Added tasks 3783-3805

Author: ThanhNIT

src/main/java/g3701_3800/s3783_mirror_distance_of_an_integer/Solution.java

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+package g3701_3800.s3783_mirror_distance_of_an_integer;
+
+// #Easy #Math #Mid_Level #Weekly_Contest_481 #2026_05_22_Time_1_ms_(99.88%)_Space_42.74_MB_(18.96%)
+
+public class Solution {
+    private int rev(int n) {
+        int a = 0;
+        while (n > 0) {
+            a = a * 10 + (n % 10);
+            n /= 10;
+        }
+        return a;
+    }
+
+    public int mirrorDistance(int n) {
+        int m = rev(n);
+        return Math.abs(m - n);
+    }
+}

src/main/java/g3701_3800/s3783_mirror_distance_of_an_integer/readme.md

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+3783\. Mirror Distance of an Integer
+
+Easy
+
+You are given an integer `n`.
+
+Define its **mirror distance** as: `abs(n - reverse(n))` where `reverse(n)` is the integer formed by reversing the digits of `n`.
+
+Return an integer denoting the mirror distance of `n`.
+
+`abs(x)` denotes the absolute value of `x`.
+
+**Example 1:**
+
+**Input:** n = 25
+
+**Output:** 27
+
+**Explanation:**
+
+*   `reverse(25) = 52`.
+*   Thus, the answer is `abs(25 - 52) = 27`.
+
+**Example 2:**
+
+**Input:** n = 10
+
+**Output:** 9
+
+**Explanation:**
+
+*   `reverse(10) = 01` which is 1.
+*   Thus, the answer is `abs(10 - 1) = 9`.
+
+**Example 3:**
+
+**Input:** n = 7
+
+**Output:** 0
+
+**Explanation:**
+
+*   `reverse(7) = 7`.
+*   Thus, the answer is `abs(7 - 7) = 0`.
+
+**Constraints:**
+
+*   <code>1 <= n <= 10<sup>9</sup></code>

src/main/java/g3701_3800/s3784_minimum_deletion_cost_to_make_all_characters_equal/Solution.java

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+package g3701_3800.s3784_minimum_deletion_cost_to_make_all_characters_equal;
+
+// #Medium #Array #String #Hash_Table #Enumeration #Senior #Weekly_Contest_481
+// #2026_05_22_Time_3_ms_(94.77%)_Space_104.58_MB_(97.09%)
+
+public class Solution {
+    public long minCost(String s, int[] cost) {
+        long[] arr = new long[26];
+        long m = Integer.MIN_VALUE;
+        long sum = 0;
+        for (int i = 0; i < s.length(); i++) {
+            arr[s.charAt(i) - 'a'] += cost[i];
+        }
+        for (long i : arr) {
+            if (i > m) {
+                m = i;
+            }
+            sum += i;
+        }
+        return sum - m;
+    }
+}

src/main/java/g3701_3800/s3784_minimum_deletion_cost_to_make_all_characters_equal/readme.md

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+3784\. Minimum Deletion Cost to Make All Characters Equal
+
+Medium
+
+You are given a string `s` of length `n` and an integer array `cost` of the same length, where `cost[i]` is the cost to **delete** the <code>i<sup>th</sup></code> character of `s`.
+
+You may delete any number of characters from `s` (possibly none), such that the resulting string is **non-empty** and consists of **equal** characters.
+
+Return an integer denoting the **minimum** total deletion cost required.
+
+**Example 1:**
+
+**Input:** s = "aabaac", cost = [1,2,3,4,1,10]
+
+**Output:** 11
+
+**Explanation:**
+
+Deleting the characters at indices 0, 1, 2, 3, 4 results in the string `"c"`, which consists of equal characters, and the total cost is `cost[0] + cost[1] + cost[2] + cost[3] + cost[4] = 1 + 2 + 3 + 4 + 1 = 11`.
+
+**Example 2:**
+
+**Input:** s = "abc", cost = [10,5,8]
+
+**Output:** 13
+
+**Explanation:**
+
+Deleting the characters at indices 1 and 2 results in the string `"a"`, which consists of equal characters, and the total cost is `cost[1] + cost[2] = 5 + 8 = 13`.
+
+**Example 3:**
+
+**Input:** s = "zzzzz", cost = [67,67,67,67,67]
+
+**Output:** 0
+
+**Explanation:**
+
+All characters in `s` are equal, so the deletion cost is 0.
+
+**Constraints:**
+
+*   `n == s.length == cost.length`
+*   <code>1 <= n <= 10<sup>5</sup></code>
+*   <code>1 <= cost[i] <= 10<sup>9</sup></code>
+*   `s` consists of lowercase English letters.

