Added tasks 3582-3585

Author: javadev

src/main/java/g3501_3600/s3582_generate_tag_for_video_caption/Solution.java

@@ -0,0 +1,32 @@
+package g3501_3600.s3582_generate_tag_for_video_caption;
+
+// #Easy #2025_06_16_Time_12_ms_(100.00%)_Space_45.47_MB_(100.00%)
+
+public class Solution {
+    public String generateTag(String caption) {
+        if (caption.trim().isEmpty()) {
+            return "#";
+        }
+        String[] arr = caption.trim().split("\\s+");
+        StringBuilder res = new StringBuilder("#");
+        String firstWord = arr[0];
+        firstWord =
+                firstWord.substring(0, 1).toLowerCase()
+                        + (firstWord.length() > 1 ? firstWord.substring(1).toLowerCase() : "");
+        res.append(firstWord);
+        for (int i = 1; i < arr.length; i++) {
+            String w = arr[i];
+            if (w.isEmpty()) {
+                continue;
+            }
+            w =
+                    w.substring(0, 1).toUpperCase()
+                            + (w.length() > 1 ? w.substring(1).toLowerCase() : "");
+            res.append(w);
+            if (res.length() >= 100) {
+                break;
+            }
+        }
+        return res.length() > 100 ? res.substring(0, 100) : res.toString();
+    }
+}

src/main/java/g3501_3600/s3582_generate_tag_for_video_caption/readme.md

@@ -0,0 +1,51 @@
+3582\. Generate Tag for Video Caption
+
+Easy
+
+You are given a string `caption` representing the caption for a video.
+
+The following actions must be performed **in order** to generate a **valid tag** for the video:
+
+1.  **Combine all words** in the string into a single _camelCase string_ prefixed with `'#'`. A _camelCase string_ is one where the first letter of all words _except_ the first one is capitalized. All characters after the first character in **each** word must be lowercase.
+    
+2.  **Remove** all characters that are not an English letter, **except** the first `'#'`.
+    
+3.  **Truncate** the result to a maximum of 100 characters.
+    
+
+Return the **tag** after performing the actions on `caption`.
+
+**Example 1:**
+
+**Input:** caption = "Leetcode daily streak achieved"
+
+**Output:** "#leetcodeDailyStreakAchieved"
+
+**Explanation:**
+
+The first letter for all words except `"leetcode"` should be capitalized.
+
+**Example 2:**
+
+**Input:** caption = "can I Go There"
+
+**Output:** "#canIGoThere"
+
+**Explanation:**
+
+The first letter for all words except `"can"` should be capitalized.
+
+**Example 3:**
+
+**Input:** caption = "hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh"
+
+**Output:** "#hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh"
+
+**Explanation:**
+
+Since the first word has length 101, we need to truncate the last two letters from the word.
+
+**Constraints:**
+
+*   `1 <= caption.length <= 150`
+*   `caption` consists only of English letters and `' '`.

src/main/java/g3501_3600/s3583_count_special_triplets/Solution.java

@@ -0,0 +1,26 @@
+package g3501_3600.s3583_count_special_triplets;
+
+// #Medium #2025_06_16_Time_250_ms_(100.00%)_Space_62.22_MB_(100.00%)
+
+import java.util.HashMap;
+import java.util.Map;
+
+public class Solution {
+    public int specialTriplets(int[] nums) {
+        int mod = 1_000_000_007;
+        int res = 0;
+        Map<Integer, Integer> left = new HashMap<>();
+        Map<Integer, Integer> right = new HashMap<>();
+        for (int num : nums) {
+            right.put(num, right.getOrDefault(num, 0) + 1);
+        }
+        for (int num : nums) {
+            right.put(num, right.get(num) - 1);
+            int ci = left.getOrDefault(num * 2, 0);
+            int ck = right.getOrDefault(num * 2, 0);
+            res = (res + ci * ck) % mod;
+            left.put(num, left.getOrDefault(num, 0) + 1);
+        }
+        return res;
+    }
+}

