

src/main/java/g0101_0200/s0152_maximum_product_subarray/readme.md

@@ -43,7 +43,7 @@ Implement the Trie class:
 ```java
 @SuppressWarnings("java:S1104")
 public class Trie {
-    private TrieNode root;
+    private final TrieNode root;
     private boolean startWith;
 
     private static class TrieNode {
@@ -82,7 +82,7 @@ public class Trie {
         return search(word, root, 0);
     }
 
-    public boolean search(String word, TrieNode root, int idx) {
+    private boolean search(String word, TrieNode root, int idx) {
         if (idx == word.length()) {
             startWith = true;
             return root.isWord;

src/main/java/g0201_0300/s0208_implement_trie_prefix_tree/readme.md

@@ -36,25 +36,18 @@ You must write an algorithm that runs in `O(n)` time and without using the divis
 ```java
 public class Solution {
     public int[] productExceptSelf(int[] nums) {
-        int product = 1;
-        int[] ans = new int[nums.length];
-        for (int num : nums) {
-            product = product * num;
-        }
+        int[] res = new int[nums.length];
+        int prefixProduct = 1;
         for (int i = 0; i < nums.length; i++) {
-            if (nums[i] != 0) {
-                ans[i] = product / nums[i];
-            } else {
-                int p = 1;
-                for (int j = 0; j < nums.length; j++) {
-                    if (j != i) {
-                        p = p * nums[j];
-                    }
-                }
-                ans[i] = p;
-            }
+            res[i] = prefixProduct;
+            prefixProduct *= nums[i];
+        }
+        int suffixProduct = 1;
+        for (int i = nums.length - 1; i >= 0; i--) {
+            res[i] *= suffixProduct;
+            suffixProduct *= nums[i];
         }
-        return ans;
+        return res;
     }
 }
 ```

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

@@ -33,22 +33,30 @@ Given a **non-empty** array `nums` containing **only positive integers**, find i
 ```java
 public class Solution {
     public boolean canPartition(int[] nums) {
-        int sums = 0;
-        for (int num : nums) {
-            sums += num;
+        int sum = 0;
+        for (int val : nums) {
+            sum += val;
         }
-        if (sums % 2 == 1) {
+        if (sum % 2 != 0) {
             return false;
         }
-        sums /= 2;
-        boolean[] dp = new boolean[sums + 1];
-        dp[0] = true;
-        for (int num : nums) {
-            for (int sum = sums; sum >= num; sum--) {
-                dp[sum] = dp[sum] || dp[sum - num];
+        sum /= 2;
+        boolean[] set = new boolean[sum + 1];
+        int[] arr = new int[sum + 2];
+        int top = 0;
+        for (int val : nums) {
+            for (int i = top; i > -1; i--) {
+                int tempSum = val + arr[i];
+                if (tempSum <= sum && !set[tempSum]) {
+                    if (tempSum == sum) {
+                        return true;
+                    }
+                    set[tempSum] = true;
+                    arr[++top] = tempSum;
+                }
             }
         }
-        return dp[sums];
+        return false;
     }
 }
 ```

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

@@ -35,6 +35,7 @@ The path does not need to start or end at the root or a leaf, but it must go dow
 
 ```java
 import com_github_leetcode.TreeNode;
+import java.util.HashMap;
 
 /*
  * Definition for a binary tree node.
@@ -52,29 +53,28 @@ import com_github_leetcode.TreeNode;
  * }
  */
 public class Solution {
-    private int count = 0;
-
     public int pathSum(TreeNode root, int targetSum) {
-        if (root == null) {
-            return 0;
-        }
-        helper(root, targetSum, 0);
-        pathSum(root.left, targetSum);
-        pathSum(root.right, targetSum);
-        return count;
+        HashMap<Long, Integer> h = new HashMap<>();
+        return dfs(root, targetSum, h, 0L);
     }
 
-    public void helper(TreeNode node, int targetSum, long currSum) {
-        currSum += node.val;
-        if (targetSum == currSum) {
-            count++;
+    int dfs(TreeNode root, int t, HashMap<Long, Integer> h, Long cs) {
+        int s = 0;
+        if (root == null) {
+            return 0;
         }
-        if (node.left != null) {
-            helper(node.left, targetSum, currSum);
+        Long k = cs + root.val;
+        if (k == t) {
+            s += 1;
         }
-        if (node.right != null) {
-            helper(node.right, targetSum, currSum);
+        if (h.getOrDefault(k - t, 0) > 0) {
+            s += h.get(k - t);
         }
+        h.put(k, h.getOrDefault(k, 0) + 1);
+        s += dfs(root.left, t, h, k);
+        s += dfs(root.right, t, h, k);
+        h.put(k, h.getOrDefault(k, 0) - 1);
+        return s;
     }
 }
 ```

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

@@ -45,35 +45,43 @@ Return the number of different **expressions** that you can build, which evaluat
 