src/main/java/g3701_3800/s3785_minimum_swaps_to_avoid_forbidden_values/Solution.java

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+package g3701_3800.s3785_minimum_swaps_to_avoid_forbidden_values;
+
+// #Hard #Array #Hash_Table #Greedy #Counting #Senior_Staff #Weekly_Contest_481
+// #2026_05_22_Time_88_ms_(86.89%)_Space_159.68_MB_(81.97%)
+
+import java.util.HashMap;
+import java.util.Map;
+
+public class Solution {
+    public int minSwaps(int[] nums, int[] forbidden) {
+        int n = nums.length;
+        Map<Integer, Integer> map = new HashMap<>();
+        for (int num : nums) {
+            map.put(num, map.getOrDefault(num, 0) + 1);
+        }
+        for (int num : forbidden) {
+            map.put(num, map.getOrDefault(num, 0) + 1);
+        }
+        for (int val : map.values()) {
+            if (val > n) {
+                return -1;
+            }
+        }
+        Map<Integer, Integer> map2 = new HashMap<>();
+        int total = 0;
+        for (int i = 0; i < n; i++) {
+            if (nums[i] == forbidden[i]) {
+                map2.put(nums[i], map2.getOrDefault(nums[i], 0) + 1);
+                total++;
+            }
+        }
+        if (total == 0) {
+            return 0;
+        }
+        int maxSwaps = 0;
+        for (int num : map2.values()) {
+            maxSwaps = Math.max(maxSwaps, num);
+        }
+        return Math.max(maxSwaps, (total + 1) / 2);
+    }
+}

src/main/java/g3701_3800/s3785_minimum_swaps_to_avoid_forbidden_values/readme.md

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+3785\. Minimum Swaps to Avoid Forbidden Values
+
+Hard
+
+You are given two integer arrays, `nums` and `forbidden`, each of length `n`.
+
+You may perform the following operation any number of times (including zero):
+
+*   Choose two **distinct** indices `i` and `j`, and swap `nums[i]` with `nums[j]`.
+
+Return the **minimum** number of swaps required such that, for every index `i`, the value of `nums[i]` is **not equal** to `forbidden[i]`. If no amount of swaps can ensure that every index avoids its forbidden value, return -1.
+
+**Example 1:**
+
+**Input:** nums = [1,2,3], forbidden = [3,2,1]
+
+**Output:** 1
+
+**Explanation:**
+
+One optimal set of swaps:
+
+*   Select indices `i = 0` and `j = 1` in `nums` and swap them, resulting in `nums = [2, 1, 3]`.
+*   After this swap, for every index `i`, `nums[i]` is not equal to `forbidden[i]`.
+
+**Example 2:**
+
+**Input:** nums = [4,6,6,5], forbidden = [4,6,5,5]
+
+**Output:** 2
+
+**Explanation:**
+
+One optimal set of swaps:
+
+*   Select indices `i = 0` and `j = 2` in `nums` and swap them, resulting in `nums = [6, 6, 4, 5]`.
+*   Select indices `i = 1` and `j = 3` in `nums` and swap them, resulting in `nums = [6, 5, 4, 6]`.
+*   After these swaps, for every index `i`, `nums[i]` is not equal to `forbidden[i]`.
+
+**Example 3:**
+
+**Input:** nums = [7,7], forbidden = [8,7]
+
+**Output:** \-1
+
+**Explanation:**
+
+It is not possible to make `nums[i]` different from `forbidden[i]` for all indices.
+
+**Example 4:**
+
+**Input:** nums = [1,2], forbidden = [2,1]
+
+**Output:** 0
+
+**Explanation:**
+
+No swaps are required because `nums[i]` is already different from `forbidden[i]` for all indices, so the answer is 0.
+
+**Constraints:**
+
+*   <code>1 <= n == nums.length == forbidden.length <= 10<sup>5</sup></code>
+*   <code>1 <= nums[i], forbidden[i] <= 10<sup>9</sup></code>

src/main/java/g3701_3800/s3786_total_sum_of_interaction_cost_in_tree_groups/Solution.java