src/main/java/g3501_3600/s3583_count_special_triplets/readme.md

@@ -0,0 +1,67 @@
+3583\. Count Special Triplets
+
+Medium
+
+You are given an integer array `nums`.
+
+A **special triplet** is defined as a triplet of indices `(i, j, k)` such that:
+
+*   `0 <= i < j < k < n`, where `n = nums.length`
+*   `nums[i] == nums[j] * 2`
+*   `nums[k] == nums[j] * 2`
+
+Return the total number of **special triplets** in the array.
+
+Since the answer may be large, return it **modulo** <code>10<sup>9</sup> + 7</code>.
+
+**Example 1:**
+
+**Input:** nums = [6,3,6]
+
+**Output:** 1
+
+**Explanation:**
+
+The only special triplet is `(i, j, k) = (0, 1, 2)`, where:
+
+*   `nums[0] = 6`, `nums[1] = 3`, `nums[2] = 6`
+*   `nums[0] = nums[1] * 2 = 3 * 2 = 6`
+*   `nums[2] = nums[1] * 2 = 3 * 2 = 6`
+
+**Example 2:**
+
+**Input:** nums = [0,1,0,0]
+
+**Output:** 1
+
+**Explanation:**
+
+The only special triplet is `(i, j, k) = (0, 2, 3)`, where:
+
+*   `nums[0] = 0`, `nums[2] = 0`, `nums[3] = 0`
+*   `nums[0] = nums[2] * 2 = 0 * 2 = 0`
+*   `nums[3] = nums[2] * 2 = 0 * 2 = 0`
+
+**Example 3:**
+
+**Input:** nums = [8,4,2,8,4]
+
+**Output:** 2
+
+**Explanation:**
+
+There are exactly two special triplets:
+
+*   `(i, j, k) = (0, 1, 3)`
+    *   `nums[0] = 8`, `nums[1] = 4`, `nums[3] = 8`
+    *   `nums[0] = nums[1] * 2 = 4 * 2 = 8`
+    *   `nums[3] = nums[1] * 2 = 4 * 2 = 8`
+*   `(i, j, k) = (1, 2, 4)`
+    *   `nums[1] = 4`, `nums[2] = 2`, `nums[4] = 4`
+    *   `nums[1] = nums[2] * 2 = 2 * 2 = 4`
+    *   `nums[4] = nums[2] * 2 = 2 * 2 = 4`
+
+**Constraints:**
+
+*   <code>3 <= n == nums.length <= 10<sup>5</sup></code>
+*   <code>0 <= nums[i] <= 10<sup>5</sup></code>

src/main/java/g3501_3600/s3584_maximum_product_of_first_and_last_elements_of_a_subsequence/Solution.java

@@ -0,0 +1,17 @@
+package g3501_3600.s3584_maximum_product_of_first_and_last_elements_of_a_subsequence;
+
+// #Medium #2025_06_16_Time_4_ms_(87.17%)_Space_61.31_MB_(100.00%)
+
+public class Solution {
+    public long maximumProduct(int[] nums, int m) {
+        long ma = nums[0];
+        long mi = nums[0];
+        long res = (long) nums[0] * nums[m - 1];
+        for (int i = m - 1; i < nums.length; ++i) {
+            ma = Math.max(ma, nums[i - m + 1]);
+            mi = Math.min(mi, nums[i - m + 1]);
+            res = Math.max(res, Math.max(mi * nums[i], ma * nums[i]));
+        }
+        return res;
+    }
+}

src/main/java/g3501_3600/s3584_maximum_product_of_first_and_last_elements_of_a_subsequence/readme.md

@@ -0,0 +1,43 @@
+3584\. Maximum Product of First and Last Elements of a Subsequence
+
+Medium
+
+You are given an integer array `nums` and an integer `m`.
+
+Return the **maximum** product of the first and last elements of any ****subsequences**** of `nums` of size `m`.
+
+**Example 1:**
+
+**Input:** nums = [-1,-9,2,3,-2,-3,1], m = 1
+
+**Output:** 81
+
+**Explanation:**
+
+The subsequence `[-9]` has the largest product of the first and last elements: `-9 * -9 = 81`. Therefore, the answer is 81.
+
+**Example 2:**
+
+**Input:** nums = [1,3,-5,5,6,-4], m = 3
+
+**Output:** 20
+
+**Explanation:**
+
+The subsequence `[-5, 6, -4]` has the largest product of the first and last elements.
+
+**Example 3:**
+
+**Input:** nums = [2,-1,2,-6,5,2,-5,7], m = 2
+
+**Output:** 35
+
+**Explanation:**
+
+The subsequence `[5, 7]` has the largest product of the first and last elements.
+
+**Constraints:**
+
+*   <code>1 <= nums.length <= 10<sup>5</sup></code>
+*   <code>-10<sup>5</sup> <= nums[i] <= 10<sup>5</sup></code>
+*   `1 <= m <= nums.length`