 ```java
 public class Solution {
-    public int findTargetSumWays(int[] nums, int s) {
-        int sum = 0;
-        s = Math.abs(s);
-        for (int num : nums) {
-            sum += num;
+    public int findTargetSumWays(int[] nums, int target) {
+        int totalSum = 0;
+        int n = nums.length;
+        for (int i = 0; i < n; i++) {
+            totalSum += nums[i];
         }
-        // Invalid s, just return 0
-        if (s > sum || (sum + s) % 2 != 0) {
+        int sum = totalSum - target;
+        if (sum < 0 || sum % 2 == 1) {
             return 0;
         }
-        int[][] dp = new int[(sum + s) / 2 + 1][nums.length + 1];
-        dp[0][0] = 1;
-        // empty knapsack must be processed specially
-        for (int i = 0; i < nums.length; i++) {
-            if (nums[i] == 0) {
-                dp[0][i + 1] = dp[0][i] * 2;
-            } else {
-                dp[0][i + 1] = dp[0][i];
-            }
+        return solve(nums, sum / 2);
+    }
+
+    private int solve(int[] nums, int target) {
+        int[] prev = new int[target + 1];
+        if (nums[0] == 0) {
+            prev[0] = 2;
+        } else {
+            prev[0] = 1;
+        }
+        if (nums[0] != 0 && nums[0] <= target) {
+            prev[nums[0]] = 1;
         }
-        for (int i = 1; i < dp.length; i++) {
-            for (int j = 0; j < nums.length; j++) {
-                dp[i][j + 1] += dp[i][j];
-                if (nums[j] <= i) {
-                    dp[i][j + 1] += dp[i - nums[j]][j];
+        int n = nums.length;
+        for (int i = 1; i < n; i++) {
+            int[] curr = new int[target + 1];
+            for (int j = 0; j <= target; j++) {
+                int taken = 0;
+                if (j >= nums[i]) {
+                    taken = prev[j - nums[i]];
                 }
+                int nonTaken = prev[j];
+                curr[j] = taken + nonTaken;
             }
+            prev = curr;
         }
-        return dp[(sum + s) / 2][nums.length];
+        return prev[target];
     }
 }
 ```

src/main/java/g0401_0500/s0494_target_sum/readme.md

@@ -0,0 +1,64 @@
+[![](https://img.shields.io/github/stars/javadev/LeetCode-in-All?label=Stars&style=flat-square)](https://github.com/javadev/LeetCode-in-All)
+[![](https://img.shields.io/github/forks/javadev/LeetCode-in-All?label=Fork%20me%20on%20GitHub%20&style=flat-square)](https://github.com/javadev/LeetCode-in-All/fork)
+
+## 1\. Two Sum
+
+Easy
+
+Given an array of integers `nums` and an integer `target`, return _indices of the two numbers such that they add up to `target`_.
+
+You may assume that each input would have **_exactly_ one solution**, and you may not use the _same_ element twice.
+
+You can return the answer in any order.
+
+**Example 1:**
+
+**Input:** nums = [2,7,11,15], target = 9
+
+**Output:** [0,1]
+
+**Explanation:** Because nums[0] + nums[1] == 9, we return [0, 1]. 
+
+**Example 2:**
+
+**Input:** nums = [3,2,4], target = 6
+
+**Output:** [1,2] 
+
+**Example 3:**
+
+**Input:** nums = [3,3], target = 6
+
+**Output:** [0,1] 
+
+**Constraints:**
+
+*   <code>2 <= nums.length <= 10<sup>4</sup></code>
+*   <code>-10<sup>9</sup> <= nums[i] <= 10<sup>9</sup></code>
+*   <code>-10<sup>9</sup> <= target <= 10<sup>9</sup></code>
+*   **Only one valid answer exists.**
+
+**Follow-up:** Can you come up with an algorithm that is less than <code>O(n<sup>2</sup>)</code> time complexity?
+
+## Solution
+
+```javascript
+/**
+ * @param {number[]} nums
+ * @param {number} target
+ * @return {number[]}
+ */
+function twoSum(nums, target) {
+    const indexMap = new Map()
+    for (let i = 0; i < nums.length; i++) {
+        const requiredNum = target - nums[i]
+        if (indexMap.has(requiredNum)) {
+            return [indexMap.get(requiredNum), i]
+        }
+        indexMap.set(nums[i], i)
+    }
+    return [-1, -1]
+}
+
+export { twoSum }
+```