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+package g3701_3800.s3786_total_sum_of_interaction_cost_in_tree_groups;
+
+// #Hard #Array #Tree #Senior_Staff #Weekly_Contest_481 #Depth_First_Search
+// #2026_05_22_Time_82_ms_(90.67%)_Space_296.78_MB_(21.33%)
+
+import java.util.ArrayList;
+import java.util.List;
+
+public class Solution {
+    private long totalCost = 0;
+    private int[][] counts;
+    private int[] totalInGroup;
+    private List<List<Integer>> adj;
+
+    public long interactionCosts(int n, int[][] edges, int[] group) {
+        adj = new ArrayList<>();
+        for (int i = 0; i < n; i++) {
+            adj.add(new ArrayList<>());
+        }
+        for (int[] edge : edges) {
+            adj.get(edge[0]).add(edge[1]);
+            adj.get(edge[1]).add(edge[0]);
+        }
+        totalInGroup = new int[21];
+        for (int g : group) {
+            totalInGroup[g]++;
+        }
+        counts = new int[n][21];
+        dfs(0, -1, group);
+        return totalCost;
+    }
+
+    private void dfs(int u, int p, int[] group) {
+        counts[u][group[u]] = 1;
+        for (int v : adj.get(u)) {
+            if (v == p) {
+                continue;
+            }
+            dfs(v, u, group);
+            for (int g = 1; g <= 20; g++) {
+                if (totalInGroup[g] < 2) {
+                    continue;
+                }
+                long inSubtree = counts[v][g];
+                long outsideSubtree = totalInGroup[g] - inSubtree;
+                totalCost += inSubtree * outsideSubtree;
+                counts[u][g] += counts[v][g];
+            }
+        }
+    }
+}

src/main/java/g3701_3800/s3786_total_sum_of_interaction_cost_in_tree_groups/readme.md

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+3786\. Total Sum of Interaction Cost in Tree Groups
+
+Hard
+
+You are given an integer `n` and an undirected tree with `n` nodes numbered from 0 to `n - 1`. This is represented by a 2D array `edges` of length `n - 1`, where <code>edges[i] = [u<sub>i</sub>, v<sub>i</sub>]</code> indicates an undirected edge between nodes <code>u<sub>i</sub></code> and <code>v<sub>i</sub></code>.
+
+You are also given an integer array `group` of length `n`, where `group[i]` denotes the group label assigned to node `i`.
+
+*   Two nodes `u` and `v` are considered part of the same group if `group[u] == group[v]`.
+*   The **interaction cost** between `u` and `v` is defined as the number of edges on the unique path connecting them in the tree.
+
+Return an integer denoting the **sum** of interaction costs over all **unordered** pairs `(u, v)` with `u != v` such that `group[u] == group[v]`.
+
+**Example 1:**
+
+**Input:** n = 3, edges = [[0,1],[1,2]], group = [1,1,1]
+
+**Output:** 4
+
+**Explanation:**
+
+**![](https://assets.leetcode.com/uploads/2025/09/24/screenshot-2025-09-24-at-50538-pm.png)**
+
+All nodes belong to group 1. The interaction costs between the pairs of nodes are:
+
+*   Nodes `(0, 1)`: 1
+*   Nodes `(1, 2)`: 1
+*   Nodes `(0, 2)`: 2
+
+Thus, the total interaction cost is `1 + 1 + 2 = 4`.
+
+**Example 2:**
+
+**Input:** n = 3, edges = [[0,1],[1,2]], group = [3,2,3]
+
+**Output:** 2
+
+**Explanation:**
+
+*   Nodes 0 and 2 belong to group 3. The interaction cost between this pair is 2.
+*   Node 1 belongs to a different group and forms no valid pair. Therefore, the total interaction cost is 2.
+
+**Example 3:**
+
+**Input:** n = 4, edges = [[0,1],[0,2],[0,3]], group = [1,1,4,4]
+
+**Output:** 3
+
+**Explanation:**
+
+![](https://assets.leetcode.com/uploads/2025/09/24/screenshot-2025-09-24-at-51312-pm.png)
+
+Nodes belonging to the same groups and their interaction costs are:
+
+*   Group 1: Nodes `(0, 1)`: 1
+*   Group 4: Nodes `(2, 3)`: 2
+
+Thus, the total interaction cost is `1 + 2 = 3`.
+
+**Example 4:**
+
+**Input:** n = 2, edges = [[0,1]], group = [9,8]
+
+**Output:** 0
+
+**Explanation:**
+
+All nodes belong to different groups and there are no valid pairs. Therefore, the total interaction cost is 0.
+
+**Constraints:**
+
+*   <code>1 <= n <= 10<sup>5</sup></code>
+*   `edges.length == n - 1`
+*   <code>edges[i] = [u<sub>i</sub>, v<sub>i</sub>]</code>
+*   <code>0 <= u<sub>i</sub>, v<sub>i</sub> <= n - 1</code>
+*   `group.length == n`
+*   `1 <= group[i] <= 20`
+*   The input is generated such that `edges` represents a valid tree.