src/main/java/g3501_3600/s3585_find_weighted_median_node_in_tree/Solution.java

@@ -0,0 +1,122 @@
+package g3501_3600.s3585_find_weighted_median_node_in_tree;
+
+// #Hard #2025_06_16_Time_162_ms_(100.00%)_Space_141.58_MB_(100.00%)
+
+import java.util.ArrayList;
+import java.util.Arrays;
+import java.util.List;
+
+public class Solution {
+    private int log;
+    private long[] dist;
+    private int[] depth;
+    private int[][] up;
+
+    public int[] findMedian(int n, int[][] edges, int[][] queries) {
+        List<List<int[]>> adj = new ArrayList<>();
+        for (int i = 0; i < n; i++) {
+            adj.add(new ArrayList<>());
+        }
+        for (int[] edge : edges) {
+            adj.get(edge[0]).add(new int[] {edge[1], edge[2]});
+            adj.get(edge[1]).add(new int[] {edge[0], edge[2]});
+        }
+        dist = new long[n];
+        depth = new int[n];
+        log = 0;
+        while (1 << log < n) {
+            log++;
+        }
+        up = new int[n][log];
+        for (int[] u : up) {
+            Arrays.fill(u, -1);
+        }
+        dfs(0, -1, adj, 0, 0);
+        int[] ans = new int[queries.length];
+        for (int i = 0; i < queries.length; i++) {
+            int[] query = queries[i];
+            int first = query[0];
+            int second = query[1];
+            long distance = getDistance(first, second);
+            long needed = (distance + 1) / 2;
+            int mid = lca(first, second);
+            if (getDistance(first, mid) < needed) {
+                needed -= getDistance(first, mid);
+                first = mid;
+            } else {
+                second = mid;
+            }
+            if (depth[first] <= depth[second]) {
+                long curDistance = getDistance(first, second);
+                for (int j = log - 1; j >= 0; j--) {
+                    if (up[second][j] == -1
+                            || curDistance - getDistance(up[second][j], second) < needed) {
+                        continue;
+                    }
+                    curDistance -= getDistance(up[second][j], second);
+                    second = up[second][j];
+                }
+                ans[i] = second;
+            } else {
+                long curDistance = 0;
+                for (int j = log - 1; j >= 0; j--) {
+                    if (up[first][j] == -1
+                            || curDistance + getDistance(up[first][j], first) >= needed) {
+                        continue;
+                    }
+                    curDistance += getDistance(up[first][j], first);
+                    first = up[first][j];
+                }
+                ans[i] = up[first][0];
+            }
+        }
+        return ans;
+    }
+
+    private long getDistance(int from, int to) {
+        if (from == to) {
+            return 0;
+        }
+        int ancesor = lca(from, to);
+        return dist[from] + dist[to] - 2 * dist[ancesor];
+    }
+
+    private int lca(int first, int second) {
+        if (depth[first] < depth[second]) {
+            return lca(second, first);
+        }
+        for (int i = log - 1; i >= 0; i--) {
+            if (depth[first] - (1 << i) >= depth[second]) {
+                first = up[first][i];
+            }
+        }
+        if (first == second) {
+            return second;
+        }
+        for (int i = log - 1; i >= 0; i--) {
+            if (depth[first] != -1 && up[first][i] != up[second][i]) {
+                first = up[first][i];
+                second = up[second][i];
+            }
+        }
+        first = up[first][0];
+        return first;
+    }
+
+    private void dfs(int current, int parent, List<List<int[]>> adj, long curDist, int curDepth) {
+        up[current][0] = parent;
+        for (int i = 1; i < log; i++) {
+            if (up[current][i - 1] != -1) {
+                up[current][i] = up[up[current][i - 1]][i - 1];
+            }
+        }
+        dist[current] = curDist;
+        depth[current] = curDepth;
+        for (int[] next : adj.get(current)) {
+            if (next[0] == parent) {
+                continue;
+            }
+            dfs(next[0], current, adj, curDist + next[1], curDepth + 1);
+        }
+    }
+}